# A question about operator norm

I hope this is not an obviously stupid question, I'm quite tired and hence extra slow today. I don't understand the following proof:

Given the context I think $X = L^2 (\mathbb T)$, so $T_n : L^2 (\mathbb T) \to \mathbb R$. (The context is: Fourier series and uniform boundedness principle)

I don't see how we get the last line from the penultimate one. The operator norm is the $\sup$ over $f$ with norm equal to $1$. But the domain is $L^2$ so it comes with the $L^2$ norm. I'm aware that $\|f\|_1 \leq \|f\|_2$ and that I can apply Hölder to either get $\|T_n\| \leq \|f\|_2 \|D_n\|_2 = \|D_n\|_2$ or $\|T_n\| \leq \|f\|_\infty \|D_n\|_1$. But I want $\|T_n\| \leq \|D_n\|_1$.

I suspect that $X$ is the continuous functions on $[0,1]$ with the $\sup$-norm. – copper.hat Aug 10 '12 at 16:31
I'm not very knowledgeable on this topic, but looking at the wiki suggest that $\Vert f \Vert_\infty \leq \Vert f \Vert_2$ since $\Vert f \Vert_{p+a} \leq \Vert f \Vert_p$ and $\Vert f \Vert_\infty = \lim_{p \to \infty}\Vert f \Vert_p$. Does this not give the desired result? – nullUser Aug 10 '12 at 16:32
I am guessing $D_n$ is (are?) the Dirichlet kernel, and the goal is to show the existence of a continuous function whose Fourier series diverges at $0$? – copper.hat Aug 10 '12 at 16:35
This looks like a part of the proof that there is a continuous function whose Fourier series diverges. Each Dirichlet kernel $D_n$ gives you a linear functional $T_n$ on $X = C[0,1]$ (I suppose) simply by integrating against it and the Lemma is about computing its norm as the $L^1$-norm of $D_n$. The next step will probably be to show that $\lVert D_n \rVert_1 \xrightarrow{n\to\infty} \infty$, then divergence of the Fourier series at $0$ using uniform boundedness and the fact that $S_N(0) = \int f D_n$, then strengthen that like Banach-Steinhaus did in their original paper, etc. – t.b. Aug 10 '12 at 22:47
The domain of $T_n$ cannot be $L^2$ for the proof to work. There are $f\in L^2$ such that $\|f\|_{\infty}=\infty$ and then the proof is flawed.