On limit of convolution of function with a summability kernel

[psie]

TL;DR

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]

I is throghout assumed to be of the form (-a,a).

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

 

[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.