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Big group 3rd of February: 3.3:9; 3.3:10b,c; 3.4:3; räkna själv: 3.3:1; 3.3:2

Small group 4th and 7th of February: 3.5:4; Eö 23; 4.2:1

\subsection*{3.3:9}
Suppose that $\{\phi_n\}_1^\infty$ is an orthonormal basis for $L^2(a,b)$.  Show that for any functions $f,g\in L^2(a,b)$, 
\[
  \langle f,g\rangle = \sum_{n=1}^\infty \langle f,\phi_n\rangle \overline{\langle g,\phi_n\rangle}
\]

\textit{Solution:} We know from this chapter that functions $f,g\in L^2(a,b)$ can be written as linear combinations of basis functions, i.e., as (generalized) Fourier series:
\[
  f=\sum_{n=1}^\infty \langle f,\phi_n\rangle \phi_n,\qquad g=\sum_{m=1}^\infty \langle g,\phi_m\rangle \phi_m
\]
Use the properties of the inner product ($\S 3.1$ in Folland):
\begin{align*}
  \langle f,g\rangle & =  \langle f, \sum \langle g,\phi_m\rangle \phi_m\rangle & (\mathrm{sum\ representation\ of}\ g) \\
  & = \overline{\langle \sum \langle g,\phi_m\rangle \phi_m, f\rangle} & \langle f_1,f_2\rangle = \overline{\langle f_2,f_1\rangle}\\
  & = \sum \overline{\langle g,\phi_m\rangle}\overline{\langle \phi_m, f\rangle} & \\ 
  & =  \sum \overline{\langle g,\phi_m\rangle}\langle f,\phi_m\rangle & (\mathrm{sesquilinearity}) \\
  %& =  \sum \overline{\langle g,\phi_m\rangle}\langle \sum \langle f,\phi_n\rangle \phi_n,   \phi_m\rangle & (\mathrm{sum\ representation\ of}\ f) \\
  %& =  \sum_n\sum_m \overline{\langle g,\phi_m\rangle} \langle f,\phi_n\rangle\langle   \phi_n,   \phi_m\rangle & (\mathrm{sesquilinearity})
\end{align*}
%Now use that the $\phi_n$'s are orthonormal:
%%
%\begin{equation*}
%  \langle \phi_n,\phi_m\rangle=
%  \begin{cases}
%    1, & m=n \\
%    0, & m\neq n
%  \end{cases}
%\end{equation*}
%%
%The result is \(\langle f,g\rangle = \sum \overline{\langle g,\phi_n\rangle} \langle f,\phi_n\rangle\).

\subsection*{3.3:10a,b,c}
Evaluate the following series by applying Parseval’s equation to certain of the Fourier expansions in Table 1 of $\S$2.1.
\[
   \mathrm{a.}\quad \sum_1^\infty \frac{1}{n^4}\qquad \qquad \mathrm{b.}\quad \sum_1^\infty \frac{1}{(2n-1)^6} \qquad
   \mathrm{c.}\quad \sum_1^\infty \frac{n^2}{(n^2+1)^2}
\]

\textit{Solution:} From the previous exercise, we deduce the following:
\begin{teo}[Parts of Theorem 3.4]
	Let $\{\phi_n\}_1^\infty$ be an orthonormal set in $L^2(a,b)$.
	For every $f\in L^2(a,b)$,
	\[ \|f\|^2=\sum_1^\infty |\langle f,\phi_n\rangle|^2\quad \iff\quad f=\sum_1^\infty \langle f,\phi_n\rangle \phi_n \]
\end{teo}

Let $(a,b)=(-\pi,\pi)$. We find in Table 1 of $\S$2.1 that
\[
  f(t)=t^2 = \frac{\pi^2}{3}+\sum_1^\infty \frac{4(-1)^n}{n^2}\cos(nt)
\]
Recall that a basis for $L^2(-\pi,\pi)$ is
\[
\{\cos nx\}_{n=0}^\infty\cup\{\sin nx\}_{n=1}^\infty
\]
Writing 
\[
f(x)=\frac{1}{2}a_0+\sum_1^\infty (a_n\cos nx + b_n \sin nx)
\]
Parseval's equality takes the form
\[
\|f\|^2=\frac{1}{2}|a_0|^2+\sum_1^\infty \left(a_n^2+ b_n^2\right)
\]
We identify
\[
a_0=\frac{2\pi^2}{3},\quad a_n=\frac{4(-1)^n}{n^2},\quad b_n=0
\]
so
\[
\|f\|^2 = \frac{1}{2}\left(\frac{2\pi^2}{3}\right)^2 +\sum_1^\infty \left(\frac{4(-1)^n}{n^2}\right)^2= 2\frac{\pi^4}{9}+16\sum_1^\infty \frac{1}{n^4}
\]
Since
\[
\|f\|^2 = \frac{1}{\pi}\int_{-\pi}^\pi |f(x)|^2dx = \frac{1}{\pi}\int_{-\pi}^\pi x^4dx = \frac{2}{\pi}\frac{\pi^5}{5}=2\frac{\pi^4}{5}
\]
Equating
\[
2\frac{\pi^4}{5}= 2\frac{\pi^4}{9}+16\sum_1^\infty \frac{1}{n^4}
\]
we get
\[
4\frac{\pi^4}{45}= 8\sum_1^\infty \frac{1}{n^4} \iff \frac{\pi^4}{90}= \sum_1^\infty \frac{1}{n^4}
\]


When $f(\theta)=\theta(\pi-|\theta|)$
\[
  f(\theta)=\sum_1^\infty \frac{8}{\pi(2n-1)^3}\sin(2n-1)\theta 
\]
so $a_n=0$ and for even $n$ we have $b_n=0$. For odd $n$, $$b_n=\frac{8}{\pi n^3}\quad \mathrm{i.e.}\quad b_{2n-1}=\frac{8}{\pi (2n-1)^3}.$$

\[
  \|f\|^2=\sum_{n=1}^\infty \left|\frac{8}{\pi (2n-1)^3}\right|^2=\frac{64}{\pi^2}\sum_{n=1}^\infty \frac{1}{(2n-1)^6}
\]
Now
\[
\|f\|^2= \frac{1}{\pi}\int_{-\pi}^\pi |f(x)|^2dx =\frac{1}{\pi}\int_{-\pi}^\pi x^2(\pi-|x|)^2dx = \frac{2}{\pi}\int_0^\pi x^2(\pi-x)^2dx=\frac{\pi^4}{15}
\]
so
\[
\frac{\pi^4}{15}=\frac{64}{\pi^2}\sum_{n=1}^\infty \frac{1}{(2n-1)^6}\qquad \mathrm{or}\qquad \frac{\pi^6}{960}=\sum_{n=1}^\infty \frac{1}{(2n-1)^6}
\]

For c., we are looking for something with $(n^2+1)$ in the denominator. On $L^2(-\pi,\pi)$,
\[
  f(t)=\sinh t = \sum_1^\infty \frac{2\sinh \pi}{\pi}\frac{(-1)^{n+1}n}{n^2+1}\sin nt
\]
so for $f(t)=\sinh t$,
\[
  \|f\|^2=\sum_1^\infty \left|\frac{2\sinh \pi}{\pi}\frac{(-1)^{n+1}n}{n^2+1}\right|^2
  =\left(\frac{2\sinh \pi}{\pi}\right)^2 \sum_1^\infty \frac{n^2}{(n^2+1)^2}
\]
Now
\[
\|f\|^2=\frac{1}{\pi}\int_{-\pi}^\pi \sinh^2(t)dt=\frac{1}{\pi}\int_{-\pi}^\pi \left( \frac{1}{2} \cosh(2t)-\frac{1}{2}\right)dt=\frac{\sinh(\pi)\cosh(\pi)-\pi}{\pi} 
\]
Equating:
\[
\frac{\sinh(\pi)\cosh(\pi)-\pi}{\pi}=\left(\frac{2\sinh \pi}{\pi}\right)^2 \sum_1^\infty \frac{n^2}{(n^2+1)^2}
\]
\[
\pi\frac{\sinh\pi\cosh\pi-\pi}{4\sinh^2\pi}= \sum_1^\infty \frac{n^2}{(n^2+1)^2}
\]

\subsection*{3.4.3}
Let $D$ be the unit disk $\{x,y\in\R :x^2+y^2\leq 1\}$ and let $f_n(x,y)=(x+iy)^n$. Show that $\{f_n\}_0^\infty$ is an orthogonal set in $L^2(D)$ and compute $\|f_n\|$ for all $n$.

\textit{Solution:} Write $x+iy=e^{i\theta}r$ with $r=\sqrt{x^2+y^2}$. We know from this chapter that an inner product on $D$ is
\[
  \langle f,g\rangle = \int_D f(x,y)\overline{g(x,y)} dxdy = \int_0^1\int_0^{2\pi} f(r,\theta)\overline{g(r,\theta)} r d\theta dr
\]

\[
  f_n(x,y)=(x+iy)^n=r^ne^{in\theta},\qquad \overline{f_m(x,y)}=(x-iy)^m=r^ne^{-im\theta}
\]

\[
  \langle f_n,f_m \rangle = \int_0^1\int_0^{2\pi} r^{n+m}e^{i(n-m)\theta} r d\theta dr = \int_0^1 r^{n+m+1} \left( \int_0^{2\pi} e^{i(n-m)\theta}d\theta \right)dr
\]
If $n=m$
\[
\int_0^{2\pi} e^{i(n-m)\theta}d\theta = \int_0^{2\pi}d\theta=2\pi
\]
and if $n\neq m$
\[
\int_0^{2\pi} e^{i(n-m)\theta}d\theta =\frac{1}{i(n-m)}\left(e^{i(n-m)2\pi}-1\right)=\frac{i}{n-m}\left(1-e^{i(n-m)2\pi}\right)=0
\]
using periodicity. Thus we arrive at
 \begin{align*}
   \langle f_n,f_m \rangle \neq 0, & \qquad \mathrm{if}\ n\neq m\\
   \langle f_n,f_m \rangle =0, & \qquad \mathrm{if}\ n=m
 \end{align*}
%
What is $\|f_n\|^2$?

\subsection*{3.1:1}
Cauchy-Schwarz' inequality and norm convergence gives
\[
|\langle f_n-f,g\rangle|\leq \|f_n-f\|\|g\| \to 0\quad \mathrm{as}\ n\to \infty
\]
The left hand side is
\[
|\langle f_n-f,g\rangle| = |\langle f_n,g\rangle - \langle f ,g\rangle|
\]
So $\|f_n-f\|\to 0$ implies that $|\langle f_n,g\rangle - \langle f ,g\rangle|\to 0$, and we are done.

\subsection*{3.1:2}
Notice that both $|\|f\|-\|g\||$ and $\|f-g\|$ are non-negative.
\[
\|f-g\|^2 = \langle f-g,f-g\rangle = \langle f,f\rangle - \langle f,g\rangle -\langle g,f\rangle +\langle g,g\rangle
\]
\[
\langle f,g\rangle +\langle g,f\rangle = 2\mathrm{Re}\{\langle f,g\rangle\}
\]
\[
\|f-g\|^2 = \|f\|^2-2\mathrm{Re}\{\langle f,g\rangle\}+\|g\|^2
\]
\[
\left|\|f\|-\|g\|\right|^2=\|f\|^2-2\|f\|\|g\| + \|g\|^2
\]
Cauchy-Schwarz' inequality and complex algebra gives
\[
 \mathrm{Re}\{\langle f,g\rangle\}\leq|\langle f,g\rangle|\leq \|f\|\|g\|
\]
Collecting results:
\[
 \left|\|f\|-\|g\|\right|^2 \leq \|f-g\|^2
\]
If $\|f_n-f\|\to 0$ then $|\|f_n\|-\|f\|| \to 0$.



\newpage
\section*{Open questions}
Going back to sum computations, why is the following not working? Write
\[
t^2 -\frac{\pi^2}{3} = \sum_1^\infty \frac{4(-1)^n}{n^2}\cos(nt)
\]
so by Parseval's equality
\[
\|t^2 -\frac{\pi^2}{3}\|^2 = \sum_1^\infty \left|\frac{4(-1)^n}{n^2}\right|^2
\]
Now
\[
\|t^2 -\frac{\pi^2}{3}\|^2 = \int_{-\pi}^\pi (t^2-\frac{\pi}{3})^2dx=\int_{-\pi}^\pi(t^4-2t^2\frac{\pi}{3}+\frac{\pi^2}{9})dx=\frac{2}{45}\pi^3(5-10\pi+9\pi^2)
\]
and
\[
\sum_1^\infty \left|\frac{4(-1)^n}{n^2}\right|^2=16\sum_1^\infty \frac{1}{n^4}
\]
so Parseval's equality says that
\[
 \sum_1^\infty \frac{1}{n^4} = \frac{1}{16}\frac{2}{45}\pi^3(5-10\pi+9\pi^2)= \frac{1}{360}\pi^3(5-10\pi+9\pi^2)
\]




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