Pointwise version of Fejer's Theorem

Question

In Principles of Mathematical Analysis Rudin asks that if $f$ is integrable in $[-\pi,\pi]$ of period $2\pi$ and $f(x+)$ and $f(x-)$ exist for some $x$, then $$\lim_{N\rightarrow \infty}\sigma_N(f;x)=\frac{1}{2}[f(x+)+f(x-)].$$

How is this done?

Answer 1

Hint: adapt the proof of Fejer's theorem.

If $F_N$ denotes Fejer's kernel we have: $$ 2\pi\,\sigma_N (f;x) = \int_{-\pi}^{\pi} f(x-t)F_N(t)\,dt = \int_0^{\pi}f(x-t)F_N(t)\,dt + \int_0^{\pi}f(x+t)F_N(t)\,dt $$

You should be able to prove that the first integral converges to $\pi\, f(x-)$ and the second to $\pi\, f(x+)$

Answer 2

You may assume $x=0$. The function $$g(t):=\cases{{f(t)+f(-t)\over2}\quad&$(t\ne0)$\cr &\cr \lim_{t\to 0+}{f(t)+f(-t)\over2}&$(t=0)$ \cr}\ .$$ has period $2\pi$ and is continuous at $0$. Therefore by Fejer's theorem $$\lim_{N\to\infty}\sigma_{g,N}(0)=g(0)=\lim_{t\to 0+}{f(t)+f(-t)\over2}\ .$$ On the other hand, for almost all $t$ we have $g(t)=f(t)+u(t)$ with $u(t):={f(t)- f(-t)\over2}$. Since $u$ is odd it follows that $\sigma_{u,N}$ is odd for all $N$. Therefore $$\lim_{N\to\infty}\sigma_{f,N}(0)=\lim_{N\to\infty}\sigma_{g,N}(0)=\lim_{t\to 0+}{f(t)+f(-t)\over2}\ ,$$ as stated.

Answer 3

This is a follow-up to Julien's answer.

Step 1 : localization

Let us fix some $0< \delta < \pi$. By dominated convergence theorem (remember that $0 \leq F_n(x) \leq (n\sin(x/2)^2)^{-1}$), we have $$ \limsup_{n\to\infty}  \left|\int_0^\pi f(x+t)F_n(t)\,dt - \int_0^\delta f(x+t)F_n(t)\,dt\right| \leq \limsup_{n\to\infty}\int_\delta^\pi|f(x+t)|F_n(t)\,dt = 0 $$

and in the same way, since $\int_0^\pi F_n(t)\,dt = \frac{1}{2}\int_{-\pi}^\pi F_n(t)\,dt = \pi$, $$ \limsup_{n\to\infty}\left|\pi\, f(x+)-\int_0^\delta f(x+)F_n(t)\,dt\right| = 0 $$

Hence, $$ \limsup_{n\to\infty} \left|\pi\, f(x+)-\int_0^\pi f(x+t)F_n(t)\,dt\right| \leq \limsup_{n\to\infty} \int_0^\delta|f(x+t)-f(x+)|F_n(t)\,dt $$

Step 2 : approximation of $f(x+t)$ for small $t$

Let $\epsilon > 0$. By definition, there exists $\delta > 0$ such that $$ \sup_{0 < t < \delta}|f(x+t)-f(x+)| \leq \epsilon $$

Using step 1, we finally have $$ \limsup_{n\to\infty} \left|\pi\, f(x+)-\int_0^\pi f(x+t)F_n(t)\,dt\right| \leq \epsilon \,\limsup_{n\to\infty} \int_0^\delta F_n(t)\,dt \leq \pi\epsilon $$ Since the later stand for every $\epsilon > 0$, we have $$ \lim_{n\to\infty} \int_0^\pi f(x+t)F_n(t)\,dt = \pi\,f(x+) $$

In the same way, $$ \lim_{n\to\infty} \int_0^\pi f(x-t)F_n(t)\,dt = \pi\,f(x-) $$

Step 3 : conclusion

Use Julien's argument.