MATH 414 Lecture 11

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Point-Wise Convergence of Fourier Series

1. Riemann-Lebesgue
2. Partial Sumes
3. Find Error
4. Prove Convergence

Theorem.

Proof.

Lemma. [Riemann-Lebesgue]. Let $f$ be $L^{\prime }(\underset{a}{\overset{b}{\int }}|f(t)|\mathrm{d}<\mathrm{\infty })$ in $[a,b]$. Then

$\begin{array}{rl}\underset{\lambda \to \mathrm{\infty }}{\lim }\underset{a}{\overset{b}{\int }}f(t){\mathrm{e}}^{i\lambda t}\mathrm{d}t& =0\\ \underset{\lambda \to \mathrm{\infty }}{\lim }\underset{a}{\overset{b}{\int }}f(t)\left\{\begin{array}{c}\cos t\\ \sin t\end{array}\right\}& =0\end{array}$

Proof. (special case). Let $f\in C^{(1)}[a,b]$. Integrating $\underset{a}{\overset{b}{\int }}f(t){\mathrm{e}}^{i\lambda t}\mathrm{d}t$ by parts gives

$\underset{a}{\overset{b}{\int }}f(t){\mathrm{e}}^{i\lambda t}\mathrm{d}t=\frac{f(b){\mathrm{e}}^{i\lambda b-f(a){\mathrm{e}}^{i\lambda a}}}{i\lambda }-\frac{1}{i\lambda }\underset{a}{\overset{b}{\int }}{f}^{\prime }(t){\mathrm{e}}^{i\lambda t}\mathrm{d}t$

Taking the absolute value of the above yields

$|\underset{a}{\overset{b}{\int }}f(t){\mathrm{e}}^{i\lambda t}\mathrm{d}t|\le \left|\frac{f(b){\mathrm{e}}^{i\lambda b}-f(a){\mathrm{e}}^{i\lambda a}}{i\lambda }\right|+\frac{1}{\left|\lambda \right|}|\underset{a}{\overset{b}{\int }}{f}^{\prime }(t){\mathrm{e}}^{i\lambda t}\mathrm{d}t|$

1. $\left|\frac{f(b){\mathrm{e}}^{i\lambda b}-f(a){\mathrm{e}}^{i\lambda a}}{i\lambda }\right|\le \frac{|f(b){\mathrm{e}}^{i\lambda b}|+|f(a){\mathrm{e}}^{i\lambda a}|}{\left|\lambda \right|}\le \frac{|f(b)|+|f(a)|}{\left|\lambda \right|}$
2. We know $|\underset{a}{\overset{b}{\int }}g(t)\mathrm{d}t|\le \underset{a}{\overset{b}{\int }}|g(t)|\mathrm{d}t$ for both real and complex numbers, so $|\underset{a}{\overset{b}{\int }}{f}^{\prime }(t){\mathrm{e}}^{i\lambda t}\mathrm{d}t|\le \underset{a}{\overset{b}{\int }}|f^{\prime }(t)|\mathrm{d}t$

Therefore we are left with

$|\underset{a}{\overset{b}{\int }}f(t){\mathrm{e}}^{i\lambda t}\mathrm{d}t|\le \frac{|f(b)|+|f(a)|}{\left|\lambda \right|}+\frac{1}{\left|\lambda \right|}\underset{a}{\overset{b}{\int }}|f^{\prime }(t)|\mathrm{d}t$

Observe that the integral does not depend on $\lambda$, and in fact, as $\lambda \to \mathrm{\infty }$, $\underset{a}{\overset{b}{\int }}f(t){\mathrm{e}}^{i\lambda t}\mathrm{d}t\to 0$.

quod erat demonstrandum

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Partial sums. Let $f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}{a}_n\cos (nx)+b_n\sin (nx)$ be a $2\pi$-periodic, piecewise-smooth function. Then

$\begin{array}{rl}{S}_N(x)& =a_0+\sum _{n=1}^N{a}_n\cos (nx)+b_n\sin (nx)\\ a_n& =\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(t)\cos (nt)\mathrm{d}t\\ b_n& =\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(t)\sin (nt)\mathrm{d}t\end{array}$

Therefore

$\begin{array}{rl}{a}_n\cos (nx)+b_n\sin (nx)& =\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(t)(\cos (nt)\cos (nx)+\sin (nt)\sin (nx))\mathrm{d}t\\ & =\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(t)\cos (n(t-x))\mathrm{d}\end{array}$

Now our partial sum is of the form

$S_N(x)=a_0+\sum _{n=1}^N{a}_n\cos (nx)+b_n\sin (nx)=\underset{-\pi }{\overset{\pi }{\int }}f(x)P_N(t-x)\mathrm{d}t$

where $P_N(u)=\frac{1}{2\pi }+\frac{1}{\pi }\sum _{n=1}^N\cos (nu)$ is called the Fourier kernel or the Dirichlet kernel

Properties of the Fourier Kernel:

1. $P_N(u)$ is $2\pi$-periodic
2. $P_N(-u)=P_N(u)$ (it is even)
3. $\underset{-\pi }{\overset{\pi }{\int }}{P}_N(u)\mathrm{d}u=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}\mathrm{d}u=1$
4. $\underset{0}{\overset{\pi }{\int }}{P}_N(u)\mathrm{d}u=\frac{1}{2}$
5. $P_N(u)=\frac{1}{2\pi }\frac{\sin \left((N+\frac{1}{2})u\right)}{\sin \left(\frac{u}{2}\right)}$

Let's change index on the integral:

$S_N(x)=\underset{-\pi }{\overset{\pi }{\int }}f(t)P_N(t-x)\mathrm{d}t=\underset{-\pi -x}{\overset{\pi -x}{\int }}f(u+x)P_N(u)\mathrm{d}u$

Observe that the entire integrand is $2\pi$-periodic, so we can shift the index back to $[-\pi ,\pi ]$:

$S_N(x)=\underset{-\pi }{\overset{\pi }{\int }}f(u+x)P_N(u)\mathrm{d}u$

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Error Estimation. Let $E_N(x)=S_N(x)-f(x)$, where $x$ is a point of continuity. Then

$E_N(x)=S_N(x)-f(x)\underset{-\pi }{\overset{\pi }{\int }}{P}_N(u)\mathrm{d}u=\underset{-\pi }{\overset{\pi }{\int }}(f(x+u)-f(x))P_N(u)\mathrm{d}u$

Using property 5, we can rewrite $E_N(x)$ as

$E_N(x)=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}\frac{f(u+x)-f(x)}{\sin \left(\frac{u}{2}\right)}\sin \left((N+\frac{1}{2})u\right)\mathrm{d}u$

Assume $f$ is differentiable at $x$. By L'Hôspital's rule,

$\underset{u\to 0}{\lim }\frac{f(u+x)-f(x)}{\sin \left(\frac{u}{2}\right)}=2f^{\prime }(x)$

This means that the integrand of $E_N(x)$ is continuous at $u=0$. If we actually calculate the integral, we find that $E_N(x)=0$.

quod erat demonstrandum

Important Identity: $(x-1)\left(\sum _{k=0}^n{x}^k\right)=z^{n+1}-1$