Show that there is $f\in \mathcal C^0(S^1)$ s.t. $\lim_{n\to \infty }\|S_nf-f\|_{L^\infty }\neq 0$.

Question

I have to show that there is $f\in \mathcal C^0(S^1)$ s.t. $$\lim_{n\to \infty }\|S_nf-f\|_{L^\infty }\neq 0.$$ The proof goes as follow : we know that $\|D_n\|_{L^1}\geq c\log(n)$ where $D_n$ is the Dirichelet kernel. Therefore, $$\sup_{\|f\|_{L^\infty }\leq 1}|D_n*f(0)|=\|D_N\|_{L^1}\geq c\log(n).$$

I just don't understand why $$\sup_{\|f\|_{L^\infty }\leq 1}|D_n*f(0)|=\|D_N\|_{L^1}.$$

Can you compute $D_n * f(0)$ ? Remember $\sup |f(x)| \leq 1$. — , Jul 07 '16 at 15:47
@N.H.: thank you for your very constructive answer, I got it :) — user349449, Jul 07 '16 at 15:50
@N.H. In fact, should it be $\sup_{|f|{L^\infty }\leq 1}|D?*f(0)|\leq |D_N|_{L^1}$ ? — user349449, Jul 07 '16 at 15:52
No, just that inequality would not be enough; we need to show that $\sup_fD_n*f(0)|\to\infty$ as $n\to\infty$, and that does not follow from the inequality in that direction. — David C. Ullrich, Jul 07 '16 at 15:57
@DavidC.Ullrich: I know the inequality is not enough. I understand $\leq$ but not $\geq$. Could you give me more explanation ? — user349449, Jul 07 '16 at 15:58

score 0 · Accepted Answer · answered Jul 07 '16 at 16:04

You've seen one inequality is obvious but we need the other direction. You can get this directly, constructing $f$ with $|f|\le1$ such that $\int fD_n$ is close to $||D_n||_1$. (Draw a picture of $D_n$ and imagine what $f$ should look like; you want $f=1$ on "most" of the set where $D_n>0$ and $f=-1$ on most of the set where $D_n<0$.)

Or you can get it from some fancy theorems. The Riesz Representation Theorem says that the norm of a complex measure $\mu$ is equal to the supremum of $|\int f\,d\mu|$ for $f$ continuous with $||f||_\infty\le1$, and something or other shows that the norm of $D_n$, regarded as a complex measure, is just $||D_n||_1$.

Show that there is $f\in \mathcal C^0(S^1)$ s.t. $\lim_{n\to \infty }\|S_nf-f\|_{L^\infty }\neq 0$.

1 Answers1