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Let $f,g: \mathbb{R} \to \mathbb{C}$ $2\pi$ periodic , Riemann integrable in $[0, 2\pi]$. I need to prove that if $f(x)=0$ for every $x$ around $x_0$ so $S_nf(x_0) \to 0$ when $n \to \infty$.

We define $S_nf(x_0)=\sum_{-n}^{n}\hat f(n)e^{inx_o}$ and $\hat f(n)==\frac{1}{2\pi}\int_{0}^{2\pi}f(t)dt$, so I can't really see what makes the series 0 when n tends to infinity since $\hat f(n)$ does not depend on $x_0$ or other $x$'s around it.

Thank you very much

2 Answers 2

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This is a direct consequence of theorem 8.14 in Rudin's 'Principles of Mathematical Analysis' and is actually made explicit in a corollary stated in the same section. (It's formulated there for $f:[-\pi,\pi] \rightarrow \mathbb{R}$, but that's just the same).

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    @Jozef ? - in my copy is a proof, if I recall correctly. I'll check when I get home.2012-01-10
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For the favor of people who will be interested for an answer to this question:

According to Dirichlet theorem: $f$ applies Lipshcitz conditions in $x_0$, it is Differentiable and f'(x_0)=0 so there's a pointwise convergent to the average of the limits in both sides, which is $f(x_0)=0$.

Another way to look at it is just by the theorem says that $||S_nF(x)-F(x)|| \to 0$.

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    A proof for the theorem. As explained in my comment, the corrolary is an _absolutely trivial_ consequence of the theorem. Just check the hypothesis.2012-01-11