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I just took my real analysis qualifying exam yesterday and this problem showed up:

Prove $F_n(x) = \int_0^x f(\sin(nt))dt$ where $f$ is Riemann integrable on $[-1,1]$ converges uniformly on compact subsets of $[0,\infty)$ and find its limit.

There were two previous parts to the problem, showing that $F_n$ is uniformly Lipschitz and that on every compact set of $\mathbb{R}$ $F_n$ has a uniformly convergent subsequence. I believe I proved those two parts correctly but I had no idea how to do this last part.

Mittens
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Aphyd
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  • I think convergence is uniform in compact subsets of $[0,\infty)$ only since $F_n(x)\xrightarrow{n\rightarrow\infty}x\frac{1}{2\pi}\int^{2\pi}_0 f\circ\sin$. The limit can be found using this posting – Mittens Sep 23 '20 at 20:07
  • What is the argument as for why it converges to $x\frac{1}{2\pi}\int_0^{2\pi}f\circ\sin$? – Aphyd Sep 23 '20 at 20:15
  • Hmmm measure theory was not on the quals syllabus and therefore we were not allowed to use it. – Aphyd Sep 23 '20 at 20:20
  • The actual full question was to show that it converged uniformly on $\mathbb{R}$ but they said they would give full credit for showing it on $[0,\infty)$ – Aphyd Sep 23 '20 at 20:28
  • No, I said I proved that on any compact subset of $\mathbb{R}$ it has a uniformly convergent subsequence. The third part was to prove the entire sequence converged uniformly on $\mathbb{R}$. – Aphyd Sep 23 '20 at 20:32
  • No worries, thanks for your input! – Aphyd Sep 23 '20 at 20:34

1 Answers1

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Hint:

  • $f\circ\sin $ is $2\pi$-periodic, and the integrability of $f$ over $[-1,1]$ implies that $f\circ \sin$ is in integrable over $[0,2\pi]$.

  • The rest is based on this posting: The change of variables $u=n\,t$ and the $2\pi$-periodicity of $f\circ\sin $ gives $$ \begin{align} F_n(x)&=\frac{1}{n}\int^{nx}_0 f\circ\sin (u)\,du=\frac{1}{n}\sum^{[nx/2\pi]}_{k=1}\int^{2\pi k}_{2\pi(k-1)}f\circ \sin +\frac{1}{n}\int^{nx}_{2\pi[nx/2\pi]}f\circ\sin\\ &=\frac{1}{n}\Big[\frac{nx}{2\pi}\Big]\int^{2\pi}_0f\circ\sin + \frac{1}{n}\int^{nx}_{2\pi[nx/2\pi]}f\circ\sin \end{align} $$ The term $E_n=\int^{nx}_{2\pi[nx/2\pi]}f\circ\sin$ is bounded: $|E_n|\leq 2\pi \sup\{|f(x)|:x\in[-1,1]\}$ independently of $x$. Hence $\lim_nF_n(x)=x\frac{1}{2\pi}\int^{2\pi}_0 f\circ\sin$.

With a little extra effort one can tweak the arguments above and show that convergence is indeed uniform in compacts sets of $[0,\infty)$.

Martin Sleziak
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Mittens
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  • Thank you. Perhaps one would then use this result to prove it converges uniformly on [0,infty). I'll give it a try. – Aphyd Oct 05 '20 at 06:06