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I'm trying to prove that if $|\widehat{f}(k)| = |a_k| < \frac{M}{|k|^{2 + \varepsilon}}$ then $f$ is continuously differentiable. I'm not quite sure how to do this. There was another question on the site (rate of decay of fourier coefficients vs smoothness) but it makes an incorrect assumption: that we can termwise differentiate the Fourier series. Here is what I've done so far.

It's obvious that $\sum a_k e^{ikx}$ converges uniformly to some function $g$ and that $g \equiv f$ almost everywhere, with $g$ continuous. So we need to show that $g$ is continuously differentiable. The continuity part is easy once we have differentiability, so we just need to show differentiability. To that end we have

$\lim_{y \to x} \frac{g(x)-g(y)}{x-y} = \lim_{y \to x} \frac{\sum a_k e^{ikx} - \sum a_k e^{iky}}{x-y} = \lim_{y \to x} \sum a_k \left( \frac{e^{ikx} - e^{iky}}{x-y} \right) = \sum a_k \lim_{y \to x} \left( \frac{e^{ikx} - e^{iky}}{x-y} \right) = \sum a_k ike^{ikx}$

which converges based on our bound on $a_k$. However, I don't really know how to prove the third equality. I don't think that the inner functions are monotone in absolute value, and I can't think of any dominating function. Can someone point me in the right direction?

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See if the answer to a question I posed along similar lines helps.… – Ron Gordon Apr 5 '13 at 6:46
Um... isn't your third inequality exactly the same problem you had with the other problem? You are actually termwise differentiating the series, just in more cumbersome notation. – Willie Wong Apr 5 '13 at 8:02
You're right, I realized later that it would imply that. – Julien Clancy Apr 5 '13 at 14:53
up vote 2 down vote accepted

Hint: Use the theorem

If $g_n \to g$ pointwise, and $g_n$ differentiable, and that $g'_n \to h$ uniformly, then $g$ is differentiable with derivative $h$.

Let $g_n$ in this case be the Fourier partial sums $\sum_{|k| \leq n} a_k e^{ikx}$.

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I was looking for some theorem like this but whenever I searched "uniform convergence derivative" the hits were always "not differentiable". Thanks so much! And what a useful theorem. – Julien Clancy Apr 5 '13 at 15:02
In case you are looking for a source of this statement, it is in Chapter 7 of Rudin's Principles of Mathematical Analysis. In the edition in my office it is Theorem 7.17. – Willie Wong Apr 5 '13 at 15:29

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