strong textLet $\phi$ be a $C^{\infty}$ function on $\mathbb R^{n}$ with
$ \operatorname{supp} \phi \subset \{\xi \in \mathbb R^{n}: |\xi|\leq 2, \phi(\xi)=1$ if $|\xi|\leq 1.$
My Question is: How to show $2^{jn}\hat{\phi}(2^{j}x)\to \delta$ in $\mathcal{S'}(\mathbb R^{n})$(= the space of tempered distributions) as $j\to \infty$ where $\delta$ stands for the $\delta-$distribution ?
(A $\delta-$ distribution is , defined by $\left\langle \delta, \varphi \right\rangle = \varphi(0),$ meaning that $\delta$ evaluates a test function at 0.)
Edit: My attempt (after looking the following hint by RE): For $f\in \mathcal{S},$ it is suffice to prove: $|\int_{\mathbb R} [(2^{in} \hat{\phi}(2^{j}x)f(x)-\hat{f}(x)]dx|$ tends to 0 as $j\to \infty.$ Now I guess, we need to use kind of approximate identity but I don't know how ?
Hint: The standard way to prove a limit in ${\mathcal S}'$ is to see how these distributions act on a member of $\mathcal S$. If $f \in \mathcal S$, you want $$ \int 2^{jn} \hat{\phi}(2^j x) f(x)\; dx \to f(0)$$ First step: express the integral on the left in terms of $\phi$ and the inverse Fourier transform of $f$.