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 $L'\left(\int _{a}^{b}\left|f(t)\right|\,\mathrm {d} <\infty \right)$ in $\left[a,b\right]$. Then

${\begin{aligned}\lim _{\lambda \to \infty }\int _{a}^{b}f(t)\,\mathrm {e} ^{i\,\lambda \,t}\,\mathrm {d} t&=0\\\lim _{\lambda \to \infty }\int _{a}^{b}f(t)\,\left\{{\begin{matrix}\cos {t}\\\sin {t}\end{matrix}}\right\}&=0\end{aligned}}$

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

$\int _{a}^{b}f(t)\,\mathrm {e} ^{i\,\lambda \,t}\,\mathrm {d} t={\frac {f(b)\,\mathrm {e} ^{i\,\lambda \,b-f(a)\,\mathrm {e} ^{i\,\lambda \,a}}}{i\,\lambda }}-{\frac {1}{i\,\lambda }}\,\int _{a}^{b}f'(t)\,\mathrm {e} ^{i\,\lambda \,t}\,\mathrm {d} t$

Taking the absolute value of the above yields

$\left|\int _{a}^{b}f(t)\,\mathrm {e} ^{i\,\lambda \,t}\,\mathrm {d} t\right|\leq \left|{\frac {f(b)\,\mathrm {e} ^{i\,\lambda \,b}-f(a)\,\mathrm {e} ^{i\,\lambda \,a}}{i\,\lambda }}\right|+{\frac {1}{\left|\lambda \right|}}\,\left|\int _{a}^{b}f'(t)\,\mathrm {e} ^{i\,\lambda \,t}\,\mathrm {d} t\right|$

1. $\left|{\frac {f(b)\,\mathrm {e} ^{i\,\lambda \,b}-f(a)\,\mathrm {e} ^{i\,\lambda \,a}}{i\,\lambda }}\right|\leq {\frac {\left|f(b)\,\mathrm {e} ^{i\,\lambda \,b}\right|+\left|f(a)\,\mathrm {e} ^{i\,\lambda \,a}\right|}{\left|\lambda \right|}}\leq {\frac {\left|f(b)\right|+\left|f(a)\right|}{\left|\lambda \right|}}$
2. We know $\left|\int _{a}^{b}g(t)\,\mathrm {d} t\right|\leq \int _{a}^{b}\left|g(t)\right|\,\mathrm {d} t$ for both real and complex numbers, so $\left|\int _{a}^{b}f'(t)\,\mathrm {e} ^{i\,\lambda \,t}\,\mathrm {d} t\right|\leq \int _{a}^{b}\left|f'(t)\right|\,\mathrm {d} t$

Therefore we are left with

$\left|\int _{a}^{b}f(t)\,\mathrm {e} ^{i\,\lambda \,t}\,\mathrm {d} t\right|\leq {\frac {\left|f(b)\right|+\left|f(a)\right|}{\left|\lambda \right|}}+{\frac {1}{\left|\lambda \right|}}\,\int _{a}^{b}\left|f'(t)\right|\,\mathrm {d} t$

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

quod erat demonstrandum

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Partial sums. Let $f(x)=a_{0}+\sum _{n=1}^{\infty }a_{n}\,\cos {(n\,x)}+b_{n}\,\sin {(n\,x)}$ be a $2\pi$-periodic, piecewise-smooth function. Then

${\begin{aligned}S_{N}(x)&=a_{0}+\sum _{n=1}^{N}a_{n}\,\cos {(n\,x)}+b_{n}\,\sin {(n\,x)}\\a_{n}&={\frac {1}{\pi }}\,\int _{-\pi }^{\pi }f(t)\,\cos {(n\,t)}\,\mathrm {d} t\\b_{n}&={\frac {1}{\pi }}\,\int _{-\pi }^{\pi }f(t)\,\sin {(n\,t)}\,\mathrm {d} t\end{aligned}}$

Therefore

${\begin{aligned}a_{n}\,\cos {(n\,x)}+b_{n}\,\sin {(n\,x)}&={\frac {1}{\pi }}\,\int _{-\pi }^{\pi }f(t)\,\left(\cos {(n\,t)}\,\cos {(n\,x)}+\sin {(n\,t)}\,\sin {(n\,x)}\right)\,\mathrm {d} t\\&={\frac {1}{\pi }}\,\int _{-\pi }^{\pi }f(t)\,\cos {(n\,(t-x))}\,\mathrm {d} \end{aligned}}$

Now our partial sum is of the form

$S_{N}(x)=a_{0}+\sum _{n=1}^{N}a_{n}\,\cos {(n\,x)}+b_{n}\,\sin {(n\,x)}=\int _{-\pi }^{\pi }f(x)\,P_{N}(t-x)\,\mathrm {d} t$

where $P_{N}(u)={\frac {1}{2\pi }}+{\frac {1}{\pi }}\,\sum _{n=1}^{N}\cos {(n\,u)}$ 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. $\int _{-\pi }^{\pi }P_{N}(u)\,\mathrm {d} u={\frac {1}{2\pi }}\,\int _{-\pi }^{\pi }\,\mathrm {d} u=1$
4. $\int _{0}^{\pi }P_{N}(u)\,\mathrm {d} u={\frac {1}{2}}$
5. $P_{N}(u)={\frac {1}{2\pi }}\,{\frac {\sin {\left(\left(N+{\frac {1}{2}}\right)\,u\right)}}{\sin {\left({\frac {u}{2}}\right)}}}$

Let's change index on the integral:

$S_{N}(x)=\int _{-\pi }^{\pi }f(t)\,P_{N}(t-x)\,\mathrm {d} t=\int _{-\pi -x}^{\pi -x}f(u+x)\,P_{N}(u)\,\mathrm {d} u$

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

$S_{N}(x)=\int _{-\pi }^{\pi }f(u+x)\,P_{N}(u)\,\mathrm {d} u$

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Error Estimation. Let $E_{N}(x)=S_{N}(x)-f(x)$, where $x$ is a point of continuity. Then

$E_{N}(x)=S_{N}(x)-f(x)\,\int _{-\pi }^{\pi }P_{N}(u)\,\mathrm {d} u=\int _{-\pi }^{\pi }\left(f(x+u)-f(x)\right)\,P_{N}(u)\,\mathrm {d} u$

Using property 5, we can rewrite $E_{N}(x)$ as

$E_{N}(x)={\frac {1}{2\pi }}\,\int _{-\pi }^{\pi }{\frac {f(u+x)-f(x)}{\sin {\left({\frac {u}{2}}\right)}}}\,\sin {\left(\left(N+{\frac {1}{2}}\right)\,u\right)}\,\mathrm {d} u$

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

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

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