Show that $\lim_{n\rightarrow \infty} \sqrt{\frac{8}{\pi}}\int_0^\infty n\sin(n^2x^2)f(x)dx=f(0)$ 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.
 A: 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:
$$ f_n = F_n' \to H' = \delta$$
A: 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.
