5
$\begingroup$

Suppose $f\in\mathbf{C^2}(\mathbb{R})$ and is periodic with period $2\pi$. Prove that the Fourier series of $f$ converges uniformly in any finite interval.

My attempt:

$|a_n~\cos(nx)+b_n~\sin(nx)|\le|a_n|+|b_n|$. So, by M-test, we just need to show $\displaystyle\sum_{n=1}^\infty |a_n|+|b_n|$ converges.

$$\begin{equation}\begin{aligned} a_n=\frac{1}{\pi} \int_{-\pi}^\pi~f(x)~\cos(nx)~dx &= -\frac{1}{n\pi}\int_{-\pi}^{\pi}~ f'(x)~\sin(nx)~dx \\ &=\frac{1}{n^2\pi}~ f'(x)~\cos(nx)~dx\Bigg|_{-\pi}^{\pi}-\frac{1}{n^2\pi}~f''(x)~\cos(nx)~dx\\ \end{aligned}\end{equation}$$ $$=\frac{1}{n^2\pi}[f(\pi)~\cos(nx)-f(-\pi)~\sin(nx)]+\frac{1}{n^2\pi}\int_{-\pi}^{\pi} f''(x)~\cos(nx)dx$$

$f\in\mathbf{C^2}(\mathbb{R})\Rightarrow |f|,|f''|\leq K$ on $[-\pi,\pi]$ for some $K$.

We can get $a_n\leq \frac{4K}{n^2}$. Similarly, $b_n\leq\frac{4M}{n^2}$. So, $M_n=\displaystyle\sum_{n=1}^\infty |a_n|+|b_n|$ converges.

So, the Fourier series $\displaystyle\sum_{n=1}^\infty [a_n~\cos(nx)+b_n~\sin(nx)]$ converges unifomly.

Is my proof correct? It seems that for any $x\in \mathbb{R}$, the Fourier series converges uniformly. Why we cannot conclude that the Fourier series of $f$ converges uniformly on $\mathbb{R}?$

$\endgroup$
  • 1
    $\begingroup$ @Matt The function is $2\pi$ periodic. OP is right, there is uniform convergence in the entire real line. $\endgroup$ – Aloizio Macedo Apr 20 '17 at 14:03
  • $\begingroup$ So, we can say the Fourier series converges uniformly on $\mathbb{R}$ ,right? Are there any mistakes in my proof? $\endgroup$ – User90 Apr 20 '17 at 14:25
2
$\begingroup$

First, prove that if $f \in C^{(k)}(\mathbb{R})$ then, you can prove that $\hat{f}(n)$ obeys the following: $$ \hat{f}(n) = o\left(\frac{1}{|n|^k}\right). $$

To see why this is true, note that $\hat{f^{(k)}}(n) = (in)^k \hat{f}(n)$. By Riemann-Lebesgue lemma, as $|n|\to\infty$, $\hat{f}(n)\to 0$, we conclude.

Now, given this, we will arrive at result. Let $N^{th}$ partial sum be given by: $$ S_N(f)(x) = \sum_{k=-N}^N\hat{f}(k)e^{ikx}. $$

Now, note that for any $N \neq M$, $|S_N(f)(x) - S_M(f)(x)|$ is computed via $$ |S_N(f)(x) - S_M(f)(x)| = \left|\sum_{M+1\leq |n|\leq N}\hat{f}(n)e^{inx}\right|\leq \sum_{M+1\leq |n|\leq N}|\hat{f}(n)|\underbrace{|e^{inx}|}_{=1} = \sum_{M+1\leq |n|\leq N}|\hat{f}(n)| $$

and $\sum_{M+1\leq |n|\leq N}|\hat{f}(n)|\to 0$, as $N,M\to \infty$ (this follows from the fact that $\hat{f}(n)$ is a summable sequence, due to characterization above). Hence, $S_N(f)(x)$ converges uniformly.

$\endgroup$
1
$\begingroup$

Your proof is (almost) correct - there are some computation errors (your second line does not have the integral, and your third line mysteriously exchanged the derivative by the function and popped up a sine, while also not evaluating $\pi$ and $-\pi$ in them).

And indeed, you have uniform convergence in the entire real line. Just one detail and an observation: this happens because the boundary terms on the integration by parts vanish, and this is due to the fact that the function is $2\pi$-periodic.

The observation is that it is important to note that this proves the Fourier series converges uniformly, but it doesn't prove that it converges uniformly to $f$. It indeed converges to $f$, but you need the density of the trigonometric functions on $L^2(S^1)$ or something of this nature (for example, density on $C^0(S^1)$ would be enough, since your function is $C^2$) to prove this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.