# Show $\left\lvert \frac{1}{\pi+2} \sum_{0 \le k \le 2n,k\ne n} \frac{i}{k-n} e^{ikt} \right\rvert > c \ln n, c>0$

The partial sum of the Fourier series of $f(t)=(t-\pi)\chi_{\left(0,2\pi\right)}$ can be written as \begin{align} S_n(f,t) &= \sum_{0<|k| \le n} \frac{i}{k} e^{ikt} \; (1) \\ & = -2\sum_{k=1}^n \frac{\sin kt }{k} \; (2)\\ &= 2 \int_0^t \left( \frac{\sin (n+1/2)t}{\sin t/2} - 1 \right) dx \; (3) \end{align}

and it can be shown that $$\left \lvert S_n(f,t) \right \rvert \le \pi+2$$ Define $g_n(t)=\frac{e^{int}}{\pi+2} S_n(f,t)$, show that for some $c>0$ $$\left \lvert S_n(g_n,0) \right \rvert > c \ln n$$ where $S_n(g_n,0)$ is the partial sum of the Fourier series of $g_n$ evaluated at $0$.

I found that the partial sum of the Fourier series of $g_n$ is $$(\pi+2)S_n(g_n,t)=\sum_{0\le k \le 2n, k\ne n} \frac{i}{k-n}e^{ikt}$$ Evaluated at zero, this looks like a taylor series of $\ln(x-1)$, but this won't get me near the result.
I know the Dirichlet kernel as a $L^1$ norm proportional to $\ln n$.
But I don't see how to use (3) to use this fact.

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Do you mean to introduce $S_n$ the Fourier series of $f$ and then $g_n$ and then $S_n(g_n)$? Then the title might be inaccurate. Or is the title correct and the body inaccurate? – Did Oct 15 '12 at 13:23
@did The title was inacurrate. Thank you. – Nicolas Essis-Breton Oct 15 '12 at 13:34
a) $S_n(t)$ is not a Fourier series but a Fourier sum, a partial sum of a Fourier series (as you rightly call $S_n(g_n,t)$ later on, though I find the reuse of $S_n$ slightly confusing). b) How did the number of terms get doubled from $S_n(t)$ to $S_n(g_n,t)$? You just multiplied by $\mathrm e^{\mathrm int}$; that should just shift all the coefficients by $n$, not produce further coefficients? I suspect you mean $0\le k\le 2n$ instead of $0\le|k|\le2n$? If so, the entire shift is irrelevant, since it doesn't change the absolute value of the result. – joriki Oct 15 '12 at 15:24
@joriki I tried to clarified the notation, and you were right for $0 \le k \le 2n$. Thank you. Sorry for the many typos, Fourier analysis is a jungle for which my machete is not sharp yet. – Nicolas Essis-Breton Oct 15 '12 at 15:44
a) You need to shift $k$, too; the denominator in $S_n(g_n,t)$ should be $k-n$. b) Do you agree that after these corrections are applied, there's no longer any point in considering $S_n(g_n,t)$ since it's just a phase factor times $S_n(f,t)$ and thus has the same absolute value? c) I think your $S_n(f,t)$ is wrong; it's an odd function whereas $f$ isn't odd. – joriki Oct 15 '12 at 16:12