# Show $\exists f \in C \ni \|f\|_\infty \le 1$, but it's Fourier series diverges

Show $\exists f \in C \ni \|f\|_\infty \le 1$, but it's Fourier series diverges.

The proof is in our textbook (Katznelson, Harmonic analysis). It uses this argument.

Let $D_n(t)=\sum_{k=-n}^ne^{ikt}$ be the Dirichlet Kernel, $g=\text{sgn}(D_n(t))$, and $E=\{f \in C: \|f\|_\infty \le 1\}$.
Choose $f$ in $E$ for which $$f(t)=g(t) \text{, except around the discountinuity of g,}$$ and for which the sum $S$ of the length of the intervals where $f$ and $D_n(t)$ differ is less than $\epsilon/2n$.

Why is choosing an $f$ with $S \le \epsilon/2n$ possible?

If the number of discontinuity of $g$ is countable, I understand.
But if it's not, I don't see how to justify this choice.
And I don't know the cardinality of the discontinuities of $g$.

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We should denote $g_n$ as $g$ depends on $n$. The sum in $D_n$ depends on $k$, so I guess it's $e^{ikt}$. What does $f\in C\ni \lVert f\rVert_{\infty}$ mean? And where do you want continuity? For the problem, the explicit formula for $D_n$ should help you to see the points of discontinuity. – Davide Giraudo Oct 9 '12 at 20:29
$\ni$ = 'such that'. – copper.hat Oct 9 '12 at 20:33
@copper.hat I didn't know that. Thanks! – Davide Giraudo Oct 9 '12 at 20:39
@copper.hat, huh, is that standard? I always took $\ni$ to just be $\in$ in reverse order. – Christopher A. Wong Oct 9 '12 at 21:06
@ChristopherA.Wong: I'm not sure how standard it is, but I have often seen it used over the last 30 years or so. Usually the horizontal line extends a little further to the right. – copper.hat Oct 9 '12 at 21:09

Since $D_n(t)= \frac{\sin (n + \frac{1}{2}) t }{\sin \frac{t}{2}}$ (taken as the limit when $\sin \frac{t}{2} = 0$), you can see that $D_n$ has discontinuities when $\sin (n + \frac{1}{2}) t = 0$, ie, when $t \in \{ \frac{k \pi}{n + \frac{1}{2}} \}_{k=-\infty}^\infty$. So the discontinuities of $g$ are nicely spaced.