Fourier Series of Continuous Functions

Bound for $\Vert D_N{\Vert }_{L^1}$

${\int }_{\mathbb{T}}|D_N(y)|dy\ge \log N$ as $N\to \infty$

Proof

Consider inequality $x\ge \sin x$, we have $\frac{1}{\sin \pi x}\ge \frac{1}{\pi x}\ge \frac{2N+1}{(i+1)\pi }$ for $x\in [\frac{i}{2N+1},\frac{i+1}{2N+1}]$ $\Vert D_N{\Vert }_{L^1(\mathbb{T})}={\sum }_{i=0}^{2N}{\int }_{[\frac{i}{2N+1},\frac{i+1}{2N+1}]}|D_N(x)|dx\ge {\sum }_{i=0}^{2N}\frac{(2N+1)}{(i+1)\pi }\frac{1}{2N+1}{\int }_{[0,1]}\sin (\pi x)dx$ with $c=\frac{{\int }_{[0,1]}\sin (\pi x)dx}{\pi }$. $\Vert D_N{\Vert }_{L^1(\mathbb{T})}\ge c(1+\frac{1}{2}+\cdots \frac{1}{2N+1}).$ So ${\int }_{\mathbb{T}}|D_N(y)|dy\ge \log N$ as $N\to \infty$

Theorem

For $f\in C(\mathbb{T})$, $S_{N}f(x)={\sum }_{n=-N}^N\hat{f}(n)e^{2\pi inx}$ doesn’t converge in $C(\mathbb{T})$.

Proof:

By proposition for convergence of fourier series of continuous functions, we need to decide whether ${\sup }_{N\ge 1}\Vert S_N{\Vert }_{C(\mathbb{T})\to C(\mathbb{T})}$ is finite or not. For context, note that by the triangle inequality, and with $S_{N}f(x)=f\ast D_N(x)$, $D_N=\frac{\sin ((2N+1)\pi x)}{\sin (\pi x)}$ $\Vert S_N{\Vert }_{C(\mathbb{T})\to C(\mathbb{T})}={\sup }_{\Vert f{\Vert }_{C(\mathbb{T})}=1}\Vert {\int }_{\mathbb{T}}f(x-y)D_N(y){\Vert }_{C(\mathbb{T})}={\sup }_{\Vert f{\Vert }_{C(\mathbb{T})}=1}{\sup }_x|{\int }_{\mathbb{T}}f(x-y)D_N(y)|$ $\le {\sup }_{\Vert f{\Vert }_{C(\mathbb{T})}=1}{\sup }_x{\int }_T|f(x-y)D_N(y)|\le \Vert D_N{\Vert }_{L^1(\mathbb{T})}.$ On the other hand, fix $\phi \in C(\mathbb{T})$ with support in $[-1/10,1/10]$. For $ϵ>0$, let ${\phi }_{ϵ}$ be the periodic function so that ${\phi }_{ϵ}(x)={ϵ}^{-1}\phi ({ϵ}^{-1}x)$ for $|x|\le 1/2.$ Also let $f_{ϵ}:=\text{sgn}(D_N)\ast {\phi }_{ϵ}$ so that $f_{ϵ}\in C(\mathbb{T})$ with $\Vert f_{ϵ}{\Vert }_{C(\mathbb{T})}=1$ for $ϵ\le \frac{1}{2(2N+1)}$. For such $ϵ$, $\Vert S_N{\Vert }_{C(\mathbb{T})\to C(\mathbb{T})}\ge \Vert S_N{f}_{ϵ}{\Vert }_{C(\mathbb{T})}\ge |S_N{f}_{ϵ}(0)|.$ Let $ϵ\to 0^{+}$ we have $S_N{f}_{ϵ}(0)\to S_N(\text{sgn}(D_N))(0)={\int }_{\mathbb{T}}{D}_N(0-y)\text{sgn}(D_N)(y)dy={\int }_{\mathbb{T}}|D_N(y)|dy.$ Thus $\Vert S_N{\Vert }_{C(\mathbb{T})\to C(\mathbb{T})}=\Vert D_N{\Vert }_{L^1(\mathbb{T})}.$ From above proposition, ${\int }_{\mathbb{T}}|D_N(y)|dy\ge \log N$ as $N\to \infty$, so ${\sup }_{N\ge 1}\Vert S_N{\Vert }_{C(\mathbb{T})\to C(\mathbb{T})}=\infty .$

One can extract a stronger result from this proof: there exists $f\in C(\mathbb{T})$ such that ${\mathrm{limsup}}_{N\to \infty }|S_{N}f(0)|=\infty .$ In fact, the above proof also gives ${\sup }_{f\in C(\mathbb{T}),\Vert f{\Vert }_{C(\mathbb{T})}=1}|S_{N}f(0)|=\Vert D_N{\Vert }_{L^1(\mathbb{T})}$ and the claimed result follows from the uniform boundedness principle.