If $f$ is Riemann-integrable, $\int_{-\pi}^{\pi}|\sigma_Nf(x)-f(x)|dx \rightarrow 0$ as $N\rightarrow\infty$

Question

Let $f$ be Riemann-integrable (not necessarily continuous). Prove $\int_{-\pi}^{\pi}|\sigma_Nf(x)-f(x)|dx \rightarrow 0$ as $N\rightarrow\infty$, where $\sigma_Nf(x)$ is $\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x-y)K_N(y)dy$ and $K_N$ is the $n$th order Fejér kernel.

Here's what I've tried thus far: $$\int_{-\pi}^{\pi}|\sigma_Nf(x)-f(x)|dx\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}\int_{-\delta}^{\delta}|f(x-y)-f(x)|K_N(y)dydx+\\\frac{1}{2\pi}\int_{-\pi}^{\pi}\int_{\delta\leq|y|\leq\pi}|f(x-y)-f(x)|K_N(y)dydx.$$

Then the second term is less than $$\frac{1}{2\pi}\int_{-\pi}^{\pi}2||f||_\infty\sup_{\delta\leq|y|\leq\pi}K_N(y)dx.$$

Since $K_N(y)$ converges uniformly to $0$ as $N\rightarrow\infty$ for $\delta\leq|y|\leq\pi$, we can exchange the integral and the limit and obtain that this term goes to $0$. However, I have no idea what to do with the first term!

I was told that I should try to approximate the function with a sequence of continuous functions, and go from there, but I have no idea which functions to choose. I also don't know how that would allow me to split up the integral into terms I could estimate.

Answer 1

First, prove the limit is 0 if $f$ is continuous. Note that in this case, $f$ is actually uniformly continuous being a continuous function defined on a compact interval, and thus you can choose $\delta$ strategically so that $|f(x-y)-f(x)|<\epsilon$ whenever $|y|<\delta$. This will make the first term arbitrarily small.

Next, for the general case when $f$ is merely Riemann integrable, use the fact that given $\epsilon>0$ you can find a continuous $g$ (defined on $[-\pi,\pi]$) such that $\int_{-\pi}^\pi |g(y)-f(y)|dy<\epsilon$. Try using triangle inequality and estimating similarly.