Need help with proof: Existence of 2-pi periodic continuous function that does not converge at 0?

Question

I have a proof in my lecture notes of the existence of a $2\pi$-periodic continuous function $f$, such that the Fourier series of f does not converge at 0, i.e. the partial sums are unbounded. Proof summarised, specific issue is highlighted below:

$n^{\text{th}}$ partial sum is $$f_n(0)=\frac{1}{2\pi}\sum_{k=-n}^{n}\Big(\int_{-\pi}^{\pi}f(t)e^{ikt}dt\Big)e^{-ikx}$$ and this eventually gives $$f_n(0)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\frac{\text{sin}((n+\frac{1}{2})t)}{\text{sin}(\frac{1}{2}t)}dt.$$

Letting $P=\lbrace f\in C([-\pi,\pi];\mathbb{R}):f(-\pi)=f(\pi)\rbrace$, $S_n:P\rightarrow \mathbb{R}$ given by $S_n(f)=f_n(0)$ and $I_n=\int_{-\pi}^{\pi}\Big{|}\frac{\text{sin}((n+\frac{1}{2})t)}{\text{sin}(\frac{1}{2}t)}\Big{|}dt$, it says that $||S_n||\leq I_n$ (should this not be $\frac{1}{2\pi}I_n$?).

This is the bit I am really stuck with: Then in order to get $||S_n||=I_n$, it says to approximate $f=\text{sign}(\frac{\text{sin}((n+\frac{1}{2})t)}{\text{sin}(\frac{1}{2}t)})$ by a sequence of functions in $P$. It just says that $f$ as just stated is not in $P$ and that this can overcome by approximating $f$ by a sequence of elements from $P$. I need to find a suitable sequence of functions and also the appropriate type of convergence.

The rest is then proving that $I_n$ is unbounded and hence $|f_n(0)|$ is unbounded by Uniform Boundedness Principle, which I think I can handle.

Any assistance with this will be much appreciated.

Answer 1

1. We know that $$G_N(x)=\sum_{n=1}^{N}\frac{\sin(nx)}{n}$$ is bounded, i.e. there exist $C>0$, such that for all $N\ge 1$, we have $\|G_N\|_\infty\le C$. Let $F_N=\sum_{n=1}^{N}\sin(nx)$, if $x=0$ or $x=2\pi$, that's trival, so let $x\in [\delta,2\pi-\delta],\delta >0$, then $$|F_N|=\left|\sum_{n=1}^{N}\sin(nx)\right|=\left|\frac{-\frac{1}{2}(\cos(N+\frac{1}{2})x-\cos\frac{1}{2}x)}{\sin\frac{1}{2}x}\right|\le \frac{1}{|\sin\frac{\delta}{2}|}=C$$ then $\forall x \in [\delta,2\pi-\delta]$ $$|G_N|=\left|\sum_{n=1}^{N}\frac{\sin(nx)}{n}\right|\le \frac{|F_N|}{N}+\sum_{n=1}^{N-1}|F_N|(\frac{1}{n}-\frac{1}{n+1})\\ \le \frac{C}{N}+\sum_{n=1}^{N-1}C(\frac{1}{n}-\frac{1}{n+1})=C(\frac{1}{N}+1-\frac{1}{N})=C$$ we get $\|G_N\|_\infty\le C$.

2. We construct two sequence, one is a sequence of positive integer $(n_k)_{k\ge 1}$, and the other one is a sequence of positive real number $(\alpha_k)_{k\ge 1}$ such that $$\inf_{k\ge 1}\frac{n_{k+1}}{n_k}>3,\ \sum_{k\ge 1}\alpha_k<\infty, \ \lim_{k\rightarrow \infty}\alpha_k\log n_k =\infty.$$ actually, let $n_k=3^{k^3}, \alpha_k = \frac{1}{k^2}$, then $$\inf_{k\ge 1}\frac{n_{k+1}}{n_k}=3^{{(k+1)}^3-k^3}>3,\ \sum_{k\ge 1}\alpha_k=\sum_{k\ge 1}\frac{1}{k^2}<\infty,\ \lim_{k\rightarrow \infty}\frac{1}{k^2}\log3^{k^3}=\lim_{k\rightarrow \infty}k\log3>\lim_{k\rightarrow \infty}k=\infty$$

3. Let $$P_k(x)=e^{2in_kx}\sum_{n=1}^{n_k}\frac{\sin(nx)}{n},$$ then by 1 $$\|P_k(x)\|_\infty = \left\|\sum_{n=1}^{n_k}\frac{\sin(nx)}{n}\right\|_\infty \le C$$ in other side, $\sum_{k=1}^{\infty}\alpha_k < \infty$ and $\alpha_k=\frac{1}{k^2}$ is decreasing, then by Abel Discriminance we know $\sum_{k\ge 1}\alpha_kP_k(x)$ is uniformly converge on $\mathbb{R}$, we definite $f=\sum_{k\ge 1}\alpha_kP_k(x)$.

4. We know the value of partial sum of $f$ on $0$ is \begin{align} S_{2n_k}(f)(0)&=\sum_{n=-2n_k}^{2n_k}\hat{f}(n)=\sum_{|n|\le 2n_k}\frac{1}{2\pi}\int_{0}^{2\pi}\sum_{l\ge 1}\alpha_l P_l(x)e^{-inx}dx\\ &=\sum_{|n|\le 2n_k}\sum_{l\ge 1}\alpha_l\sum_{r=1}^{n_l}\frac{1}{2\pi}\int_{0}^{2\pi} \frac{\sin(rx)}{r}e^{i(2n_l-n)x}dx\\ &=\sum_{|n|\le 2n_k}\sum_{l\ge 1}\alpha_l\sum_{r=1, r=2n_l-n}^{n_l}i\frac{1}{2r}. \end{align} because $\inf_{k\ge 1}\frac{n_{k+1}}{n_k}>3$, then $n_{k+1}-n_k> 2n_k$, then in above $l=k$.then $$S_{2n_k}(f)(0)=\sum_{|n|\le 2n_k}\alpha_k\sum_{r=1, r=2n_k-n}^{n_k}\frac{i}{2r}=\frac{-\alpha_k}{2i}\sum_{r=1}^{n_k}\frac{1}{r}=(\sum_{n=1}^{n_k}\frac{1}{n})(-\frac{\alpha_k}{2i})$$

5. $$\lim_{k\rightarrow \infty}S_{2n_k}(f)(0)=\lim_{k\rightarrow \infty}\sum_{n=1}^{n_k}\frac{1}{n}(-\frac{\alpha_k}{2i})=\lim_{k\rightarrow \infty}-\frac{\alpha_k}{2i}\int_{1}^{n_k}\frac{1}{x}dx=\lim_{k\rightarrow \infty}-\frac{\alpha_k}{2i}\log n_k=\infty.$$

Answer 2

Let $$ g(t)=\frac{\sin((n+\frac{1}{2})t)}{sin(\frac{1}{2}t)}$$ Then $f=sign(g)$ is a step function with finitely many jumps. To render it continuous pick $\varepsilon>0$ and interpolate by an affine function in an $\varepsilon$-nbhd of the discontinuity (straight line starting $\varepsilon$ to the left of the discontinuity and ending $\varepsilon$ to the right of the discontinuity). This continuous function $f_\varepsilon$ has still norm $1$ (we picked the interpolating line such that function still takes values between $-1$ and $1$) and converges in $L^1$ to $f$. This is enough to get for $\varepsilon \downarrow 0$ $$\vert S_n(f_\varepsilon)\vert = \Vert f_\varepsilon g\Vert_1 \rightarrow \Vert f g\Vert_1 = I_n$$ where we used that $g$ is a bounded function (such that we can take the limit) and that the sign of $f_\varepsilon$ and of $f$ coincide (for the first equality).