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How does ont prove Fejer's lemma:

If $f \in L^1(\mathbb{T})$ and $g\in L^\infty(\mathbb{T})$, then $\lim_{n \rightarrow \infty} \int f(t) g(nt) \, dt = \hat{f}(0)\hat{g}(0).$

Michael Hardy
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usere5225321
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  • Approximate f i the $L^1$ by polynomials – usere5225321 Jan 22 '18 at 14:57
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    I think that the following path could bring you to a proof. Fix an arbitrary trigonometric polynomial $f$ and an arbitrary characteristic function $g$ of an interval $(a,b)$. Prove the conclusion using brute force in this case. Then by the $\pi-\lambda$ theorem (or something like that), extend the result for an arbitrary characteristic function $g$ of a Borel-measurable set. Then, by linearity and bounded convergence, extend the result for an arbitrary $g\in L^\infty$ function. Finally, by the density of trigonometric polynomials in $L^1$, extend the result to get the conclusion – Bob Aug 21 '18 at 10:15

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Show that it's enough to prove it assuming $f$ is a trigonomtric polynomial. Figure out the Fourier coefficients of $g(nt)$ in terms of $\hat g(k)$.

David C. Ullrich
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  • C.Ullrich I have seen that continuous functions on thr circle can be uniformly approximated by trigonometric polynomials. So I assume it is enough to show it for $f$ trigonometric. Also, $\hat{g}(nt)=\hat{g} (t/n)$ if t|n and 0 otherwise. Now im stuck – usere5225321 Jan 22 '18 at 18:27
  • @Bach That's not a valid counterexample to the correct statement of what's needed here. If $f$ has period $2\pi$ then in fact $\int_0^{2\pi}f(2x)\sin(7x),dx=0$. Your example does not contradict this because there is no $2\pi$ function $f$ with $f(2x)=2x$ for all $x\in]0.2\pi]$. – David C. Ullrich May 04 '20 at 21:35
  • (where of course "$2\pi$ function" was supposed to be "$2\pi$-periodic function"...) – David C. Ullrich May 04 '20 at 21:41
  • (key to proof: $\sin(7(x+\pi))=-\sin(7x)$.) – David C. Ullrich May 04 '20 at 21:45
  • Its obvious there is no such $f$. We would have $f(2\pi)=2\pi$ and $f(0)=0$, hence $f$ is not $2\pi$-[eriodic. – David C. Ullrich May 05 '20 at 13:43
  • Doesn't the above construction work, you ask. What "construction"? You haven't said what function $f$ you have in mind! – David C. Ullrich May 05 '20 at 13:44
  • @Bach I didn't see that you'd left two comments, sorry. The function you define above does not satisfy $f(2x)=2x$ for every $x\in[0,2\pi]$. – David C. Ullrich May 05 '20 at 14:35
  • @Bach Why do I say$f(2\pi)=2\pi$? Come on now! We're assuming that $f(2x)=2x$ for $x\in[0.2\pi]$. Let $x=\pi$. – David C. Ullrich May 05 '20 at 14:36
  • @Bach Re "I define the values of f on [0,2π], then we can extend f to be periodic on R. Like a sawtooth wave. So what do you want to say? ": I want to say what I've said: Doing this cannot give you a $2\pi$-periodic $f$ with $f(2x)=2x$ on $[0,2\pi]$. – David C. Ullrich May 05 '20 at 14:40