Consider the following integral $$ \int_{-\infty}^{\infty}f(t) K(\frac{a-t}{h})dt $$ where
(1) $h>0$, $a \in \mathbb{R}$
(2) $f:\mathbb{R}\rightarrow[0,\infty)$ is such that $\int_{-\infty}^{\infty}f(t)dt=1$
(3) For fixed $a$ and $h$, the map $K:\mathbb{R}\rightarrow[0,\infty)$ is such that $\int_{-\infty}^{\infty}K(\frac{a-t}{h})dt<\infty$
(4) For fixed $a$ and $t$, $\lim_{h\rightarrow 0}K(\frac{a-t}{h})= \begin{cases} 1 \text{ if $t=a$}\\ 0 \text{ otherwise} \end{cases}$
Could you help me to show that $$ \lim_{h\rightarrow 0} \int_{-\infty}^{\infty}f(t) K(\frac{a-t}{h})dt=f(a) $$ also stating under which sufficient conditions? The difficult part for me is bringing the limit inside the integral. Any hint would be really appreciated
The only way this can be true for bounded $f$ is when $f \equiv 0$.
Letting $a = 0, h = 1$ in (3) gives that $\int_{-\infty}^\infty K(t)dt = \int_{-\infty}^\infty K(-t)dt = M$ for some $M < \infty$. So, making the substitution $u =\frac{a-t}h$, we get: $$\int_{-\infty}^\infty K\left(\frac{a-t}h\right) dt = h\int_{-\infty}^\infty K(u)du = Mh$$
If $f \le N$ everywhere, then $$ 0 \le \int_{-\infty}^\infty f(t)K\left(\frac{a-t}h\right) dt \le \int_{-\infty}^\infty NK\left(\frac{a-t}h\right) dt = NMh$$
By the squeeze theorem, $$ \lim_{h\to 0^+}\int_{-\infty}^\infty f(t)K\left(\frac{a-t}h\right) dt = 0.$$
Somewhat more generally, if we assume that both $f^2$ and $K^2$ have finite integrals, then it follows that $\int_{-\infty}^\infty K^2\left(\frac{a-t}h\right) dt = Ah$ for some $A$ by the same reasoning as above. By the Cauchy-Schwarz inequality, $$ 0 \le \int_{-\infty}^\infty f(t)K\left(\frac{a-t}h\right) dt \le \sqrt{\int_{-\infty}^\infty f^2(t)dt \int_{-\infty}^\infty K^2\left(\frac{a-t}h\right) dt} = B\sqrt h$$ for some B. Therefore once again the limit of the integral must be $0$.
So necessary conditions for this to hold for $f \not\equiv 0$ are that $f$ is unbounded, and at least one of $f^2$ or $K^2$ is not square-integrable over $\Bbb R$.
The only sufficient condition I've spotted for this to hold is when $f \equiv 0$.