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Problem

Is there a continuous function $f: \mathbb{T} \rightarrow \mathbb{R}$ such that $\lim_{n\rightarrow\infty} |S_{n}f(0)| = 1$?

Relevant Definitions

The Dirichlet Kernel, $D_{n}$, is defined as:

$D_{n}(x)=\sum_{k=-n}^{n}e^{ikx}$

and the partial sums, $S_{n}$, as

$(S_{n}f)(x)= (D_{n}*f)(x)=\int_{-\pi}^{\pi}D_{n}(x-y)f(y)dy$

and $\mathbb{T}=[-\pi,\pi]$

Actual Question

I dont know how to start to proof this type of statement. I've tried coming up with an example but can't. I've also tried assuming that such functions exists and tried to find a contradiction using the continuous property and also failed.

If someone could tell me if such a function exists or not it would help immensly!

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    $\begingroup$ You should replace $e^{ikn}$ by $e^{ikx}$ in your Dirichlet kernel. $\endgroup$ Jun 26, 2021 at 15:16

1 Answer 1

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Here is a $90\%$ hint.

Observe that $$(D_n*f)(0)=\int_{-\pi}^{\pi}D_n(-y)f(y)\,dy=\sum_{|k|\leq n}\hat{f}(k)$$

Now, choose a nice continuous function whose Fourier series converges to some non-zero number, and you are almost done.

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  • $\begingroup$ Isn't it true that for a continuous function its fourier coefficients must converge to $0$? $\endgroup$ Jul 4, 2021 at 19:41
  • $\begingroup$ Yes, it is true, but the Fourier coefficients is one thing, and the Fourier series is another. Just like $1/n$ tends to zero, but the corresponding series behaves quite differently. $\endgroup$ Jul 6, 2021 at 7:46

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