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Let $f(x)$ be a $C^1$ function of period $2\pi$. I'm trying to show that we may as well assume that $f(0)=0$ and that the fourier series converges to zero at $x = 0$ and let $$g(x) = \frac{f(x)}{e^{ix}-1}.$$ Also, I have to show that $g(x)$ is a continous function.

Since $f$ is a continous function of order $1$, then can we do something with uniform continuity to show $f(x) = 0$? And since $f(x)$ is a continous function, $e^{ix}$ is a continuous function, then $g(x)$ is a composition of two continuous functions which is continuous.

I really need help with using a $\delta-\epsilon$ proof to making my assertion rigorous in the aforementioned sentence

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The sentence "we may as well" is unclear: $f$ is given, so $f(0)$ is what it is. I suspect that you are working on the proof of some theorem, and you want to add an assumption that is not among those of the theorem itself. In this case, in my opinion, it is enough to remark that $f-f(0)$ is a $C^1$ function, with the same period as $f$, which vanishes at $0$. Moreover, being $f$ of class $C^1$, its Fourier series converges uniformly and hence pointwise to $f$. As a consequence, up to replacing $f$ with $f-f(0)$, you can think that $f(0)=0$ and that the Fourier series of $f$ converges to $f(0)=0$ at $x=0$.

Moving to your second question, the equation $e^{ix}=1$ is solved in $[0,2\pi)$ only by $x=0$. Since the map $x \mapsto e^{ix}-1$ is continuous (here you need to know something about vector-valued functions), the function $g$ is continuous away from $x=0$. You need to check that $g$ is also continuous at $x=0$. Now, $$ e^{ix}-1 = \sum_{n=1}^\infty \frac{(ix)^n}{n!}, $$ and $$ g(x) = \frac{f(x)}{\sum_{n=1}^\infty \frac{(ix)^n}{n!}} = \frac{f(x)}{x \left( i+\sum_{n=2}^\infty \frac{(ix)^n}{n!} \right)} = \frac{f(x)}{x} \frac{1}{i+\sum_{n=2}^\infty \frac{(ix)^n}{n!}} = \frac{f'(0)}{i}+o(1) $$ as $x \to 0$. We have proved that the function $$ \tilde{g}(x)=\begin{cases} \frac{f(x)}{e^{ix}-1} &\text{if $x \neq 0$} \\ \frac{f'(0)}{i} &\text{if $x=0$} \end{cases} $$ is continuous on $[0,2\pi)$, which is a more precise statement than the one you wrote.

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thanks, this was an amazing answer that really challenged how I look at things – mary Oct 28 '12 at 9:33

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