Fourier Series in L_p space

$p=\infty$

If $f\in L^{\infty }(\mathbb{T})$ then for every $N\ge 0$, $S_{N}f\in C(\mathbb{T})$. Thus the essential supremum of $S_{N}f$ is equal to the actual supremum, i.e. $\Vert S_{N}f{\Vert }_{L^{\infty }(\mathbb{T})}=\Vert S_{N}f{\Vert }_{C(\mathbb{T})}$. Since $\text{sgn}(D_N)\in L^{\infty }(\mathbb{T})$, we have $\Vert S_N{\Vert }_{L^{\infty }(\mathbb{T})\to L^{\infty }(\mathbb{T})}\ge \Vert S_N(\text{sgn}{D}_N){\Vert }_{C(\mathbb{T})}\ge |S_N(\text{sgn}(D_N))(0)|=\Vert D_N{\Vert }_{L^1(\mathbb{T})},$ so with bound for Dirichlet kernel, $\Vert S_N{\Vert }_{L^{\infty }(\mathbb{T})\to L^{\infty }(\mathbb{T})}\to \infty \text{as }N\to \infty .$ Proposition 1 implies that there exists $f\in L^{\infty }(\mathbb{T})$ such that $S_{N}f$ does not converge in $L^{\infty }(\mathbb{T})$ to any element of $L^{\infty }(\mathbb{T})$ as $N\to \infty$.

$p=1$

By Proposition 1, we need to decide whether ${\sup }_{N\ge 1}\Vert S_N{\Vert }_{L^1(\mathbb{T})\to L^1(\mathbb{T})}$ is finite or not. 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 $|x|\le 1/2$. Then $\Vert S_N{\Vert }_{L^1(\mathbb{T})\to L^1(\mathbb{T})}\ge \Vert S_N{\phi }_{ϵ}{\Vert }_{L^1(\mathbb{T})}\text{for all }ϵ>0\text{ and since }{D}_N\in C(\mathbb{T}),$ ${\lim }_{ϵ\to 0^{+}}\Vert S_N{\phi }_{ϵ}-D_N{\Vert }_{C(\mathbb{T})}=0$ ($S_N{\phi }_{ϵ}$ converges uniformly to $D_N$ as $ϵ\to 0^{+}$). In particular, ${\lim }_{ϵ\to 0^{+}}\Vert S_N{\phi }_{ϵ}{\Vert }_{L^1(\mathbb{T})}=\Vert D_N{\Vert }_{L^1(\mathbb{T})}.$ This shows $\Vert S_N{\Vert }_{L^1(\mathbb{T})\to L^1(\mathbb{T})}\ge \Vert D_N{\Vert }_{L^1(\mathbb{T})}.$ As a result, with bound for Dirichlet kernel, ${\sup }_{N\ge 1}\Vert S_N{\Vert }_{L^1(\mathbb{T})\to L^1(\mathbb{T})}\ge {\sup }_{N\ge 1}\Vert D_N{\Vert }_{L^1(\mathbb{T})}=\infty ,$ and there exists some $f\in L^1(\mathbb{T})$ for which $S_{N}f$ does not converge in the $L^1(\mathbb{T})$ norm to any function in $L^1(\mathbb{T})$.

$1<p<\infty$

The easiest case is of course $p=2$: We know $S_{N}f$ converges to $f$ in $L^2(\mathbb{T})$ whenever $f\in L^2(\mathbb{T})$. One way to see this is via Proposition 1: note that $\Vert S_N{\Vert }_{L^2(\mathbb{T})\to L^2(\mathbb{T})}=1$ for all $N$. Alternatively, recall that $S_N$ is the orthogonal projection from $L^2(\mathbb{T})$ onto ${\mathcal{P}}_N$, the space of trigonometric polynomials of degree $N$. The set of all trigonometric polynomials is a dense subset of $L^2(\mathbb{T})$. Together we see that $\Vert f-S_{N}f{\Vert }_{L^2(\mathbb{T})}={\inf }_{P\in {\mathcal{P}}_N}\Vert f-P{\Vert }_{L^2(\mathbb{T})}\to 0\text{as }N\to \infty .$ Now we tackle the case of a general $1<p<\infty$. By Proposition 1, we need to decide whether ${\sup }_{N\ge 1}\Vert S_N{\Vert }_{L^p(\mathbb{T})\to L^p(\mathbb{T})}$ is finite or not. We will prove the existence of some finite constant $C_p$, such that

$$\Vert Hf{\Vert }_{L^p(\mathbb{T})}\le C_p\Vert f{\Vert }_{L^p(\mathbb{T})}$$

for all $f\in L^p\cap L^2(\mathbb{T})$, with $Hf$ the Hilbert Transform. This will then allow us to prove that ${\sup }_{N\ge 1}\Vert S_N{\Vert }_{L^p(\mathbb{T})\to L^p(\mathbb{T})}<\infty$ and hence Proposition 1 shows that $S_{N}f$ converges to $f$ in $L^p(\mathbb{T})$ for every $f\in L^p(\mathbb{T})$.

Proposition

Define the modulations $M_{N}f(x):=e^{2\pi iNx}f(x)$ for $N\in \mathbb{Z}$ and the reflection $Rf(x):=f(-x)$.
For $f\in L^p\cap L^2(\mathbb{T})$ we have $S_{N}f(x)=M_{N}R(iH)R{M}_{-2N}(iH)M_{N}f(x)+\frac{\hat{f}(-N)}{2}{e}^{-2\pi iNx}+\frac{\hat{f}(N)}{2}{e}^{2\pi iNx}$

Proof $M_{N}f(x)={\sum }_{n=-\infty }^{\infty }\hat{f}(n-N)e^{2\pi inx}$ $(iH)M_{N}f(x)=\frac{\hat{f}(-N)}{2}+{\sum }_{n=1}^{\infty }\hat{f}(n-N)e^{2\pi inx}$ $M_{-2N}(iH)M_{N}f(x)=\frac{\hat{f}(-N)}{2}{e}^{-2\pi i(2N)x}+{\sum }_{n=1-2N}^{\infty }\hat{f}(n+N)e^{2\pi inx}$ $R(iH)R{M}_{-2N}(iH)M_{N}f(x)=\frac{\hat{f}(-N)}{2}{e}^{-2\pi i(2N)x}+\frac{\hat{f}(N)}{2}+{\sum }_{n=-(2N-1)}^{-1}\hat{f}(n+N)e^{2\pi inx}$ $M_{N}R(iH)R{M}_{-2N}(iH)M_{N}f(x)=\frac{\hat{f}(-N)}{2}{e}^{-2\pi iNx}+\frac{\hat{f}(N)}{2}{e}^{2\pi iNx}+{\sum }_{n=-(N-1)}^{N-1}\hat{f}(n)e^{2\pi inx}.$

Note that $\Vert M_N{\Vert }_{L^p(\mathbb{T})\to L^p(\mathbb{T})}=\Vert R{\Vert }_{L^p(\mathbb{T})\to L^p(\mathbb{T})}=1$ for every $N$, and $|\hat{f}(\pm N)|\le \Vert f{\Vert }_{L^1(\mathbb{T})}\le \Vert f{\Vert }_{L^p(\mathbb{T})},$ As a result, if $f\in L^p\cap L^2(\mathbb{T})$, then from this proposition and bound for Hilbert transform $\Vert Hf{\Vert }_{L^p(\mathbb{T})}\le C_p\Vert f{\Vert }_{L^p(\mathbb{T})}$, we have $\Vert S_{N}f{\Vert }_{L^p(\mathbb{T})}\le (C_p^2+1)\Vert f{\Vert }_{L^p(\mathbb{T})}$ uniformly in $N$. Now since $L^p\cap L^2(\mathbb{T})$ is dense in $L^p(\mathbb{T})$, we have $\Vert S_N{\Vert }_{L^p(\mathbb{T})\to L^p(\mathbb{T})}={\sup }_{f\in L^p\cap L^2(\mathbb{T}),\Vert f{\Vert }_{L^p(\mathbb{T})}=1}\Vert S_{N}f{\Vert }_{L^p(\mathbb{T})}\le C_p^2+1.$ This shows that ${\lim }_{N\to \infty }\Vert S_{N}f-f{\Vert }_{L^p(\mathbb{T})}=0$ for every $f\in L^p(\mathbb{T})$ when $1<p<\infty$.