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Let $f(t)=(t-\pi)\chi_{(0,2\pi)}$, $t \in [0,2\pi]$, then the partial sum of the Fourier series of $f$ is $$ S_n(t)=- \sum_{0 < |k| \le n} \frac{\sin k t}{k}. $$ Show $|S_n(t)| \le \pi+2$ for all $n$ and $t$.

I know the Fourier series of $f$ converges to $f(t)$ on $(0,2\pi)$, to $-\pi/2$ at $0$, and to $\pi/2$ at $2\pi$.
So there is an $n_0$ past which I can uniformly bound $S_n(t)$ by $\pi+2$.
But I don't know how to handle the term before $n_0$.

I considered writing $$ S_n(t)=\int_0^t \left(\frac{\sin (k+1/2) x}{\sin x/2} - 1 \right)dx $$

But I don't see how to make $\pi+2$ appear.

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I suspect a typo. Maybe $\frac{i \sin k t}{k}$ ? –  john mangual Oct 12 '12 at 19:31
    
@johnmangual You are right. Thank you. –  Nicolas Essis-Breton Oct 12 '12 at 19:36
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Have you tried using Parseval's Theorem or Bessel's Inequality? –  Pragabhava Oct 12 '12 at 19:52
    
@Pragabhava I don't see the two functions to which I should apply Parseval theorem. –  Nicolas Essis-Breton Oct 12 '12 at 19:58

1 Answer 1

up vote 6 down vote accepted

I cannot see how to derive a limitation for $\|S_n(x)\|_{\infty}$ from an approximation of the $\mathcal{L}^2$-norm of $S_n(x)$, since it is clear that exist unbounded functions belonging to $\mathcal{L}^2$!

I'd rather say that it is easy to locate the stationary points of $S_n(x)$, since:

$$\frac{d}{dx}\,S_n(x) = \frac{2\cos\left(\frac{n+1}{2}x\right)\sin\left(\frac{n}{2}x\right)}{\sin\left(\frac{x}{2}\right)}=-1+\frac{\sin\left(\frac{2n+1}{2}x\right)}{\sin\left(\frac{x}{2}\right)}$$

Now, it should be not too difficult to prove that the maximum of $S_n(X)$ is reached in the first positive stationary point, i.e. $x=\frac{\pi}{n+1}$.

This gives: $$(\heartsuit)\quad S_n(x)\leq 2\sum_{j=1}^{n}\frac{1}{j}\sin\left(\frac{j\,\pi}{n+1}\right).$$

Now, using $\sin x\leq\frac{4}{\pi^2}x(\pi-x)$ (that holds for any $x\in[0,\pi]$), we have a much stronger inequality:

$$ S_n(x) \leq \frac{4n}{n+1} < 4 < \pi+2. $$

In fact, I think that, regarding the right side of $(\heartsuit)$ as the approximation of an integral, it is also possible to prove that:

$$ |S_n(x)|<2\int_{0}^{\pi}\frac{\sin x}{x}\,dx < \pi + \frac{9}{16}. $$

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The hardest part is to prove that the first stationary point is the point of maximum, but this can be done by using the inequality $|f(x)-S_n(x)|\leq\frac{1}{n\sin(x/2)}$ - to prove the last point, I used partial summation on the Fourier series of $f-S_n$. –  Jack D'Aurizio Oct 13 '12 at 14:55
    
How do you get the bound on $\sin x$ and the bound on the last integral? –  Nicolas Essis-Breton Oct 14 '12 at 15:50
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The bound on the last integral is given by an explicit computation of $2\operatorname{SinIntegral}(\pi)$, the bound for $\sin x$ is due to the fact that $\frac{\sin x}{x(\pi-x)}$ is concave on $[0,\pi]$ and symmetric wrt to $x=\frac{\pi}{2}$. –  Jack D'Aurizio Oct 14 '12 at 16:41

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