Let's consider the following formula $$\lim_{n\to\infty}\int_{-\infty}^\infty \phi_n(x) f(x)dx=f(0)$$ where $\phi_n(x)=\frac{1}{n\pi}\frac{\sin^2(nx)}{x^2}$, which doesn't hold for all $f$ (A quick counterexample for $f$ is when $f$ equals $1$ in the rationals and zero in the irrationals, or a simpler is $f=0$ except $f(0)=1$). Does it hold for any continuous function $f$?
I thought about some function like $f(x)=x\sin^2\frac{1}{x}$ (And $f(0)=0$), and computationally it seems to be a counterexample since the integral, if it exists, seems to be greater than 0 (The integrals between $-5$ and $5$, and between $-30$ and $30$ are greater than $0.5$). In that case, I don't know how to prove it's a counterexample. And since the computer struggles to do those calculations there might be an error, or the integral might not exist (It would still be a counterexample in this last case).
Is there any way to do that? Or is there any better (simpler) counterexample?
If f is unbounded the integral on the left may not exist. But for any bounded continuous function f the formula holds: first observe that $\phi_n \geq 0$ and $\int \phi_n =1$. Also $\int_{\{x:|x|>k/n\}} \phi_n (x) dx=\int_{\{|x|>k\}} \frac {sin ^{2} y} {\pi y^{2}}dy <\epsilon$ for all n if $k$ is sufficiently large. Now consider $\int_{\{x:|x| \leq k/n \}}|\phi_n (x)\{f(x)-f(0)\}|dx$. This is same as $\int_{\{y:|y| \leq k \}}|\phi_1 (y)\{f(y/n)-f(0)\}|dy$. Since $|f(y/n)-f(0)| \to 0$ uniformly as $n \to \infty$ for $|y| \leq k$ the result follows.