On limit of convolution of function with a summability kernel

[psie]

TL;DR Summary

I'm stuck at proving a corollary regarding the limit of a convolution with a positive summability kernel and an arbitrary function.

I'm reading the following theorem in Fourier Analysis and its Applications by Vretblad.

Theorem 2.1 Let ##I=(-a,a)## be an interval (finite or infinite). Suppose that ##(K_n)_{n=1}^\infty## is a sequence of real-valued, Riemann-integrable functions defined on ##I##, with the following properties:
1) ##K_n(s)\geq 0##,
2)##\int_{-a}^a K_n(s)ds=1##, and
3) if ##\delta>0##, then ##\lim\limits_{n\to\infty}\int_{\delta<|s|<a} K_n(s)ds=0.##
If ##f:I\to\mathbb{C}## is integrable and bounded on ##I## and continuous for ##s=0##, we then have $$\lim_{n\to\infty}\int_{-a}^a K_n(s)f(s)ds=f(0).$$

Corollary 2.1 If ##(K_n)_{n=1}^\infty## is a positive summability kernel on the interval ##I##, ##s_0## is an interior point of ##I##, and ##f## is continuous at ##s=s_0##, then $$\lim_{n\to\infty}\int_I K_n(s)f(s_0-s)ds=f(s_0).$$

The proof is left as an exercise (do the change of variables ##s_0-s=u##.

It's silly, but I'd like to prove the corollary and I'm getting stuck. I'm a little unsure if ##I## in the corollary is also of the form ##(-a,a)##. Moreover, the change of variables as suggested gives us ##s=s_0-u##, so ##K_n(s)## becomes ##K_n(s_0-u)##. Is this a kernel still centered at ##0##? If I'm understanding things right, the author alludes to using $$\lim_{n\to\infty}\int_{-a}^a K_n(s)f(s)ds=f(0),$$ from theorem 2.1 to prove the corollary. Appreciate any help.

 

[pasmith]

[itex]I[/itex] is throghout assumed to be of the form [itex](-a,a)[/itex].

Easier than substitution is to define [itex]g(s) \equiv f(s_0 - s)[/itex] so that [itex]g(0) = f(s_0)[/itex] and apply the theorem to [itex]g[/itex].

 

[psie]

Thank you @pasmith.

1. Do you know if positive summability kernels (i.e. a sequence of functions satisfying 1), 2) and 3) above) are even functions?
2. It looks to me that if ##K_n(s)## is a positive summability kernel, then so is ##K_n(-s)##. Is this right?

When we make the substitution ##u=-s## in the integral in the corollary, we obtain $$\int_I K_n(-u)f(s_0+u)du,$$ where ##I=(-a,a)## remains unchanged. If ##K_n(-u)=K_n(u)## or if ##K_n(-u)## is also a positive summability kernel over ##I##, and we set ##s_0=0##, then we can apply the theorem.