MATH 414 Lecture 13

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Convergence in the Mean

(also called $L^2$ convergence)

Orthogonal projections

Set-up:

1. Inner Product Space $V$, $⟨\cdot ,\cdot ⟩$
2. Subspace with orthonormal basis $V_N=\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}\{{\hat{u}}_1,{\hat{u}}_2,\dots ,{\hat{u}}_N\}$

The orthogonal projection $\overrightarrow{p}$ of $\overrightarrow{v}\in V$ onto $V_N$ is the unique vector in $V_N$ such that $\Vert \overrightarrow{v}-\overrightarrow{p}\Vert =\underset{\overrightarrow{w}\in V_N}{\min }\Vert \overrightarrow{v}-\overrightarrow{w}\Vert$.

Properties:

1. $\overrightarrow{p}$ satisfies $⟨\overrightarrow{v}-\overrightarrow{p},\overrightarrow{w}⟩=0$ for all $\overrightarrow{w}\in V_N$.
2. $\overrightarrow{p}=\sum _{j=1}^N⟨\overrightarrow{v},{\hat{u}}_j⟩{\hat{u}}_j$
3. ${\Vert \overrightarrow{v}-\overrightarrow{p}\Vert }^2={\Vert \overrightarrow{v}\Vert }^2-{\Vert \overrightarrow{p}\Vert }^2$

Application to Fourier Series

With Fourier Series, we project a function $f$ onto the vector space $V_N=\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}\{\frac{1}{\sqrt{2\pi }},\frac{1}{\sqrt{\pi }}\cos t,\frac{1}{\sqrt{\pi }}\sin t,\dots ,\frac{1}{\sqrt{\pi }}\cos (Nt),\frac{1}{\sqrt{\pi }}\sin (Nt)\}$ (this basis is indeed orthonormal.

The $L^2$ error of this projection is ${\Vert f-S_n\Vert }^2={\Vert f\Vert }^2-2\pi a_0^2-\pi (\sum _{k=1}^N{a}_k^2+b_k^2)$

Theorem. $S_n$ converges to $f$ in the mean if and only if

$\underset{-\pi }{\overset{\pi }{\int }}{|f(t)|}^2\mathrm{d}t=2\pi (\frac{a_0^2}{2}+\sum _{n=1}^{\mathrm{\infty }}(a_n^2+b_n^2))$

Proof. $\underset{-\pi }{\overset{\pi }{\int }}{|f-S_N|}^2\mathrm{d}t=E_N^2=\underset{-\pi }{\overset{\pi }{\int }}{\left|f\right|}^2\mathrm{d}t-2\pi \{\frac{a_0^2}{2}+\sum _{k=1}^N(a_k^2+b_k^2)\}$. If $E_N^2$ converges to 0, then $\sum _{k=1}^N(a_k^2+b_k^2)$ must also converge to 0.

quod erat demonstrandum

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Analysis and Synthesis

Given $f(t)$, it's easy to come up with a Fourier series (Analysis of signal)

However, Given $a_0$, $a_n$, and $b_n$, is there a $f(t)$ such that $f(t)=a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos (nt)+b_n\sin (nt)$? (Synthesize to get a signal)

Answer. Yes! As long as $\frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}{a}_n^2+b_n^2$ is finite, but this requires $L^2$ integrals (Riesz-Fisher Theorem) because Riemann integrals don't work.