--- title: "Hilbert Spaces" book: "University Mathematics — Year 3" subject: math language: en chapter: 13 exercises: 12 source: https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces --- # Chapter 13 — Hilbert Spaces A [Hilbert space](#def-b3-hilbert-inner) is a Banach space whose norm comes from an [inner product](#def-b3-hilbert-inner) — and that single extra structure restores, in infinite dimension, almost all of Euclidean geometry: orthogonal projections exist, every [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) functional is an [inner product](#def-b3-hilbert-inner) against a fixed vector (Riesz), and orthonormal bases expand every vector in a convergent series with Pythagorean bookkeeping ([Parseval](#thm-b3-hilbert-parseval)). The chapter’s climax is a debt honored: the trigonometric system is an orthonormal basis of $L^2$, so [Parseval](#thm-b3-hilbert-parseval)’s identity holds for *every* [square-integrable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-l1) function — the statement Year 2 could only prove for piecewise $\mathcal C^1$ functions. We end with Lax–Milgram, the workhorse lemma of the variational approach to differential equations. Throughout, $H$ is a vector space over $K = \R$ or $\C$. ## 13.1 Inner products; the projection theorem **Definition 13.1.** An *inner product* is a map $\langle \cdot,\cdot\rangle \colon H\times H \to K$, linear in the second variable, with $\langle y, x\rangle = \overline{\langle x, y\rangle}$ and $\langle x, x\rangle > 0$ for $x \neq 0$. It induces the norm $\norm x = \langle x, x\rangle^{1/2}$, the *Cauchy–Schwarz inequality* $\abs{\langle x, y\rangle} \leq \norm x\norm y$ (Year 2’s proof — the discriminant — is unchanged), and the *parallelogram law* $$ \norm{x + y}^2 + \norm{x - y}^2 = 2\norm x^2 + 2\norm y^2 . $$ A *Hilbert space* is an inner-product space [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) for this norm. Examples: $\ell^2$ ([Problem 8.1](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#pb-b3-banach-1)) and, the fundamental one, $L^2(\mu)$ with $\langle f, g\rangle = \int\bar fg\,\dd\mu$ — [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) by Riesz–Fischer ([Theorem 12.4](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#thm-b3-lp-complete)); the inner product is finite by Cauchy–Schwarz ($=$ Hölder at $p = q = 2$). **Theorem 13.2 (Projection onto a closed convex set).** Let $C \neq \varnothing$ be a closed *convex* subset of the [Hilbert space](#def-b3-hilbert-inner) $H$ and $x \in H$. There is a unique $p_C(x) \in C$ with $$ \norm{x - p_C(x)} = d(x, C), $$ characterized by: $\operatorname{Re}\langle x - p_C(x),\ c - p_C(x)\rangle \leq 0$ for all $c \in C$. The map $p_C$ is $1$-Lipschitz. **Proof.** Let $d = d(x, C)$ and $(c_n) \subseteq C$ with $\norm{x - c_n} \to d$. Parallelogram on $x - c_n$ and $x - c_m$: $$ \norm{c_n - c_m}^2 = 2\norm{x - c_n}^2 + 2\norm{x - c_m}^2 - 4\,\bigl\|x - \tfrac{c_n + c_m}2\bigr\|^2 \leq 2\norm{x{-}c_n}^2 + 2\norm{x{-}c_m}^2 - 4d^2 $$ (convexity puts the midpoint in $C$): the right side tends to $0$, so $(c_n)$ is Cauchy, and its limit $p \in C$ (closed) attains $d$. Uniqueness: two minimizers give, by the same identity, $\norm{p - p'}^2 \leq 2d^2 + 2d^2 - 4d^2 = 0$. Characterization: for $c \in C$, $t \in \intoc01$, the vector $p + t(c - p) \in C$, so $$ d^2 \leq \norm{x - p - t(c-p)}^2 = d^2 - 2t\operatorname{Re}\langle x - p, c - p\rangle + t^2\norm{c-p}^2 ; $$ divide by $t \to 0^+$: $\operatorname{Re}\langle x - p, c - p\rangle \leq 0$. Conversely this inequality gives $\norm{x - c}^2 = \norm{x - p}^2 - 2\operatorname{Re}\langle x - p, c - p\rangle + \norm{p - c}^2 \geq \norm{x-p}^2$. Lipschitz: for $x, y$ with projections $p, q$, add the two variational inequalities (with $c = q$, resp. $c = p$): $\operatorname{Re}\langle x - y - (p - q), p - q\rangle \geq 0$, so $\norm{p - q}^2 \leq \operatorname{Re}\langle x - y, p - q\rangle \leq \norm{x - y}\norm{p - q}$. ∎ **Theorem 13.3 (Orthogonal decomposition).** Let $F$ be a *closed subspace* of $H$. Then $p_F$ is linear, $x - p_F(x) \perp F$ for all $x$, and $$ H = F \oplus F^\perp, \qquad F^\perp = \{y : \langle y, f\rangle = 0\ \forall f\in F\}, \qquad (F^\perp)^\perp = F . $$ For a general subspace, $(F^\perp)^\perp = \bar F$; in particular $F$ is dense iff $F^\perp = \{0\}$. **Proof.** For a subspace, the variational characterization with $c = p_F(x) \pm f$ ($f \in F$, both signs, and $\iu f$ in the complex case) forces $\langle x - p_F(x), f\rangle = 0$: the residual is orthogonal to $F$. Decomposition $x = p_F(x) + (x - p_F(x))$ with $F \cap F^\perp = \{0\}$ ($\langle y, y \rangle = 0$); linearity of $p_F$ follows from uniqueness of such decompositions (both sides linear in them). $(F^\perp) ^\perp \supseteq F$ always; conversely if $x \perp F^\perp$, write $x = f + g$: $g = x - f \in F^\perp$ and $\langle g, g\rangle = \langle x, g\rangle - \langle f, g\rangle = 0$: $x = f \in F$. For a general subspace $F$: $F^\perp = \bar F^{\,\perp}$ ([continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of the [inner product](#def-b3-hilbert-inner)), so $(F^\perp)^\perp = \bar F$ by the closed case; [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) iff $\bar F = H$ iff $F^\perp = 0$. ∎ **Theorem 13.4 (Riesz representation).** For every [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) linear functional $\varphi \in H'$ there is a unique $a \in H$ with $$ \varphi(x) = \langle a, x\rangle \quad (x \in H), \qquad \norm\varphi_{H'} = \norm a . $$ **Proof.** If $\varphi = 0$: $a = 0$. Otherwise $F = \ker\varphi$ is a closed proper subspace; pick $u \in F^\perp$, $\norm u = 1$ ([Theorem 13.3](#thm-b3-hilbert-decomposition): $F^\perp \neq 0$ since $F \neq H$). For any $x$, the vector $\varphi(x)u - \varphi(u)x \in \ker\varphi$, hence $\perp u$: $$ 0 = \langle u, \varphi(x)u - \varphi(u)x\rangle = \varphi(x) - \varphi(u)\langle u, x\rangle : \qquad \varphi(x) = \langle \overline{\varphi(u)}\,u,\ x\rangle . $$ So $a = \overline{\varphi(u)}u$ works. Uniqueness: $\langle a - a', x\rangle = 0$ for all $x$, test $x = a - a'$. Norms: $\abs{\varphi(x)} \leq \norm a\norm x$ (Cauchy–Schwarz) with equality at $x = a$. ∎ **Example 13.5 (A projection, computed to the end).** In $H = L^2(\intcc01)$, what is the best approximation of $f(x) = x^2$ by an affine function? The subspace $F = \operatorname{Vect}(1, x)$ is closed (finite-dimensional), and $p_F(f) = a + bx$ is characterized by orthogonality of the residual to $1$ and to $x$: $$ \int_0^1(x^2 - a - bx)\,\dd x = 0, \qquad \int_0^1x\,(x^2 - a - bx)\,\dd x = 0, $$ i.e. $\frac13 = a + \frac b2$ and $\frac14 = \frac a2 + \frac b3$: $a = -\frac16$, $b = 1$. So $p_F(x^2) = x - \frac16$, and the error is $$ d(f, F)^2 = \int_0^1\Bigl(x^2 - x + \frac16\Bigr)^2\dd x = \frac1{180}, \qquad d(f, F) = \frac1{6\sqrt5} . $$ Two remarks worth internalizing. First, the computation is nothing but a $2\times2$ linear system — the *normal equations*; for the monomial basis their matrix $\bigl(\frac1{i+j+1}\bigr)$ is the notoriously ill-conditioned Hilbert matrix, and orthogonalizing first (Legendre polynomials, [Problem 13.1](#pb-b3-hilbert-1)) is the cure. Second, the best *uniform* approximation of $x^2$ by affine functions is different ($x - \frac18$, by equioscillation): each norm has its own geometry, and only the Hilbertian one answers with a linear system. ## 13.2 Orthonormal bases **Definition 13.6.** A family $(e_i)_{i\in I}$ is *orthonormal* if $\langle e_i, e_j\rangle = \delta_{ij}$, and a *Hilbert basis* (orthonormal basis) if moreover its finite linear combinations are dense in $H$ (the family is *total*). We treat the countable case $I = \N$, which by Gram–Schmidt covers every *separable* $H$ ([Proposition 13.8](#prop-b3-hilbert-gramschmidt)). **Theorem 13.7 (Bessel, Parseval).** Let $(e_n)_{n\in\N}$ be orthonormal in $H$, and $c_n(x) = \langle e_n, x\rangle$. 1. (Bessel) $\sum_n\abs{c_n(x)}^2 \leq \norm x^2$ , and the series $\sum_nc_n(x)e_n$ converges in $H$ , with sum $p_F(x)$ , $F = \overline{\operatorname{Vect}}(e_n)$ . 2. The following are equivalent: (i) $(e_n)$ is a [Hilbert basis](#def-b3-hilbert-onb) ; (ii) $x = \sum_nc_n(x)e_n$ for every $x$ ; (iii) *Parseval* : $\norm x^2 = \sum_n\abs{c_n(x)}^2$ for every $x$ ; (iv) the only vector orthogonal to all $e_n$ is $0$ . 3. If $(e_n)$ is a [Hilbert basis](#def-b3-hilbert-onb) , $x \mapsto (c_n(x))_n$ is an isometric isomorphism $H \to \ell^2$ ( *every* infinite-dimensional separable [Hilbert space](#def-b3-hilbert-inner) “is” $\ell^2$ ), and $\langle x, y\rangle = \sum_n\overline{c_n(x)}c_n(y)$ . **Proof.** (1) For finite $N$: $x - \sum_{n\leq N}c_ne_n \perp e_k$ ($k \leq N$), so Pythagoras gives $\norm x^2 = \sum_{n\leq N}\abs{c_n}^2 + \norm{x - \sum_{n\leq N}c_ne_n}^2$: Bessel. The partial sums $S_N = \sum_{n\leq N}c_ne_n$ are Cauchy: $\norm{S_N - S_M}^2 = \sum_{M 0$). Then for every $\varphi \in H'$ there is a unique $u \in H$ with $$ a(u, v) = \varphi(v) \qquad \text{for all } v \in H . $$ **Proof.** For fixed $u$, $v \mapsto a(u, v)$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) linear: Riesz gives a unique $Au \in H$ with $a(u,v) = \langle Au, v\rangle$; $A$ is linear with $\norm{Au} \leq M\norm u$ (uniqueness of representatives, then bound). Coercivity: $\alpha\norm u^2 \leq a(u,u) = \langle Au, u\rangle \leq \norm{Au}\norm u$, so $\norm{Au} \geq \alpha\norm u$: $A$ is injective with closed range (a Cauchy image sequence $Au_n$ forces $u_n$ Cauchy). The range is dense: $w \perp \operatorname{im}A$ gives $0 = \langle Aw, w\rangle \geq \alpha\norm w^2$. Closed and dense: $A$ is bijective. Given $\varphi$, let $f$ represent it (Riesz) and $u = A^{-1}f$: $a(u, v) = \langle f, v\rangle = \varphi(v)$, uniquely ($a(u - u', \cdot) = 0$ and coercivity). ∎ **Remark 13.11.** When $a$ is symmetric, Lax–Milgram’s solution is the unique minimizer of the *energy* $J(v) = \frac12a(v,v) - \varphi(v)$ ([Exercise 13.9](#exo-b3-hilbert-9)): existence of solutions to variational problems in one stroke. Applied to suitable function spaces (the Sobolev spaces of a later course), this solves boundary value problems for differential equations — the modern entry point to partial differential equations. ## 13.3 Exercises **Exercise 13.1 ★.** (a) Prove the polarization identities (real: $4\langle x, y\rangle = \norm{x+y}^2 - \norm{x-y}^2$; complex: the four-term version). (b) Show that $\norm\cdot_1$ on $L^1(\intcc01)$ and $\norm\cdot_\infty$ on $\mathcal C(\intcc01)$ violate the parallelogram law: these norms come from no [inner product](#def-b3-hilbert-inner). **Solution of Exercise 13.1.** (a) Real: expand $\norm{x \pm y}^2 = \norm x^2 \pm 2\langle x,y\rangle + \norm y^2$ and subtract. Complex ([inner product](#def-b3-hilbert-inner) linear in the second slot): expanding as above, $$ \langle x, y\rangle = \frac14\sum_{k=0}^{3} \iu^k\,\bigl\|\iu^kx + y\bigr\|^2, $$ each term contributing $\iu^k\cdot2\operatorname{Re}\bigl( (-\iu)^k\langle x,y\rangle\bigr)$, whose sum is $4\langle x,y\rangle$ (check the four values of $k$; the $\sum\iu^k (\norm x^2 + \norm y^2) = 0$). (b) $L^1$: $f = \mathbf 1_{\intcc0{1/2}}$, $g = \mathbf 1_{\intcc{1/2}1}$: $\norm{f\pm g}_1^2 = 1$ each, sum $2$; $2\norm f_1^2 + 2\norm g_1^2 = 1 \neq 2$. Sup norm: $f = \mathbf 1$, $g(t) = t$ on $\intcc01$: $\norm{f + g}_\infty^2 + \norm{f-g}_\infty^2 = 4 + 1 = 5 \neq 4 = 2 + 2$. Failing the parallelogram law, these norms are induced by no [inner product](#def-b3-hilbert-inner) (which would force it by direct expansion). **Exercise 13.2 ★.** In $H = L^2(\intcc01)$ (real): (a) compute the projection of $f$ onto the subspace of constant functions, and interpret; (b) compute the projection onto $\{g : g = 0 \text{ a.e.\ on } \intcc0{1/2}\}$; (c) compute $d\bigl(x \mapsto x,\ \operatorname{Vect}(\mathbf 1)\bigr)$. **Solution of Exercise 13.2.** (a) $p(f) = \bigl(\int_0^1f\bigr)\mathbf 1$: indeed $f - \int f \perp$ constants ($\int(f - \int f)c = 0$). The best constant approximation of $f$ in mean square is its *average* — the first instance of conditional expectation ([Chapter 22](https://one-course.com/books/math/5/en/chapter/22-probability-foundations-and-the-law-of-large-numbers#ch-b3-probability)). (b) $p(f) = f\,\mathbf 1_{\intcc{1/2}1}$: the difference $f\mathbf 1_{\intcc0{1/2}}$ is orthogonal to every $g$ vanishing on $\intcc0{1/2}$. (c) $d^2 = \bigl\|x - \tfrac12\bigr\|_2^2 = \int_0^1(x - \tfrac12)^2\dd x = \tfrac1{12}$: $d = \frac1{2\sqrt3}$. **Exercise 13.3 ★★.** (a) Show that for a subspace $F$: $F$ dense $\iff$ $F^\perp = \{0\}$, and give an example in $\ell^2$ of a *proper* dense subspace (so $F^\perp = 0$ without $F = H$: the decomposition theorem genuinely needs $F$ closed). (b) Show that if $x_n \to x$ and $y_n \to y$ in norm, then $\langle x_n, y_n\rangle \to \langle x, y\rangle$, and locate two places where the chapter used this [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). **Solution of Exercise 13.3.** (a) The equivalence is [Theorem 13.3](#thm-b3-hilbert-decomposition) ($\bar F = (F^\perp)^\perp$, and $\bar F = H \iff F^\perp = \{0\}$). Example: the space $F$ of finite sequences is dense in $\ell^2$ (truncation) and proper: $F^\perp = \{0\}$ yet $F \neq \ell^2$ — for a non-closed subspace, $H = F \oplus F^\perp$ fails blatantly ($F \oplus \{0\} \neq H$). (b) $\abs{\langle x_n, y_n\rangle - \langle x, y\rangle} \leq \abs{\langle x_n - x, y_n\rangle} + \abs{\langle x, y_n - y\rangle} \leq \norm{x_n - x}\sup_n\norm{y_n} + \norm x\,\norm{y_n - y} \to 0$ (convergent sequences are bounded). Used: in [Theorem 13.7](#thm-b3-hilbert-parseval)(1) to see $x - \lim S_N \perp e_k$, and in [Theorem 13.3](#thm-b3-hilbert-decomposition) to see $F^\perp = \bar F^{\,\perp}$. **Exercise 13.4 ★★.** Apply Gram–Schmidt to $1, x, x^2$ in $L^2(\intcc{-1}1)$ ([Lebesgue measure](https://one-course.com/books/math/5/en/chapter/9-measure-theory#def-b3-measure-lebesgueouter)): obtain the first three normalized *Legendre polynomials*, and verify they match $\sqrt{n + \frac12}\,P_n$ for the Rodrigues polynomials $P_n$ of [Problem 13.1](#pb-b3-hilbert-1). **Solution of Exercise 13.4.** $e_0 = \frac1{\sqrt2}$. Next, $x \perp \mathbf 1$ already ($\int_{-1}^1x = 0$), and $\int_{-1}^1x^2 = \frac23$: $e_1 = \sqrt{\tfrac32}\,x$. Then $x^2 - \langle e_0, x^2\rangle e_0 = x^2 - \frac13$ (and $\perp e_1$ by parity), with $$ \int_{-1}^1\Bigl(x^2 - \frac13\Bigr)^2\dd x = \frac25 - \frac49 + \frac29 = \frac{8}{45}: \qquad e_2 = \sqrt{\tfrac{45}8}\,\Bigl(x^2 - \frac13\Bigr). $$ Comparison: $P_0 = 1$, $P_1 = x$, $P_2 = \frac{3x^2 - 1}2$, and $\sqrt{n + \tfrac12}\,P_n$ gives $\frac1{\sqrt2}$, $\sqrt{\frac32}x$, $\sqrt{\frac52}\,\frac{3x^2-1}2 = \sqrt{\frac{45}8}\bigl(x^2 - \frac13\bigr)$: exactly $e_0, e_1, e_2$. **Exercise 13.5 ★★.** Apply [Parseval](#thm-b3-hilbert-parseval) ([Theorem 13.9](#thm-b3-hilbert-fourier)) to $f(t) = t$ and $f(t) = t^2$ on $\intcc{-\pi}\pi$ — now legitimately for these ([continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), but previously the identity needed piecewise-$\mathcal C^1$ care at the wrap-around discontinuity): recover $$ \sum_{n\geq1}\frac1{n^2} = \frac{\pi^2}6, \qquad \sum_{n\geq1}\frac1{n^4} = \frac{\pi^4}{90} . $$ **Solution of Exercise 13.5.** For $f(t) = t$: $c_0 = 0$ and, integrating by parts, $c_n = \frac{\iu(-1)^n}{n}$ for $n \neq 0$: $\abs{c_n}^2 = \frac1{n^2}$. [Parseval](#thm-b3-hilbert-parseval): $$ \frac1{2\pi}\int_{-\pi}^\pi t^2\dd t = \frac{\pi^2}3 = \sum_{n\neq0}\frac1{n^2} = 2\sum_{n\geq1}\frac1{n^2} \ \Longrightarrow\ \sum_{n\geq1}\frac1{n^2} = \frac{\pi^2}6 . $$ For $f(t) = t^2$: $c_0 = \frac{\pi^2}3$, $c_n = \frac{2(-1)^n}{n^2}$ ($n \ne 0$). [Parseval](#thm-b3-hilbert-parseval): $$ \frac1{2\pi}\int_{-\pi}^{\pi}t^4\dd t = \frac{\pi^4}5 = \frac{\pi^4}9 + \sum_{n\neq0}\frac4{n^4} \ \Longrightarrow\ \sum_{n\geq1}\frac1{n^4} = \frac18\Bigl(\frac{\pi^4}5 - \frac{\pi^4}9\Bigr) = \frac{\pi^4}{90} . $$ No piecewise-$\mathcal C^1$ caveats are needed: [Theorem 13.9](#thm-b3-hilbert-fourier) covers every $L^2$ function. **Exercise 13.6 ★★.** (a) Find $a \in L^2(\intcc01)$ with $\int_0^{1/2}f = \langle a, f\rangle$ for all $f$; compute $\norm\varphi$ for this functional. (b) Show that the evaluation $f \mapsto f(\frac12)$, defined on the subspace $\mathcal C(\intcc01) \subseteq L^2(\intcc01)$, is *not* [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) for $\norm\cdot_2$: no Riesz representative exists (evaluation is not an $L^2$ notion). **Solution of Exercise 13.6.** (a) $\varphi(f) = \int_0^{1/2}f = \langle\mathbf 1_{\intcc0{1/2}},\ f\rangle$: the representative is $a = \mathbf 1_{\intcc0{1/2}}$, and $\norm\varphi = \norm a_2 = \frac1{\sqrt2}$ ([Theorem 13.4](#thm-b3-hilbert-riesz)). (b) Take the tent functions $f_n$ with peak $1$ at $\frac12$ and support of width $\frac2n$: $f_n(\tfrac12) = 1$ while $\norm{f_n}_2^2 \leq \frac2n \to 0$: no constant $C$ can give $\abs{f(\frac12)} \leq C\norm f_2$. Point evaluation is meaningless in $L^2$ — elements are classes modulo null sets — and this computation is the quantitative reason. **Exercise 13.7 ★★★.** Let $H$ be separable with [Hilbert basis](#def-b3-hilbert-onb) $(e_n)$, and $(x_k)$ a bounded sequence. (a) Show that some subsequence converges *weakly*: there is $x$ with $\langle y, x_{k_j}\rangle \to \langle y, x\rangle$ for every $y \in H$. *(Diagonal extraction on the coefficients $\langle e_n, x_k\rangle$; assemble $x$ via Bessel and uniform boundedness of norms.)* (b) Show $e_n \rightharpoonup 0$ but $\norm{e_n} = 1$: weak limits can lose norm. Show $\norm x \leq \liminf\norm{x_{k_j}}$ in (a). **Solution of Exercise 13.7.** (a) Let $M = \sup_k\norm{x_k}$. The scalar sequences $(\langle e_n, x_k\rangle)_k$ are bounded by $M$: a diagonal extraction yields $x_{k_j}$ with $\langle e_n, x_{k_j}\rangle \to \gamma_n$ for every $n$. For each $N$: $\sum_{n\leq N}\abs{\gamma_n}^2 = \lim_j\sum_{n\leq N}\abs{\langle e_n, x_{k_j}\rangle}^2 \leq M^2$ (Bessel), so $(\gamma_n) \in \ell^2$ and $x = \sum_n\gamma_ne_n \in H$ ([Theorem 13.7](#thm-b3-hilbert-parseval)(3)). For $y \in H$: $$ \abs{\langle y, x_{k_j} - x\rangle} \leq \Bigl|\sum_{n\leq N}\overline{c_n(y)}\bigl(\langle e_n, x_{k_j}\rangle - \gamma_n\bigr)\Bigr| + 2M\Bigl(\sum_{n>N}\abs{c_n(y)}^2\Bigr)^{1/2}, $$ using the expansion $\langle y, z\rangle = \sum\overline{c_n(y)}c_n(z)$ and Cauchy–Schwarz on the tail; choose $N$ then $j$: weak convergence to $x$. (b) $\langle y, e_n\rangle = c_n(y) \to 0$ for every $y$ ($\ell^2$ tails): $e_n \rightharpoonup 0$, yet $\norm{e_n} = 1$: the norm is not weakly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). In (a): $\norm x^2 = \sum\abs{\gamma_n}^2 \leq \liminf_j\norm{x_{k_j}}^2$ (finite sections and Bessel again): weak limits can only lose norm. **Exercise 13.8 ★★.** (Adjoints) For $T \in \mathcal L(H)$, show there is a unique $T^* \in \mathcal L(H)$ with $\langle Tx, y\rangle = \langle x, T^*y\rangle$ (Riesz), and $\vertiii{T^*} = \vertiii T$. Compute the adjoint of the shift $S$ on $\ell^2$, and prove $\ker T^* = (\operatorname{im}T)^\perp$ — deduce $\overline{\operatorname{im}T} = (\ker T^*)^\perp$. **Solution of Exercise 13.8.** For fixed $y$, $x \mapsto \langle y, Tx\rangle$ is a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) linear functional; Riesz gives a unique $T^*y$ with $\langle y, Tx\rangle = \langle T^*y, x\rangle$ for all $x$ — conjugating, $\langle Tx, y\rangle = \langle x, T^*y\rangle$. Uniqueness makes $T^*$ linear; $$ \norm{T^*y} = \sup_{\norm x = 1}\abs{\langle T^*y, x\rangle} = \sup_{\norm x=1}\abs{\langle y, Tx\rangle} \leq \vertiii T\,\norm y, $$ so $\vertiii{T^*} \leq \vertiii T$, and $T^{**} = T$ gives equality. Shift: $\langle Sx, y\rangle = \sum_{n\geq1} x_n\bar y_{n+1} = \langle x, S^*y\rangle$ with $(S^*y)_n = y_{n+1}$: the backward shift. Kernel–image: $T^*y = 0$ iff $\langle x, T^*y\rangle = 0$ for all $x$ iff $\langle Tx, y\rangle = 0$ for all $x$ iff $y \perp \operatorname{im}T$: $\ker T^* = (\operatorname{im}T)^\perp$; taking $\perp$ and using [Theorem 13.3](#thm-b3-hilbert-decomposition), $\overline{\operatorname{im}T} = (\ker T^*)^\perp$. **Exercise 13.9 ★★.** Let $a$ be as in Lax–Milgram and moreover *symmetric*. Show that $u$ solves $a(u, \cdot) = \varphi$ iff $u$ minimizes $J(v) = \frac12a(v, v) - \varphi(v)$, and that the minimum is attained at exactly one point. *([Complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) the square: $J(u + w) - J(u) = \frac12a(w,w) \geq \frac\alpha2\norm w^2$.)* Application: re-derive the projection theorem for closed subspaces from Lax–Milgram. **Solution of Exercise 13.9.** If $a(u, \cdot) = \varphi$: for any $w$, $$ J(u + w) - J(u) = a(u, w) - \varphi(w) + \tfrac12a(w,w) = \tfrac12a(w,w) \geq \tfrac\alpha2\norm w^2, $$ strictly positive for $w \neq 0$: $u$ is the unique minimizer. Conversely, at a minimizer the function $t \mapsto J(u + tw)$ (a quadratic polynomial in $t$) has vanishing derivative at $0$: $a(u, w) - \varphi(w) = 0$ for every $w$. Projection re-derived: for a closed subspace $F$, apply Lax–Milgram on the [Hilbert space](#def-b3-hilbert-inner) $F$ with $a(u,v) = \langle u, v\rangle$ ($M = \alpha = 1$) and $\varphi(v) = \langle x, v\rangle$: a unique $p \in F$ with $\langle p, v\rangle = \langle x, v\rangle$ for all $v \in F$, i.e. $x - p \perp F$ — and by the symmetric case, $p$ minimizes $\frac12\norm v^2 - \langle x, v\rangle = \frac12\norm{v - x}^2 - \frac12\norm x^2$ over $F$: the projection. **Exercise 13.10 ★★★.** (The Haar system) On $\intcc01$, let $h_{0} = \mathbf 1$, and for $n = 2^j + k$ ($j \geq 0$, $0 \leq k < 2^j$): $$ h_n = 2^{j/2}\Bigl(\mathbf 1_{[k2^{-j},\,(k + \frac12)2^{-j})} - \mathbf 1_{[(k+\frac12)2^{-j},\,(k+1)2^{-j})} \Bigr). $$ Show that $(h_n)_{n\geq0}$ is orthonormal in $L^2(\intcc01)$, and total. *(Orthogonality: disjoint or nested supports; totality: finite spans contain all dyadic step functions, which are dense — via [Theorem 12.6](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#thm-b3-lp-density)(1) and dyadic approximation of intervals.)* The Haar system is the ancestor of wavelets. **Solution of Exercise 13.10.** Normalization: $\int h_n^2 = 2^j\cdot 2^{-j} = 1$. Orthogonality: two distinct Haar functions have either disjoint ([interiors](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) of) supports (product zero a.e.), or the support of the finer is contained in a half-interval where the coarser is constant — then the integral of the product is that constant times $\int h_{\text{finer}} = 0$; against $h_0 = \mathbf 1$, again $\int h_n = 0$. Totality: the span of $\{h_0, \dots, h_{2^J-1}\}$ consists of step functions on the dyadic grid of step $2^{-J}$; both spaces have dimension $2^J$ and the Haar functions are independent (orthonormal): the span is *all* such step functions. Dyadic step functions are dense in $L^2(\intcc01)$: [simple functions](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-simple) are dense ([Theorem 12.6](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#thm-b3-lp-density)(1)), [measurable](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#def-b3-lebesgue-measurable) sets are approximated by finite unions of intervals ([Exercise 9.7](https://one-course.com/books/math/5/en/chapter/9-measure-theory#exo-b3-measure-7)), and intervals by dyadic ones (endpoints move by $\leq 2^{-J}$). By [Theorem 13.7](#thm-b3-hilbert-parseval), the Haar system is a [Hilbert basis](#def-b3-hilbert-onb). **Exercise 13.11 ★★.** (Orthogonal projections, characterized) Let $H$ be a [Hilbert space](#def-b3-hilbert-inner) and $P \in \mathcal L(H)$ with $P^2 = P$, $P \neq 0$. Show the equivalence of: (i) $P$ is the orthogonal projection onto $\operatorname{im}P$; (ii) $P = P^*$ ([Exercise 13.8](#exo-b3-hilbert-8)); (iii) $\vertiii P = 1$. *(For (iii) $\Rightarrow$ (i): if some $x \in (\ker P)^\perp$ had $Px \neq x$, consider $x + t(Px - x)$ — or directly: for $u \in \operatorname{im}P$ and $v \in \ker P$, expand $\norm{P(u + tv)}^2 \leq \norm{u + tv}^2$ for all $t \in \R$ and conclude $\langle u, v\rangle = 0$.)* Exhibit a non-orthogonal projection on $\R^2$ and compute its norm. **Solution of Exercise 13.11.** (i) $\Rightarrow$ (ii): for the orthogonal projection, $\langle Px, y\rangle = \langle Px, Py\rangle = \langle x, Py\rangle$ (insert the decompositions $x = Px + (x - Px)$ etc. and kill cross terms). (ii) $\Rightarrow$ (iii): $\norm{Px}^2 = \langle P^2x, x\rangle = \langle Px, x\rangle \leq \norm{Px}\norm x$, so $\vertiii P \leq 1$, and $Pu = u$ on the nonzero image: $= 1$. (iii) $\Rightarrow$ (i): $H = \operatorname{im}P \oplus \ker P$ ([algebraically](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic), from $P^2 = P$); take $u = Pu \in \operatorname{im}P$, $v \in \ker P$, $t \in \R$: $\norm{P(u + tv)}^2 = \norm u^2$ must be $\leq \norm{u + tv}^2 = \norm u^2 + 2t\operatorname{Re}\langle u, v\rangle + t^2\norm v^2$ for every $t$, forcing $\operatorname{Re}\langle u, v\rangle = 0$ (compare the linear terms as $t \to 0^\pm$); replacing $v$ by $\iu v$ kills the imaginary part too: $\operatorname{im}P \perp \ker P$, which is exactly orthogonality of the projection. Example: $P(x, y) = (x + y, 0)$ on $\R^2$: $P^2 = P$, image the $x$-axis, kernel the line $y = -x$, and $\vertiii P = \sup\frac{\abs{x+y}}{\norm{(x,y)}} = \sqrt2$ (attained at $(1,1)/\sqrt2$): an oblique projection has norm $> 1$. (For the record, (ii) also gives (i) directly: $\ker P = \ker P^* = (\operatorname{im}P)^\perp$ by [Exercise 13.8](#exo-b3-hilbert-8).) **Exercise 13.12 ★★★.** (Von Neumann’s ergodic theorem) Let $U \in \mathcal L(H)$ be *unitary* ($U^*U = UU^* = I$), $F = \ker(U - I)$ the fixed space, $P$ the orthogonal projection onto $F$, and $A_n = \frac1n\sum_{k=0}^{n-1}U^k$. (a) Show $\ker(U - I) = \ker(U^* - I)$ *(from $\norm{Ux - x}^2 = 2\norm x^2 - 2\operatorname{Re}\langle Ux, x\rangle$ and unitarity)*, and deduce $\overline{\operatorname{im}(U - I)} = F^\perp$. (b) Show that $A_nx \to x$ for $x \in F$, and $A_nx \to 0$ for $x \in \operatorname{im}(U - I)$ *(telescoping)*, then for $x \in \overline{\operatorname{im}(U - I)}$ (uniform bound $\vertiii{A_n} \leq 1$). (c) Conclude: $A_nx \to Px$ for *every* $x \in H$ — time averages converge to the projection on the invariants. (d) Spell it out for $H = L^2(\R/\Z)$ and $Uf = f(\cdot + \alpha)$ with $\alpha$ irrational: identify $F$ (use Fourier series, [Theorem 13.9](#thm-b3-hilbert-fourier)) and deduce that $\frac1n\sum_{k 0$ a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) *weight* on the [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) of $I$ such that $\int_I \abs t^nw(t)\dd t < \infty$ for all $n$; work in $H = L^2(I, w\,\dd\lambda)$ with $\langle f, g\rangle = \int_I \bar fg\,w$. Gram–Schmidt applied to $1, t, t^2, \dots$ produces the *[orthogonal polynomials](#pb-b3-hilbert-1)* $(p_n)$ for $w$ (monic normalization: $p_n = t^n + \cdots$). **Part I — General theory.** 1. Show that $p_n$ is orthogonal to every polynomial of degree $< n$ , and that $(p_0, \dots, p_n)$ is a basis of $\R_n[t]$ . 2. (Three-term recurrence) Show there are reals $a_n, b_n$ with $$p_{n+1}(t) = (t - a_n)\,p_n(t) - b_n\,p_{n-1}(t), \qquad b_n = \frac{\norm{p_n}^2}{\norm{p_{n-1}}^2} > 0 .$$ *(Expand $t\,p_n$ in the basis $(p_k)_{k \leq n+1}$ and kill coefficients by orthogonality, using $\langle tp_n, p_k\rangle = \langle p_n, tp_k\rangle$.)* 3. (Roots) Show that $p_n$ has $n$ *distinct* roots, all [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) to $I$ . *(Let $t_1 < \dots < t_m$ be the [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) sign changes of $p_n$; if $m < n$, test $p_n$ against $\prod_{i\leq m}(t - t_i)$ and contradict orthogonality.)* **Part II — Legendre ($I = \intcc{-1}1$, $w = 1$).** Define $P_n(t) = \frac{1}{2^nn!}\,\frac{\dd^n}{\dd t^n}\bigl[(t^2 - 1)^n\bigr]$ (Rodrigues). 4. Show $\deg P_n = n$ with leading coefficient $\frac{(2n)!}{2^n(n!)^2}$ , and, integrating by parts $n$ times, that $\langle P_n, Q\rangle = 0$ for every polynomial $Q$ of degree $< n$ : the $P_n$ are (up to normalization) the [orthogonal polynomials](#pb-b3-hilbert-1) for $w = 1$ . 5. Compute $\norm{P_n}_2^2 = \frac{2}{2n+1}$ *(integrate by parts $n$ times against itself and reduce to a Beta/Wallis integral, [Exercise 11.8](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#exo-b3-product-8))* . 6. Show that the normalized Legendre polynomials form a [Hilbert basis](#def-b3-hilbert-onb) of $L^2(\intcc{-1}1)$ *(Weierstrass, [Corollary 7.16](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#cor-b3-complete-weierstrass), plus [density](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#ex-b3-lebesgue-gamma) of $\mathcal C$ in $L^2$)* , and expand $f(t) = \abs t$ up to degree $2$ : compute the best quadratic $L^2$ -approximation of $\abs t$ . **Part III — Hermite ($I = \R$, $w(t) = \eu^{-t^2}$).** Define $H_n(t) = (-1)^n\eu^{t^2}\frac{\dd^n}{\dd t^n}\eu^{-t^2}$. 7. Show that $H_n$ is a polynomial of degree $n$ with leading coefficient $2^n$ , that $H_{n+1} = 2tH_n - H_n'$ , and that $\langle H_m, H_n\rangle_w = \delta_{mn}\,2^nn!\sqrt\pi$ *(parts again)* . 8. Show that the Hermite family is total in $L^2(\R, \eu^{-t^2}\dd t)$ , admitting one result from [Chapter 14](https://one-course.com/books/math/5/en/chapter/14-the-fourier-transform#ch-b3-fouriertransform) : if $g \in L^1(\R)$ has $\int g(t)\eu^{-\iu\xi t}\dd t = 0$ for all $\xi$ , then $g = 0$ a.e. *(For $f \perp$ all $H_n$, i.e. $\perp$ all polynomials: show $z \mapsto \int f(t)\eu^{-t^2}\eu^{-\iu zt}\dd t$ is well defined, expand the exponential as a series, justify the interchange by domination, and conclude that the Fourier transform of $f\eu^{-t^2}$ vanishes.)* **Part IV — Gauss quadrature.** Fix $n$, let $t_1 < \dots < t_n$ be the roots of $p_n$ (Part I), and define the weights $w_i = \int_I \ell_i(t)\,w(t)\dd t$ where $\ell_i$ are the Lagrange interpolation basis polynomials at the $t_i$. 9. Show that the quadrature rule $Q(f) = \sum_iw_if(t_i)$ is exact on all polynomials of degree $\leq n - 1$ (interpolation), and in fact — the miracle — on all polynomials of degree $\leq 2n - 1$ : write $P = qp_n + r$ and use orthogonality on the quotient $q$ . 10. Show that the weights are positive *(apply the rule to $\ell_i^2$, of degree $2n - 2$)* , and deduce from Polya’s theorem ( [Exercise 8.9](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#exo-b3-banach-9) ) that Gauss quadrature converges: $Q_n(f) \to \int_I fw$ for every [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f$ on a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $I$ . 11. For $n = 2$ , $I = \intcc{-1}1$ , $w = 1$ : compute the nodes $\pm\frac1{\sqrt3}$ and weights $1, 1$ , and verify exactness on $1, t, t^2, t^3$ by hand. Compare with the trapezoid rule on the same two evaluation points. **Part V — Chebyshev: the polynomials that oscillate best.** Now $I = \intcc{-1}1$ and $w(t) = \frac1{\sqrt{1 - t^2}}$. 12. Show that $T_n(\cos\theta) = \cos n\theta$ defines a polynomial $T_n$ of degree $n$ (establish $T_{n+1} = 2t\,T_n - T_{n-1}$ from a trigonometric identity), with leading coefficient $2^{n-1}$ for $n \geq 1$; and that the substitution $t = \cos\theta$ gives $$\langle T_m, T_n\rangle_w = \int_0^\pi\cos m\theta\,\cos n\theta\,\dd\theta = 0 \ (m \neq n), \qquad \norm{T_0}_w^2 = \pi,\ \ \norm{T_n}_w^2 = \frac\pi2 :$$ the $T_n$ are the [orthogonal polynomials](#pb-b3-hilbert-1) for this weight, and Chebyshev expansions *are* Fourier cosine series in disguise. 13. Locate explicitly the $n$ roots $t_k = \cos\frac{(2k-1)\pi}{2n}$ and the $n + 1$ extrema $s_j = \cos\frac{j\pi}n$ of $T_n$ on $\intcc{-1}1$ , where $T_n(s_j) = (-1)^j$ : the graph *equioscillates* between $\pm1$ . 14. (Minimax) Show that among all *monic* polynomials of degree $n$ , the polynomial $2^{1-n}T_n$ has the smallest sup-norm on $\intcc{-1}1$ , namely $2^{1-n}$ — and it is the unique minimizer. *(If a monic $P$ had $\sup\abs P < 2^{1-n}$, the difference $2^{1-n}T_n - P$, of degree $\leq n-1$, would alternate in sign at the $n+1$ equioscillation points.)* 15. Application to interpolation: for nodes $t_1 < \dots < t_n$ in $\intcc{-1}1$ , the error of Lagrange interpolation of a $\mathcal C^n$ function involves $\omega(t) = \prod_i(t - t_i)$ . Show that choosing Chebyshev roots as nodes minimizes $\sup_{\intcc{-1}1}\abs\omega$ , and give the resulting bound $\norm{f - L_nf}_\infty \leq \frac{\norm{f^{(n)}}_\infty} {2^{n-1}\,n!}$ — compare with equally spaced nodes (state Runge’s phenomenon as the cautionary tale). 16. Verify $\abs{T_n'(\pm1)} = n^2$ *(differentiate $T_n(\cos\theta) = \cos n\theta$ and take limits $\theta \to 0, \pi$)* : polynomials bounded by $1$ on $\intcc{-1}1$ can have derivative as large as $n^2$ at the edge (Markov’s inequality says no larger — statement only). Where in the interval is the derivative bound only $O(n)$ ? 17. (Chebyshev–Gauss quadrature) Show that the Gauss rule for the weight $w$ at the $n$ Chebyshev roots has *equal* weights $w_i = \frac\pi n$ *(exactness on $T_0, \dots, T_{n-1}$ plus the trigonometric sums $\sum_{k=1}^n\cos\bigl(j\tfrac{(2k-1)\pi}{2n}\bigr) = 0$ for $1 \leq j \leq n - 1$)* : the most uniform of all quadratures. Write it out for $n = 3$ . **Part VI — Christoffel–Darboux, interlacing, and the Jacobi matrix.** Back to a general weight; $h_k = \norm{p_k}^2$ (monic $p_k$), $b_k = h_k/h_{k-1}$. 18. (Least norm) Show that among all *monic* polynomials of degree $n$ , the orthogonal $p_n$ is the unique one of minimal $L^2(w)$ -norm — identify the minimization as an orthogonal projection onto $\R_{n-1}[t]$ ( [Theorem 13.2](#thm-b3-hilbert-projection) or the finite-dimensional projection of Year 2). The minimax property of question 14 is the same statement with $L^\infty$ in place of $L^2$ : same hero, two norms. 19. (Christoffel–Darboux) Prove, by induction on $n$ using the three-term recurrence, the identity $$\sum_{k=0}^{n}\frac{p_k(x)\,p_k(y)}{h_k} = \frac{p_{n+1}(x)\,p_n(y) - p_n(x)\,p_{n+1}(y)}{h_n\,(x - y)} \qquad (x \neq y),$$ and its confluent form ($y \to x$): $\sum_{k\leq n}\frac{p_k(x)^2}{h_k} = \frac{p_{n+1}'(x)p_n(x) - p_n'(x)p_{n+1}(x)}{h_n}$. 20. Deduce that $p_n$ and $p_{n+1}$ have no common root, and that at every root $x_0$ of $p_{n+1}$ : $p_n(x_0)\,p_{n+1}'(x_0) > 0$ . Conclude the *interlacing* of roots: between two consecutive roots of $p_{n+1}$ lies exactly one root of $p_n$ . 21. (Jacobi matrix) Let $J_n$ be the $n\times n$ symmetric tridiagonal matrix with diagonal $a_0, \dots, a_{n-1}$ and off-diagonal entries $\sqrt{b_1}, \dots, \sqrt{b_{n-1}}$ . Show by induction that $\det(tI_n - J_n) = p_n(t)$ , so the roots of $p_n$ are the eigenvalues of a real symmetric matrix — re-proving in one line that they are real, and (with the interlacing above) tying [orthogonal polynomials](#pb-b3-hilbert-1) to the spectral world of [Chapter 15](https://one-course.com/books/math/5/en/chapter/15-compact-operators-and-the-spectral-theorem#ch-b3-spectral) . 22. (Synthesis) Assemble the dictionary for the three classical families (Legendre, Hermite, Chebyshev): interval, weight, defining formula, three-term recurrence, norm, and the natural habitat of each (quadrature and approximation on compacta; Gaussian analysis; minimax and Fourier-cosine methods). One sentence on what the general theory (Parts I, VI) gave that no individual computation could. **Part VII — The error term, and the kernel behind the weights.** Here $I$ is [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) and $f \in \mathcal C^{2n}(I)$. 23. (Gauss error formula) Let $Hf$ be the Hermite interpolant of degree $\leq 2n - 1$ matching $f$ and $f'$ at the nodes $t_1, \dots, t_n$ (prove its existence and the pointwise error $$f(t) - Hf(t) = \frac{f^{(2n)}(\xi_t)}{(2n)!}\;p_n(t)^2$$ by the usual auxiliary-function argument). Deduce, by integrating this identity against $w$ and squeezing between the extrema of $f^{(2n)}$, that $$\int_I f\,w - Q_n(f) = \frac{f^{(2n)}(\xi)}{(2n)!}\,h_n \qquad\text{for some } \xi \in I,$$ with $h_n = \norm{p_n}^2$ as in Part VI: Gauss quadrature errs by one $2n$-th derivative, weighted by the squared norm of the monic orthogonal polynomial. 24. (The weights are Christoffel values) Using the reproducing kernel $K_n(x, y) = \sum_{k=0}^{n-1} \frac{p_k(x)p_k(y)}{h_k}$ of $\R_{n-1}[t]$ and the exactness of $Q_n$ up to degree $2n - 2$, prove $$w_i \;=\; \Bigl(\,\sum_{k=0}^{n-1} \frac{p_k(t_i)^2}{h_k}\Bigr)^{\!-1} :$$ each weight is the value at its node of the *Christoffel function* — positivity of the weights (question 10) again, now with an exact formula. Verify it recovers $w_1 = w_2 = 1$ for $n = 2$, $I = \intcc{-1}1$, $w = 1$. 25. (Everything checks on one integral) For the Chebyshev weight and $n = 3$ nodes, compute both sides of $$\int_{-1}^{1}\frac{t^6}{\sqrt{1 - t^2}}\,\dd t = \frac{5\pi}{16}, \qquad Q_3(t^6) = \frac{9\pi}{32},$$ so the quadrature error is exactly $\frac{\pi}{32}$; then check that the error formula of question 23 predicts precisely this value (here $f^{(6)} = 6!$ is constant, and $h_3 = \norm{2^{-2}T_3}_w^2 = \frac\pi{32}$): theory and computation agree to the last digit. **Solution of Problem 13.1.** **1.** Gram–Schmidt guarantees $\operatorname{Vect}(p_0, \dots, p_n) = \operatorname{Vect}(1, \dots, t^n) = \R_n[t]$ and $p_n \perp p_k$ ($k < n$), hence $p_n \perp \R_{n-1}[t]$. The $p_k$, of strictly increasing degrees, are independent: a basis. **2.** $t\,p_n$ is monic of degree $n + 1$: expand $t\,p_n = p_{n+1} + \sum_{k\leq n}c_kp_k$ with $c_k = \langle p_k, tp_n\rangle/\norm{p_k}^2$. For $k \leq n - 2$: $\langle p_k, tp_n\rangle = \langle tp_k, p_n\rangle = 0$ (degree $k + 1 < n$). So $tp_n = p_{n+1} + a_np_n + b_np_{n-1}$, the stated recurrence, with $$ b_n = \frac{\langle p_{n-1}, tp_n\rangle}{\norm{p_{n-1}}^2} = \frac{\langle tp_{n-1}, p_n\rangle}{\norm{p_{n-1}}^2} = \frac{\langle p_n + (\text{lower}),\ p_n\rangle}{\norm{p_{n-1}}^2} = \frac{\norm{p_n}^2}{\norm{p_{n-1}}^2} > 0 . $$ **3.** Let $t_1 < \dots < t_m$ be the points [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) to $I$ where $p_n$ changes sign, and $q = \prod_{i\leq m}(t - t_i)$ (with $q = 1$ if $m = 0$). Then $p_nq$ has constant sign on $I$ and is not a.e. zero: $\int_Ip_nq\,w \neq 0$. If $m < n$, this contradicts $p_n \perp \R_{n-1}[t]$. So $m = n$: $p_n$ has $n$ distinct [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) roots (it has at most $n$ roots in all). **4.** $(t^2 - 1)^n$ has degree $2n$; $n$ derivatives leave degree $n$, with leading coefficient $\frac{(2n)(2n-1)\cdots(n+1)}{2^nn!} = \frac{(2n)!}{2^n(n!)^2}$. For $\deg Q < n$, integrate by parts $n$ times: all boundary terms contain a derivative of order $< n$ of $(t^2-1)^n$, which vanishes at $\pm1$ (root of order $n$); after $n$ steps the integrand carries $Q^{(n)} = 0$. **5.** With $u = (t^2 - 1)^n$: $$ (2^nn!)^2\norm{P_n}^2 = \int_{-1}^1(u^{(n)})^2 = (-1)^n\int_{-1}^1 u\,u^{(2n)} = (2n)!\int_{-1}^1(1 - t^2)^n\dd t , $$ ($u^{(2n)} = (2n)!$; boundary terms vanish as in question 4). And $\int_{-1}^1(1-t^2)^n\dd t = B(\tfrac12, n+1) = \frac{\Gamma(\frac12)\Gamma(n+1)}{\Gamma(n + \frac32)} = \frac{2\cdot4^n(n!)^2}{(2n+1)!}$ ([Exercise 11.8](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#exo-b3-product-8)). Combining: $\norm{P_n}^2 = \frac{2}{2n + 1}$. **6.** Polynomials are $\norm\cdot_\infty$-dense in $\mathcal C(\intcc{-1}1)$ (Weierstrass, [Corollary 7.16](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#cor-b3-complete-weierstrass)), [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) functions are $L^2$-dense ([Theorem 12.6](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#thm-b3-lp-density)), and $\norm\cdot_2 \leq \sqrt2\norm\cdot_\infty$: polynomial spans are total, so the normalized $P_n$ form a [Hilbert basis](#def-b3-hilbert-onb). Expansion of $\abs t$: the coefficient against $P_0$ is $\frac{\langle P_0, \abs t\rangle}{\norm{P_0}^2} = \frac12$; against $P_1$: $0$ (parity); against $P_2$: $\frac{\int_{-1}^1\abs t\,\frac{3t^2-1}2\dd t}{2/5} = \frac{1/4}{2/5} = \frac58$. Best quadratic approximation: $$ \abs t \approx \frac12 + \frac58\,P_2(t) = \frac{3}{16} + \frac{15}{16}\,t^2 . $$ **7.** From $\frac{\dd^{n+1}}{\dd t^{n+1}}\eu^{-t^2} = \frac{\dd^n}{\dd t^n}(-2t\,\eu^{-t^2})$ and Leibniz, $H_{n+1} = 2tH_n - H_n'$; induction gives degree $n$ and leading coefficient $2^n$. For $m < n$, integrate by parts $n$ times in $\int H_m H_n\eu^{-t^2} = (-1)^n\int H_m\,\bigl(\eu^{-t^2} \bigr)^{(n)}$: boundary terms (polynomial $\times$ $\eu^{-t^2}$) vanish at $\pm\infty$, leaving $\int H_m^{(n)}\,\eu^{-t^2} = 0$. For $m = n$: $H_n^{(n)} = 2^nn!$, so $\norm{H_n}_w^2 = 2^nn!\int\eu^{-t^2} = 2^nn!\sqrt\pi$. **8.** Let $f \in L^2(\R, \eu^{-t^2}\dd t)$ be orthogonal to every polynomial, and $g = f\eu^{-t^2}$. Then $g \in L^1$: $\int\abs f\eu^{-t^2} \leq \bigl(\int\abs f^2\eu^{-t^2}\bigr)^{1/2}\bigl(\int\eu^{-t^2}\bigr)^{1/2}$ (Cauchy–Schwarz). For $\xi \in \R$, expand $\eu^{-\iu\xi t}$: the partial sums are dominated since $$ \sum_k\frac{\abs\xi^k}{k!}\int\abs f\,\abs t^k\eu^{-t^2}\dd t \leq \Bigl(\int \abs f^2\eu^{-t^2}\Bigr)^{1/2} \sum_k\frac{\abs\xi^k}{k!}\Bigl(\int t^{2k}\eu^{-t^2}\Bigr)^{1/2} < \infty $$ (the last series converges: $\int t^{2k}\eu^{-t^2} = \Gamma(k+\frac12) \leq k!\,\sqrt\pi$, so the terms are $O(\abs\xi^k/\sqrt{k!})$). Term-by-term integration ([Corollary 10.7](https://one-course.com/books/math/5/en/chapter/10-the-lebesgue-integral#cor-b3-lebesgue-additivity) applied to the absolute series, then Fubini for series) gives $$ \int_\R g(t)\,\eu^{-\iu\xi t}\dd t = \sum_k\frac{(-\iu\xi)^k}{k!}\int f(t)\,t^k\,\eu^{-t^2}\dd t = 0 , $$ each integral being $\langle t^k, f\rangle_w$-type $= 0$. By the admitted injectivity of the Fourier transform ([Chapter 14](https://one-course.com/books/math/5/en/chapter/14-the-fourier-transform#ch-b3-fouriertransform)), $g = 0$ a.e., so $f = 0$ a.e.: the Hermite family (whose spans are the polynomials) is total. **9.** Exactness to degree $n - 1$: for such $P$, $P = \sum_iP(t_i)\ell_i$ exactly, so $\int Pw = \sum_iP(t_i)\int \ell_iw = Q(P)$. Degree $\leq 2n - 1$: divide $P = qp_n + r$, $\deg q \leq n - 1$, $\deg r \leq n-1$; then $\int Pw = \int qp_nw + \int rw = 0 + Q(r)$ ($p_n \perp \R_{n-1}[t]$), while $Q(P) = \sum_iw_i\bigl(q(t_i)\,p_n(t_i) + r(t_i)\bigr) = Q(r)$ since the nodes are the roots of $p_n$. Equal. **10.** $\ell_i^2$ has degree $2n - 2 \leq 2n - 1$ and $\ell_i^2(t_j) = \delta_{ij}$: $0 < \int\ell_i^2w = Q(\ell_i^2) = w_i$. Polya ([Exercise 8.9](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#exo-b3-banach-9), transported to $I$ with weight): condition (i) holds — each polynomial is integrated exactly once $2n - 1 \geq$ its degree; condition (ii): $\sum_i\abs{w_{i}} = \sum_iw_i = Q(\mathbf 1) = \int_Iw$, bounded: $Q_n(f) \to \int fw$ for every $f \in \mathcal C(I)$, $I$ [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact). **11.** Monic $p_2 = t^2 - \frac13$ (from [Exercise 13.4](#exo-b3-hilbert-4)): nodes $\pm\frac1{\sqrt3}$. Weights: $\ell_1(t) = \frac{t - \frac1{\sqrt3}}{-\frac2{\sqrt3}}$, and $w_1 = \int_{-1}^1\ell_1 = 1$; by symmetry $w_2 = 1$. Exactness: $\int 1 = 2 = 1 + 1$; $\int t = 0 = -\frac1{\sqrt3} + \frac1{\sqrt3}$; $\int t^2 = \frac23 = \frac13 + \frac13$; $\int t^3 = 0$. The two-point trapezoid rule (nodes $\pm1$, weights $1, 1$) is exact only to degree $1$: on $t^2$ it returns $2$ instead of $\frac23$. Same cost, two extra degrees of exactness: the payoff of orthogonal nodes. **12.** From $\cos(n{+}1)\theta + \cos(n{-}1)\theta = 2\cos\theta\cos n\theta$: $T_{n+1} = 2tT_n - T_{n-1}$ with $T_0 = 1$, $T_1 = t$; induction gives polynomials of degree $n$ with leading coefficient $2^{n-1}$ ($n \geq 1$). Substituting $t = \cos\theta$ ($w(t)\dd t \mapsto \dd\theta$): $\langle T_m, T_n\rangle_w = \int_0^\pi\cos m\theta\cos n\theta\,\dd\theta = 0$ for $m \neq n$, $= \pi$ for $m = n = 0$, $= \frac\pi2$ otherwise (product-to-sum). Degrees and pairwise orthogonality identify the $T_n$ with the Gram–Schmidt output up to scalars; a Chebyshev expansion of $f$ is exactly the Fourier cosine series of $\theta \mapsto f(\cos\theta)$. **13.** $T_n(t) = 0$ iff $\cos n\theta = 0$ iff $\theta = \frac{(2k-1)\pi}{2n}$: the $n$ distinct roots $t_k = \cos\frac{(2k-1)\pi}{2n} \in \intoo{-1}1$. Extrema: $\abs{T_n} \leq 1$ on $\intcc{-1}1$, with $T_n(s_j) = (-1)^j$ at the $n + 1$ points $s_j = \cos\frac{j\pi}n$: [perfect](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#prop-b3-galois-perfect) equioscillation. **14.** $2^{1-n}T_n$ is monic with sup-norm $2^{1-n}$. If a monic $P$ of degree $n$ had $\sup\abs P < 2^{1-n}$, the difference $D = 2^{1-n}T_n - P$ would have degree $\leq n - 1$ (leading terms cancel) yet alternate in sign at $s_0 > \dots > s_n$ (there $2^{1-n}T_n = \pm2^{1-n}$ dominates $P$): at least $n$ zeros — $D \equiv 0$, contradiction. For uniqueness at equality, the same $D$ satisfies $(-1)^jD(s_j) \geq 0$; a nonzero polynomial of degree $\leq n-1$ cannot have $n$ weakly alternating extremal constraints without $n$ roots counted properly (if $D(s_j) = 0$ for some [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) $s_j$, that zero is double in the counting since $D$ keeps a sign locally): again $D \equiv 0$. **15.** The Lagrange error formula (Rolle, Year 2) gives $f - L_nf = \frac{f^{(n)}(\xi_t)}{n!}\,\omega(t)$, so the uniform error is at most $\frac{\norm{f^{(n)}}_\infty} {n!}\,\sup\abs\omega$, and $\omega$ is monic of degree $n$: by question 14, $\sup_{\intcc{-1}1}\abs\omega \geq 2^{1-n}$ with equality iff the nodes are the Chebyshev roots. Hence the optimal bound $\norm{f - L_nf}_\infty \leq \frac{\norm{f^{(n)}}_\infty}{2^{n-1}n!}$. With equally spaced nodes, $\sup\abs\omega$ is exponentially larger near the endpoints, and interpolating even $\frac1{1 + 25t^2}$ diverges there as $n \to \infty$ (Runge’s phenomenon); Chebyshev nodes are the cure. **16.** Differentiating $T_n(\cos\theta) = \cos n\theta$: $T_n'(\cos\theta) = \frac{n\sin n\theta} {\sin\theta}$, which tends to $n^2$ as $\theta \to 0$ and to $(-1)^{n+1}n^2$ as $\theta \to \pi$: $\abs{T_n'(\pm1)} = n^2$. At [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) points, $\abs{T_n'(t)} \leq \frac{n}{\sqrt{1 - t^2}} = O(n)$: the quadratic blow-up lives only at the edges (Bernstein’s [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) bound versus Markov’s global one). **17.** Let $\theta_k = \frac{(2k-1)\pi}{2n}$ and $S_j = \sum_{k=1}^n\cos(j\theta_k)$ for $1 \leq j \leq n-1$. Then $$ S_j = \operatorname{Re}\Bigl[\eu^{\iu j\pi/2n} \sum_{k=0}^{n-1}\eu^{\iu jk\pi/n}\Bigr] = \operatorname{Re}\Bigl[\eu^{\iu j\pi/2n}\, \frac{\eu^{\iu j\pi} - 1}{\eu^{\iu j\pi/n} - 1}\Bigr] . $$ For $j$ even the numerator vanishes: $S_j = 0$. For $j$ odd the numerator is $-2$, and $\eu^{\iu j\pi/n} - 1 = \eu^{\iu j\pi/2n}\cdot2\iu\sin\frac{j\pi}{2n}$, so the whole expression is $\frac{-2}{2\iu\sin(j\pi/2n)} = \frac{\iu}{\sin(j\pi/2n)}$: purely imaginary, $S_j = 0$ again. Hence the equal-weight rule $\frac\pi n\sum_kf(t_k)$ integrates $T_0$ ($\sum w_i = \pi = \int w$) and kills $T_1, \dots, T_{n-1}$ exactly as $\int T_jw = 0$ does: it is exact to degree $n - 1$. Weights exact to degree $n-1$ at given nodes are unique (Lagrange basis): the Gauss weights are all $\frac\pi n$. For $n = 3$: nodes $\pm\frac{\sqrt3}2, 0$ and $$ \int_{-1}^1\frac{f(t)}{\sqrt{1 - t^2}}\,\dd t \approx \frac\pi3\Bigl[f\Bigl(\tfrac{\sqrt3}2\Bigr) + f(0) + f\Bigl(-\tfrac{\sqrt3}2\Bigr)\Bigr], $$ exact through degree $5$. **18.** For monic $P$ of degree $n$: $P = p_n + r$ with $r \in \R_{n-1}[t]$, and $p_n \perp \R_{n-1}[t]$ (question 1), so $\norm P^2 = \norm{p_n}^2 + \norm r^2 \geq \norm{p_n}^2$, with equality iff $r = 0$: $p_n$ is the orthogonal projection residue of $t^n$ onto $\R_{n-1}[t]^\perp$, i.e. the monic polynomial closest to the subspace it must avoid. Chebyshev’s $2^{1-n}T_n$ answers the same question for the sup-norm: least deviation from zero, once in $L^2(w)$, once in $L^\infty$. **19.** Write $K_n(x, y) = \sum_{k=0}^n\frac{p_k(x)p_k(y)}{h_k}$. Base $n = 0$: $(x - y)\frac1{h_0} = \frac{p_1(x)\cdot1 - 1\cdot p_1(y)}{h_0}$ since $p_1 = t - a_0$. Step: assuming the identity for $n - 1$, $$ (x - y)\,K_n(x,y) = \frac{p_n(x)p_{n-1}(y) - p_{n-1}(x)p_n(y)}{h_{n-1}} + \frac{(x - y)\,p_n(x)p_n(y)}{h_n} ; $$ substitute $x\,p_n(x) = p_{n+1}(x) + a_np_n(x) + b_np_{n-1}(x)$ and $y\,p_n(y) = p_{n+1}(y) + a_np_n(y) + b_np_{n-1}(y)$ in the second term: the $a_n$ contributions cancel, and the $b_n = \frac{h_n}{h_{n-1}}$ contributions cancel the induction term; what survives is $\frac{p_{n+1}(x)p_n(y) - p_n(x)p_{n+1}(y)}{h_n}$. The confluent form follows by letting $y \to x$ (both sides are polynomials in $y$). **20.** The confluent form gives $p_{n+1}'p_n - p_n'p_{n+1} = h_n\sum_{k\leq n}\frac{p_k^2}{h_k} \geq \frac{h_n}{h_0} > 0$ everywhere. At a root $x_0$ of $p_{n+1}$: $p_{n+1}'(x_0)\,p_n(x_0) > 0$, so $p_n(x_0) \neq 0$ (no common roots). Between consecutive roots $x_0 < x_1$ of $p_{n+1}$ (all simple, Part I), $p_{n+1}'$ has opposite signs, hence so does $p_n$: a root of $p_n$ lies in each of the $n$ gaps — and that exhausts its $n$ roots: interlacing. **21.** Expanding $D_n(t) = \det(tI_n - J_n)$ along the last row: $D_n = (t - a_{n-1})D_{n-1} - b_{n-1}D_{n-2}$, with $D_0 = 1$, $D_1 = t - a_0$: the recurrence and seeds of the monic $p_n$, so $D_n = p_n$. Roots of $p_n$ = eigenvalues of the symmetric $J_n$: real, and simple by question 19 — Gauss quadrature is the spectral theory of a tridiagonal matrix in disguise, the finite-dimensional shadow of [Chapter 15](https://one-course.com/books/math/5/en/chapter/15-compact-operators-and-the-spectral-theorem#ch-b3-spectral). **22.** Dictionary: | | Legendre | Hermite | Chebyshev | | --- | --- | --- | --- | | interval | $\intcc{-1}1$ | $\R$ | $\intcc{-1}1$ | | weight | $1$ | $\eu^{-t^2}$ | $(1-t^2)^{-1/2}$ | | formula | Rodrigues | $(-1)^n\eu^{t^2} \frac{\dd^n}{\dd t^n}\eu^{-t^2}$ | $\cos(n\arccos t)$ | | norm$^2$ | $\frac2{2n+1}$ | $2^nn!\sqrt\pi$ | $\pi, \frac\pi2$ | | habitat | quadrature | Gaussian calculus | minimax | (each with its three-term recurrence: general form for Legendre, $H_{n+1} = 2tH_n - 2nH_{n-1}$, $T_{n+1} = 2tT_n - T_{n-1}$). The general theory supplied what no single family shows: reality and interlacing of roots, positivity of quadrature weights, the mere existence of the recurrence and of Christoffel–Darboux — consequences of orthogonality alone, uniform in the weight. **23.** Existence: the linear map $\R_{2n-1}[t] \to \R^{2n}$, $P \mapsto (P(t_1), P'(t_1), \dots, P(t_n), P'(t_n))$, is injective (a $P$ in the kernel has $n$ double roots and degree $\leq 2n - 1$, so $P = 0$) between spaces of equal dimension $2n$: bijective. Pointwise error: fix $t$ not a node and choose $K$ so that $g(s) = f(s) - Hf(s) - K\,p_n(s)^2$ vanishes at $s = t$. Then $g$ vanishes at the $n + 1$ distinct points $t, t_1, \dots, t_n$, and $g'$ vanishes at each $t_i$ too (both $f - Hf$ and $p_n^2$ have double zeros there). Rolle gives $n$ zeros of $g'$ strictly between consecutive zeros of $g$ — distinct from the nodes — so $g'$ has $2n$ distinct zeros; applying Rolle $2n - 1$ more times produces $\xi_t$ with $g^{(2n)}(\xi_t) = 0$. Since $\deg Hf \leq 2n - 1$ and $p_n^2$ is monic of degree $2n$, $g^{(2n)} = f^{(2n)} - K\,(2n)!$, whence $K = f^{(2n)}(\xi_t)/(2n)!$ — and the identity is trivial at the nodes. Integration: $Q_n(f) = Q_n(Hf)$ ($Hf$ matches $f$ at the nodes) and $Q_n(Hf) = \int Hf\,w$ by exactness up to degree $2n - 1$ (question 9), so the quadrature error is $\int(f - Hf)\,w$. With $m, M$ the extrema of $f^{(2n)}$ on $I$, the pointwise identity squeezes $$ \frac{m\,h_n}{(2n)!} \;\leq\; \int_I(f - Hf)\,w \;\leq\; \frac{M\,h_n}{(2n)!} , $$ and the intermediate value theorem applied to the [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f^{(2n)}$ delivers $\xi$. (For Legendre with $n = 2$: $h_2 = \int_{-1}^1(t^2 - \frac13)^2\dd t = \frac8{45}$, so the error is $f^{(4)}(\xi)/135$.) **24.** The kernel reproduces $\R_{n-1}[t]$: expanding $q = \sum_k\frac{\langle p_k, q\rangle}{h_k}p_k$ gives $\int_I K_n(t_i, t)\,q(t)\,w(t)\dd t = q(t_i)$ for every $q$ of degree $\leq n - 1$. Take $q = \ell_i$: the left side equals $\ell_i(t_i) = 1$. But $t \mapsto K_n(t_i, t)\,\ell_i(t)$ is a polynomial of degree $\leq (n - 1) + (n - 1) = 2n - 2$, on which $Q_n$ is exact (question 9), and it vanishes at every node $t_j \neq t_i$ (factor $\ell_i$), so $$ 1 = \int_I K_n(t_i, t)\,\ell_i(t)\,w(t)\dd t = w_i\,K_n(t_i, t_i) = w_i\sum_{k=0}^{n-1}\frac{p_k(t_i)^2}{h_k} . $$ The sum is $> 0$ (its $k = 0$ term is $1/h_0 > 0$): the stated formula, and positivity again. Check ($n = 2$, Legendre): $p_0 = 1$, $h_0 = 2$, $p_1 = t$, $h_1 = \frac23$; at $t_i = \pm\frac1{\sqrt3}$, $$ K_2(t_i, t_i) = \frac12 + \frac{1/3}{2/3} = 1, \qquad w_i = 1, $$ as found in question 11. **25.** Substituting $t = \cos\theta$, the integral is $\int_0^\pi\cos^6\theta\,\dd\theta = \pi\,\frac{5\cdot3\cdot1}{6\cdot4\cdot2} = \frac{5\pi}{16}$ (Wallis, [Exercise 11.8](https://one-course.com/books/math/5/en/chapter/11-product-measures-fubini-change-of-variables#exo-b3-product-8)). The $n = 3$ Chebyshev–Gauss rule (question 17) has nodes $\cos\frac\pi6 = \frac{\sqrt3}2$, $\cos\frac\pi2 = 0$, $\cos\frac{5\pi}6 = -\frac{\sqrt3}2$ and equal weights $\frac\pi3$: $$ Q_3(t^6) = \frac\pi3\Bigl(2\cdot\Bigl(\frac{\sqrt3}2 \Bigr)^{6}\Bigr) = \frac\pi3\cdot\frac{54}{64} = \frac{9\pi}{32}, \qquad \frac{5\pi}{16} - \frac{9\pi}{32} = \frac\pi{32} . $$ Prediction: the monic degree-$3$ orthogonal polynomial is $2^{-2}T_3 = t^3 - \frac34t$, with $h_3 = \frac1{16}\norm{T_3}_w^2 = \frac1{16}\cdot\frac\pi2 = \frac\pi{32}$; and $f = t^6$ has constant $f^{(6)} = 720 = 6!$, so question 23 gives error $\frac{6!}{6!}\,h_3 = \frac\pi{32}$ — with no dependence on $\xi$ left, the formula is forced to be exact, and it is.