For $x\in \mathbb R$ and $a>0$ let $$\delta_a(x) := \frac{\sin(ax)}{\pi x}.$$ If $f$ is a sufficiently well-behaved test function, then $$\lim_{a\to \infty} \int_{\mathbb R} \delta_a(x) f(x) dx = f(0).$$ I would like to now what one can say about more general regions of integration: For example, I tried to prove $$\lim_{a\to \infty} \int_{-r}^r \delta_a(x) f(x) dx = f(0)$$ for $r>0$. However, the standard method of using the Fourier transform seems not to be working here. Indeed, even if $f\in C_c^\infty(\mathbb R)$, using Fubini and that the Fourier transform of $\delta_a$ is the indicator function of $[-a,a]$, we arrive (up to constants) at $$\int_{-r}^r \delta_a(x) f(x) \, dx = \int_{\mathbb R} \delta_a(x) \left[ \chi_{[-r,r]}(x)f(x) \right] \, dx = \int_{-a}^a \widehat{[\chi_{[-r,r]}f]}(k) dk.$$ But I do not know how $\widehat{[\chi_{[-r,r]}f]}$ behaves. In fact, it could have very poor integrability properties, since the function we are transforming is discontinuous if the support of $f$ is large enough.
So my questions are:
- What are sufficient conditions on $f$ for the limit $a\to \infty$ above to exist?
- If it does not converge, can one at least say that the expression is uniformly bounded in $a$? Thanks in advance!