If the Fourier serie $S_nf\longrightarrow g$ in $L^p$ then $f=g$.

Question

Suppose that $f\in L^1(\mathbb S^1)$ where $\mathbb S^1=\mathbb R/\mathbb Z$. Suppose that the sequence of partial Fourier sums $\{S_nf\}_{n\geq 1}$ converge in $L^p(\mathbb S^1)$ toward some $g\in L^p(\mathbb S^1)$ (here $p\in [1,\infty ]$). Show that $f=g$.

My work

If $\{S_nf\}_{n\in\mathbb N}$ converge do we necessarily have $S_nf$ converge also to $f$ ? If yes, the result is obvious, but it's may be what I have to show. Any help would be appreciated.

Answer 1

The easiest thing is probably to show that $\hat f(n)=\hat g(n)$.

Answer 2

Another way to do it, but with probably more material required :

Define the Fejér means : $$\sigma_n(f)=\frac{1}{n+1}(S_0(f)+\ldots+S_n(f)),$$ one know that $\sigma_n(f)$ converge to $f$ almost everywhere (theorem page 20 of introduction to harmonic analysis third edition by Y. Katznelson). But since $S_n(f)$ is converging to $g$ in $L^p$ one also has that $\sigma_n(f)$ is converging to $g$ in $L^p$. Now, up to a sub sequence, $\sigma_n(f)$ converge almost everywhere to g and also to f, and thus we get $f=g$.

Answer 3

If $\lim_n\|S_n f - g\|_{L^p}=0$ for some $g \in L^p$ with $p\in [1,\infty)$, then, for $n \ge |k|$ and for $q \in (1,\infty]$ conjugate to $p$, \begin{align} |\hat{f}(k)-\hat{g}(k)| & =|\widehat{(S_n f)}(k)-\widehat{g}(k)| \\ & \le \frac{1}{2\pi}\int_{0}^{2\pi}|S_nf-g|dx \\ & \le \frac{1}{2\pi}\|S_n f-g\|_p\|1\|_q \rightarrow 0 \mbox{ as }n\rightarrow\infty. \end{align} So it is definitely true that $\hat{f}(k)=\hat{g}(k)$ for all $k$. Both $f$ and $g$ are in $L^1$. So the problem of showing that $f=g$ a.e. is reduced to looking at the uniqueness of Fourier coefficients for functions in $L^1$. Every $2\pi$ periodic continuously-differentiable function $h$ is the uniform limit of its Fourier series, which gives $$ \int_{0}^{2\pi}(f-g)hdx=\lim_{n}\int_{0}^{2\pi}(f-g)S_{n}hdx =\lim_n 0=0. $$ Using the bounded convergence theorem, you can obtain $$ 0=\int_{0}^{2\pi}(f-g)\chi_{[0,t]}dx=\int_{0}^{t}(f-g)dx. $$ By the Lebesgue differentiation theorem, $$ 0=\frac{d}{dt}\int_{0}^{t}(f-g)dx = f(t)-g(t) \;\;\; a.e. t\in [0,2\pi]. $$ Hence, $f=g$ a.e..