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Assume $f$ is compactly supported smooth function. Show that $\lim_{n\rightarrow \infty} \sqrt{\frac{8}{\pi}}\int_0^\infty n\sin(n^2x^2)f(x)dx=f(0)$. Use the fact that $\lim_{a\rightarrow\infty}\int_0^a\sin(t^2)dt=\sqrt{\frac{\pi}{8}}$.

Here is what I have tried so far. Let $\int sin(t^2)dt=G(t)$. Then from integration by parts.

$$\begin{align} \int_0^t n\sin(n^2x^2)f(x)dx&=\int_0^{nt}\sin(x^2)f(\frac{x}{n})\\ &=G(nt)f(t)-G(0)f(0)-\frac{1}{n}\int_0^{nt}G(x)f'(\frac{x}{n})dx\\ \end{align}$$

Now here is the problem. I'm trying to maniuplate the last equation so that I have $G(t)f(0)$ in the last line so that I can use the hint, but so far I am unable to. Could anyone tell me what I should try because I'm stuck to be honest.

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  • $\begingroup$ No need for integration by parts. Use continuity of $f$. $\endgroup$ – abnry Sep 2 '17 at 3:40
  • $\begingroup$ Could you expand on that a bit? If that's true then the problem would be redundant in assuming $f$ is smooth right. I mean differentiability has to be used somewhere, right? $\endgroup$ – thegamer Sep 2 '17 at 3:43
  • $\begingroup$ Can't elaborate now, but often "f is compactly supported and smooth" is written into a problem to say "f is as nice as possible, don't worry about the details about f." $\endgroup$ – abnry Sep 2 '17 at 3:46
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The idea is to use Taylor's Theorem. So, assume that $\alpha\geq 0$ is such that $f(x)=0$ (and hence $f'(x)=0$) for $x>\alpha.$ This $\alpha $ exists because of the compact support condition.

Then we must find

$$\lim_{n\to \infty} \sqrt{\frac{8}{\pi}}\int_0^{\alpha}n \sin(n^2x^2)f(x)dx. $$

Now, apply the change of variable $t=nx.$ Then, the problem is equivalent to find

$$\lim_{n\to \infty} \sqrt{\frac{8}{\pi}}\int_0^{n\alpha}\sin(t^2)f\left(\frac{t}{n}\right)dt. $$ By the mean value theorem, for each $t >0,$ there exists $\gamma(t)\in [0, \frac{t}{n}]$ such that

$$f\left(\frac{t}{n}\right)=f(0)+ f'(\gamma(t))\frac{t}{n}.$$ Putting this into the integral we get

$$\lim_{n\to \infty} \sqrt{\frac{8}{\pi}}\int_0^{n\alpha}\sin(t^2)\left(f(0)+ f'(\gamma(t))\frac{t}{n}\right)dt=$$
$$\lim_{n\to \infty} \sqrt{\frac{8}{\pi}}\int_0^{n\alpha}\sin(t^2)f(0)dt + \lim_{n\to \infty} \sqrt{\frac{8}{\pi}}\int_0^{n\alpha}\sin(t^2)f'(\gamma(t))\frac{t}{n}dt=$$ $$f(0)+ \lim_{n\to \infty} \sqrt{\frac{8}{\pi}}\int_0^{n\alpha}\sin(t^2)f'(\gamma(t))\frac{t}{n}dt.$$ In order to finish the proof, it suffices to show that the last limit is $0.$

Since $f$ is smooth, we have that $f'$ is continuous and hence there exists a constant $M$ such that

$$|f'(x)|\leq M \;\forall \;x\in [0,\alpha].$$ Now note that since $\gamma(t)\leq \frac{t}{n},$ we must have $f'(\gamma(t))\leq M$ for $t\leq n\alpha.$ Now,

$$\left|\int_0^{n\alpha}\sin(t^2)f'(\gamma(t))\frac{t}{n}dt\right|\leq \int_0^{n\alpha}\left|\sin(t^2)f'(\gamma(t))\frac{t}{n}\right|dt \leq \frac{M}{n} \int_0^{n\alpha} t |\sin(t^2)|dt \leq \frac{MK}{n} $$ for some constant $K$.(I leave this detail to you. Hint: $\int t \sin(t^2)dt= \frac{1}{2}(-\cos(t^2))+C$) This proves that the second limit is $0$, and the result follows from this.

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  • $\begingroup$ Nice! Thanks Magnusseen. $\endgroup$ – thegamer Sep 2 '17 at 19:40
  • $\begingroup$ Nice exposition, but there is a mistake in the last few lines: $\int_0^{na} t|\sin(t^2)|\, dt$ is not bounded. Far from it, it's on the order of $n^2$ as $n\to\infty.$ $\endgroup$ – zhw. Sep 2 '17 at 22:57
  • $\begingroup$ @zhw thanks for pointing that out. I am working to prove the last part avoiding the module operation. Please feel free to complete the proof if you find it. $\endgroup$ – John D Sep 2 '17 at 23:51
  • $\begingroup$ @zhw. So far nothing. I think that Stone- Weierstrass should work here, but I am out of time. It is better if we delete this post. $\endgroup$ – John D Sep 3 '17 at 8:16
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This is a direct consequence of the following Lemma about delta sequences:

Lemma: If $f\in L^1_\text{loc}(\mathbb R)$ with $\int_{-\infty}^{+\infty} f(x) dx = 1$ then $\lim_{n\to\infty} nf(nx) = \delta(x)$

[Note that in the $d$-dimensional case it is required that $f\ge 0 $ to ensure $\lim_{n\to\infty} n^d f(nx) = \delta(x)$]

The proof is surprisingly simple with a bit of background in distribution theory: observe that

$$ F_n(x) := \int_{-\infty}^{x} nf(n\zeta)\,d\zeta = \int_{-\infty}^{xn} f(\zeta)\,d\zeta $$

converges point-wise against the Heaviside step function $H$ as $n\to\infty$. Moreover $F_n(x)$ is uniformly bounded by $\sup_{s\in\mathbb R} \int_{-\infty}^{s} f(\zeta)\,d\zeta$, hence by the dominated convergence theorem $F_n \to H$ in $L^1(\Omega)$ for any compact $\Omega\subset \mathbb R$ and so also $F_n\to H$ in $L^1_\text{loc}(\mathbb R)$. Consequently: $$ f_n = F_n' \to H' = \delta$$

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