Computing the limit of an integral 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
 A: 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 $f \equiv 0$.
