Let $K_N=\frac{1}{N}\sum_{n=0}^{N-1}D_n(x)$ be the Fejer kernel and let $\sigma_N(f)=\frac{1}{N}\sum_{n=0}^{N-1}S_N(f)$ where $S_n(f)=\frac{1}{\pi}\int_{-\pi}^{\pi}f(\tau)D_n(t-\tau)d\tau=f*D_n.$ With "*" we denote the convolution.

Then it is easy to prove that $\sigma_N(f)=f*K_N$.

My question is if $f\in L^p[-\pi,\pi]$ how to prove that $\lim_{N\to \infty}\left\lVert \sigma_N(f)-f\right\rVert_p=0?$


Hint: Fejer's Theorem says that if $f$ is a continuous periodic function then $\|\sigma_N(f) -f\|_{\infty} \to 0$. This also implies that $\|\sigma_N(f) -f\|_{p} \to 0$. Now use the fact that continuous periodic functions are dense in $L^{p}$. You will also need Young's inequality $\|f*g\|_p \leq \|f\|_p \|g\|_{1}$ and $\|g\|_{1}=1$ when $g$ is the Fejer kernel.

  • $\begingroup$ The last inequaliti comes from the fact that if $1\leq p < q$ implies $||g||_q < ||g||_p$, right? $\endgroup$ – Emo Feb 23 '20 at 11:59
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    $\begingroup$ The inequality is called Young's inequality. See en.wikipedia.org/wiki/Young%27s_convolution_inequality [ Put $q=1, r=p$ in that inequality] @Emo $\endgroup$ – Kavi Rama Murthy Feb 23 '20 at 12:07
  • $\begingroup$ Thank you. One more question: it should be obvious, but i can't see why zero convergence of supremum norm implies zero convergence of the p-norm? Thanks in advance! $\endgroup$ – Emo Feb 23 '20 at 12:43
  • $\begingroup$ @Emo $\|f\|_p \leq (2\pi)^{1/p} \|f\|_{\infty}$. $\endgroup$ – Kavi Rama Murthy Feb 23 '20 at 12:49

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