Fejer's theorem with Riemann integrable function

Question

If $f$ is integrable and $f(x+), f(x-)$ exists for some $x$, then $$ \lim_{N \rightarrow \infty} {\frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( {x - t} \right){K_N}\left( t \right)dt} } = \frac{1}{2}[f(x+) + f(x-)] $$ where $K_N(t)$ is Fejer kernel.

Here is my naive try: First observe the difference \begin{array}{l} \left| {\frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( {x - t} \right){K_N}\left( t \right)dt} - \frac{1}{2}[f(x + ) + f(x - )]} \right|\\ = \left| {\frac{1}{{2\pi }}\int_{ - \pi }^\pi {f\left( {x - t} \right){K_N}\left( t \right)dt} - \frac{1}{2}[f(x + ) + f(x - )]\frac{1}{{2\pi }}\int_{ - \pi }^\pi {{K_N}\left( t \right)dt} } \right|\\ = \frac{1}{{2\pi }}\left| {\int_{ - \pi }^\pi {\left[ {f\left( {x - t} \right) - \frac{1}{2}[f(x + ) + f(x - )]} \right]{K_N}\left( t \right)dt} } \right|\\ \le \frac{1}{{2\pi }}\int_{ - \pi }^\pi {\left| {\left[ {f\left( {x - t} \right) - \frac{1}{2}[f(x + ) + f(x - )]} \right]{K_N}\left( t \right)} \right|dt} \end{array}

Then I stuck to say more words about my approach... I think if I can rewrite $f(x+)$ and $f(x-)$ in terms of $f(x-t)$ maybe I can pursue further..

Thank you

Answer 1

The Fejer kernel $K_{N}(t)$ has several nice properties:

1. $K_{N}(t) \ge 0$
2. $K_{N}(t)=K_{N}(-t)$
3. $K_{N}(t+2\pi)=K_{N}(t)$
4. $\int_{0}^{\pi}K_{N}(t)\,dt = \pi$
5. $K_{N}(t)$ tends uniformly to $0$ for $0 < \delta \le |t| \le \pi$ as $N\rightarrow\infty$.

Those are the only properties that you need. For example, $$ \left|\frac{1}{2\pi}\int_{-\pi}^{0}K_{N}(t)f(x-t)\,dt - \frac{1}{2}f(x+0)\right| \\ = \frac{1}{2\pi}\left|\int_{-\pi}^{0}K_{N}(t)(f(x-t)-f(x+0))\,dt\right| \\ \le \frac{1}{2\pi}\int_{-\pi}^{-\delta}K_{N}(t)|f(x-t)-f(x+0))|\,dt + \frac{1}{2\pi}\int_{-\delta}^{0}K_{N}(t)|f(x-t)-f(x+0)|\,dt $$ The first term on the right is bounded by $$ \frac{1}{2\pi}\left(\sup_{t\in[-\pi,-\delta]}K_{N}(t)\right)\left(\int_{-\pi}^{\pi}|f(t)|\,dt+\frac{1}{2}|f(x+0)|\right), $$ a term which tends to $0$ as $N\rightarrow\infty$ because of property (5). The second term on the right is bounded by $$ \frac{1}{2\pi}\int_{-\pi}^{0}K_{N}(t)\,dt \sup_{t\in[x,x+\delta]}|f(t)-f(x+0)| \le \frac{1}{2}\sup_{t\in[x,x+\delta]}|f(t)-f(x+0)| $$ Therefore, for $\epsilon > 0$, there exists $\delta > 0$ such that the term on the right is bounded by $\epsilon/2$, assuming that $\lim_{t\downarrow 0}f(x+t)=f(x+0)$ exists. Then, for that fixed $\delta$, there exists $N_{0}$ large enough that $N \ge N_{0}$ guarantees that the previous term is bounded by $\epsilon/2$. Therefore, for $N \ge N_{0}$, $$ \left|\frac{1}{2\pi}\int_{-\pi}^{0}K_{N}(t)f(x-t)\,dt - \frac{1}{2}f(x+0)\right| < \epsilon. $$ By definition of the limit, $$ \lim_{N\rightarrow\infty}\frac{1}{2\pi}\int_{-\pi}^{0}K_{N}(t)f(x-t)\,dt=f(x+0). $$ The other half is handled in a similar manner.