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Given the operator $F_n : C(\mathbb{T}) \rightarrow \mathbb{C}, f \mapsto S_nf (0)$ where $C(\mathbb{T})$ are continuous functions on $\mathbb{T} = (-\pi, \pi)$ and $S_n$ is the Fourier Sum operator $S_n f(x) := \frac{1}{\sqrt{2\pi}} \sum_{k=-n}^n \hat{f}(k) e^{i k x}$, I want to calculate the operator norm $\|F_n\|$.

I already got that $|F_n f| = |S_n f(0)| = \frac{1}{2\pi} | \int_{-\pi}^\pi D_n(x) f(x) dx| \leq \frac{1}{2\pi} \|D_n\|_1 \|f\|_\infty$ with the Dirichlet kernel $D_n(x) := \sum_{k=-n}^n e^{ikx}$.

So, $\|F_n\| \leq \frac{1}{2\pi} \|D_n\|_1$.

How do I now get equality? I tried letting $f(x):=\frac{\overline{D_n(x)}}{|D_n(x)|}$ which would work apart from the fact that this f is not continuous.

I would appreciate any help and ideas, thanks.

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1 Answer 1

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Fix $\def\e{\varepsilon}\e>0$. Since $f(x)=\frac{\overline{D_n(x)}}{|D_n(x)|}$ is measurable, by Lusin's Theorem there exists $g\in C(\def\TT{\mathbb T}\TT)$ and $E\subset\TT$ with $\|g\|_\infty=1$, $g=f$ on $E$ and $m(E^c)<\e$. Then$\def\abajo{\\[0.3cm]}$ \begin{align} 2\pi|F_ng|&≥\int_E|D_n|-\int_{E^c}D_n g≥\int_E|D_n|-\e\|D_n\|_1\abajo &=\|D_n\|_1-\int_{E^c}|D_n|-\e\|D_n\|_1\abajo &≥\|D_n\|_1-m(E^c)^{1/2}\|D_n\|_2-\e\|D_n\|_1\abajo &≥\|D_n\|_1-\e^{1/2}\|D_n\|_2-\e\|D_n\|_1. \end{align} As $\e$ was arbitrary, this shows that $$ \|F_n\|≥\frac1{2\pi}\|D_n\|_1. $$

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