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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: enter image description here

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 $\|T_n\| \leq \|D_n\|_1$.

Thanks for your help.

  • 3
    I suspect that $X$ is the continuous functions on $[0,1]$ with the $\sup$-norm.2012-08-10
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    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?2012-08-10
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    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 $0$?2012-08-10
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    Please provide us with the reference to this article/textbook2012-08-10
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    @nullUser Since $L^\infty \subset L^2$, your suggestion cannot be true. Otherwise, $L^\infty = L^2$, with equivalent norms.2012-08-10
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    I believe this is a fairly standard exercise in applying the Banach Steinhaus theorem, see Rudin's "Real & Complex Analysis", Theorem 5.12 (or rather the discussion beforehand).2012-08-10
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    If $D_n$ denotes Dirichlet kernel, then $D_n$ is bounded, hence in $L^2$ and the norm of $T_n$ is $\lVert D_n\rVert_{L^2}$.2012-08-10
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    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.2012-08-10
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    @t.b. Yes, to everything except $X = C[0,1]$. I am trying to compute its norm. We're computing Fourier series of $L^2$ functions so $T_n$ should be linear functionals $L^2 \to \mathbb R$. No?2012-08-11
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    Dear @Norbert, the lecture notes can be found here: http://www.math.ethz.ch/~einsiedl/FA-lecture.pdf However, I am using the previous version of it.2012-08-11
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    Dear @t.b. Does this work as follows: prove lemma for dense subspace then continuously extend the operator to all of $L^2$? ($C$ is dense in $L^2$ w.r.t. $\|\cdot\|_2$ ) That would answer my question. I'm very sorry for not being clear enough in the OP : (2012-08-11
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    I don't think $X = L^2$ is intended here. The displayed excerpt gives a bound for the functional $T_N \colon f \mapsto \int f D_N$ with respect to *something equipped with the* $\sup$-*norm*. You can define this functional on every $L^p$ (by extending continuously, but then you'd have to use Hölder (or Cauchy-Schwarz for $p=2$) in the form $\lvert \int f D_N \rvert \leq \lVert f \rVert_p \rVert D_n \rVert_q$. The truncated Fourier series is computed by $$\sum_{N}^N \hat{f}(k) e^{2\pi i\, k\, y} = S_N(f)(y) = (f \ast D_N)(y) = \int f(y-x) D_N(x)\,dx.$$ (a convolution), not by $T_N(f)$.2012-08-11
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    @t.b. I know how the truncated Fourier series is computed by convolution with $D_N$ but the functional is $f \mapsto D_N \ast f (0)$. I'd understand the proof if $X$ was indeed a space equipped with the $\sup$-norm. But we get Fourier series for $f \in L^2$. So I expect the operator to be defined on $L^2$. Did you see my previous comment about extending $T_n$ to all of $L^2$?2012-08-11
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    Yes, I saw that comment. Hence the part about $L^p$ and Hölder. Do they mention $L^2$ *anywhere* in that proof: Does an $L^2$-norm appear anywhere in the subsequent argument? Otherwise there's no reason to think about $L^2$ here (and if it's about divergence of Fourier series then only $C[0,1]$ matters). Is boundedness or continuity of $f$ mentioned or used later on? Note: Fourier stuff is not confined to $L^2$, you can define the FT on about every sensible function space on $[0,1]$ (sensible includes invariant under shifts $f \mapsto f(\cdot-x)$).2012-08-11
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    @t.b. No there is no $L^2$ mentioned anywhere. While I can compute a Fourier series of almost any function, the thing I want is $L^2$-convergence of that series. And for that I want my $f$ to be in $L^2$. The proof of existence of a continuous $f$, which this $T_n$ is used in, asserts that while we *might* have $L^2$-convergence, there are $f$ in $L^2$ for which we don't have pointwise convergence. Now I thought that I need $T_n$ to be a functional defined on $L^2$. Then I realised that *maybe* we can show all the properties for $C(X)$ and then I got stuck because the $\sup$-norm was used.2012-08-11
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    But that's most definitely not what that lemma is about. If you want to prove convergence of the $N$th truncated Fourier approximation $S_N f \to f$, you'd need to estimate $\lVert f - S_N(f) \rVert = \lVert f - D_N \ast f \rVert = $ in the relevant norm. Only the evaluation of $S_N(f)$ at $0$ is discussed in that lemma. The functional $T_n$ is already defined on all $L^p$'s, since $D_n \in L^\infty \subset L^q$, no need to extend continuously.2012-08-11
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    @t.b. My confusion was completely beside the point: I can endow $C(\mathbb T)$ with any norm, in particular with $\|\cdot\|_\infty$.2012-08-11
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    @t.b. Thank you very much.2012-08-11

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