ANA2 Quiz 5

Problem 5

Part a

Let $S=\{\sqrt{2}\cos (nt/2):n\in \mathbb{Z}\}$. $S$ is orthonormal:

$$\begin{array}{rl}⟨\sqrt{2}\cos (nt/2),\sqrt{2}\cos (mt/2)⟩& =\frac{1}{2\pi }{\int }_0^{2\pi }2\cos (nt/2)\cos (mt/2)dt\\ & =\frac{1}{2\pi }{\int }_0^{2\pi }\cos \left(\frac{(n-m)t}{2}\right)+\cos \left(\frac{(n+m)t}{2}\right)dt\\ & =\{\begin{array}{ll}1& n=m\\ 0& n\ne m.\end{array}\end{array}$$

Observe that $\mathcal{A}:=\text{span}S$ is a unital subalgebra of $(C([0,2\pi ],\mathbb{R}),\Vert \cdot {\Vert }_{\infty })$ which separates points. By Thm 315.6, $\mathcal{A}$ is dense in $(C([0,2\pi ],\mathbb{R}),\Vert \cdot {\Vert }_{\infty })$. Thus, for every $f\in C([0,2\pi ],\mathbb{R})$, there exists $\{f_n\}\subseteq \mathcal{A}$ such that $\{f_n\}\rightrightarrows f$. It follows that $|f-f_n{|}^2\rightrightarrows 0$. By Thm 155.7,

$$\begin{array}{r}\underset{n\to \infty }{\lim }{\int }_0^{2\pi }|f-f_n{|}^{2}dx=0,\end{array}$$

so $\{f_n\}\to f$ under $\Vert \cdot {\Vert }_2$. It follows that $\mathcal{A}$ is dense in $(C([0,2\pi ],\mathbb{R}),\Vert \cdot {\Vert }_2)$. The embedding of $(C([0,2\pi ],\mathbb{R}),\Vert \cdot {\Vert }_2)$ in $X$ is dense in $X$, since Riemann integrable functions can be approximated arbitrarily well by continuous functions when using $\Vert \cdot {\Vert }_2$. It follows that (the image under the quotient map of) $C([0,2\pi ],\mathbb{R})$ remains dense in $L$, and hence in $L^2$. Thus, $\mathcal{A}$ is dense in $L^2([0,2\pi ],\mathbb{R})$.

Part b

Note that

$$\frac{1}{2\pi }\sum _{m=1}^k\left(\frac{8m((-1{)}^n-1)}{n^2-4m^2}\right)\sqrt{2}\sin \left(mx\right)\to \sqrt{2}\cos (nt/2)\text{as}k\to \infty$$

for all $n$ ¹. It follows that $T$ is dense in $L^2([0,2\pi ],\mathbb{R})$.

Suppose, for contradiction, there exists $p\in T$ with

$$\phi (f)=⟨p,f⟩=\frac{1}{2\pi }{\int }_0^{2\pi }p(x)f(x)dx$$

for every $f\in T$. Then for every $f\in T$

$$\frac{1}{2\pi }{\int }_0^{2\pi }(p(x)-x)f(x)dx=0,$$

so $p-x$ is orthogonal to every element of $T$. Since $T$ is dense in $L^2([0,2\pi ])$ and the inner product is continuous, $p-x$ is orthogonal to all of $L^2$, which forces $p-x=0$ in $L^2$. But $x$ is not a trigonometric polynomial, so this is impossible.

This does not violate Riesz, since $T$ is not closed and hence not a complete space.

Yes, $\phi$ is representable as a functional on $L^2([0,2\pi ])$. $x\in L^2([0,2\pi ])$, and for every $f\in L^2([0,2\pi ])$, the map

$$f\mapsto \frac{1}{2\pi }{\int }_0^{2\pi }xf=⟨x,f⟩$$

is a bounded linear functional, since $|⟨x,f⟩|⩽\Vert x{\Vert }_2\Vert f{\Vert }_2$ and is represented by $[x]\in L^2([0,2\pi ])$ via Riesz.

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Problem 2

Exercise 362.1(Exercise).

Find the Fourier series of the function $f:[-\pi ,\pi ]\to \mathbb{R}$ given by $f(x)=x^2$. Deduce that

$$\begin{array}{r}\sum _{n=1}^{\infty }\frac{1}{n^4}=\frac{{\pi }^4}{90}.\end{array}$$

$$\begin{array}{r}\hat{f}(n)=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }{x}^2{e}^{-inx}dx=\{\begin{array}{ll}\frac{2(-1{)}^n}{n^2}& n\ne 0\\ & \\ \frac{{\pi }^2}{3}& n=0\end{array}\end{array}$$

We now have

$$\frac{{\pi }^4}{5}=\Vert f{\Vert }_2^2=\frac{{\pi }^4}{9}+\sum _{n=-\infty }^{-1}|\hat{f}(n){|}^2+\sum _{n=1}^{\infty }|\hat{f}(n){|}^2=\frac{{\pi }^4}{9}+\sum _{n=1}^{\infty }\frac{8}{n^4}.$$

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Problem 3

Exercise 362.2(Exercise).

Suppose that $\{f_k{\}}_{k=1}^{\infty }$ is a sequence of Riemann integrable functions on the interval $[0,1]$ such that

$${\int }_0^1|f_k(x)-f(x)|dx\to 0.$$

Show that ${\hat{f}}_k(n)\to \hat{f}(n)$ uniformly in $n$ as $k\to \infty$.

$$\begin{array}{rl}|{\hat{f}}_k(n)-\hat{f}(n)|& =\left|{\int }_0^1(f_k(x)-f(x))e^{-inx}dx\right|\\ & ⩽{\int }_0^1|f_k(x)-f(x)|dx\\ & \\ ⟹\underset{n\in \mathbb{N}}{\sup }|{\hat{f}}_k(n)-\hat{f}(n)|& ⩽{\int }_0^1|f_k(x)-f(x)|dx\to 0\text{ as }k\to 0\end{array}$$

Thus, ${\hat{f}}_k(n)\to \hat{f}(n)$ uniformly.

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Problem 4

Footnotes

1. At this point, I realized that $S^{\prime }=\{1\}\cup \{\sqrt{2}\cos nt,\sqrt{2}\sin nt:n\in \mathbb{N}\}$ is also an ONB, and one from which the density of $T$ in $L^2([0,2\pi ])$ is evident. Instead of redoing my work from part a, I instead show that elements of $S$ can be approximated by trigonometric polynomials using the Fourier series of $\sqrt{2}\cos (nt/2)$ in $S^{\prime }$. ↩