# Justifying taking limit into an integral

I want to write the following equality: $$\lim_{h\to0}\left[\int\limits _{a}^{b}\frac{1}{h}\left[g\left(x-y+h\right)-g\left(x-y\right)\right]f\left(y\right)dy\right]=\left[\int\limits _{a}^{b}\lim_{h\to0}\frac{1}{h}\left[g\left(x-y+h\right)-g\left(x-y\right)\right]f\left(y\right)dy\right]$$

But i'm not sure what are sufficient terms for it to be correct and how to properly justify it. In my specific case $f$ is continuous and $g$ is continuously differentiable.

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– Jean-Claude Arbaut Mar 14 '13 at 12:42

If you replace $h$ by $1/n$ the question boils down whether $$\lim_{n\to\infty}\int \phi_n(y)dy = \int \lim_{n\to\infty} \phi_n(y)dy$$ with $\phi_n(y)=n[g(x-y+1/n)-g(x-y)]f(y)$. This is the case when $\phi_n(y)$ converges uniformly to some $\phi(x)$ on $[a,b]$. This is the case if for all $\delta>0$ you can find some $N$, such that $|\phi_n(y)-\phi(y)|<\delta$ for all $n\geq N$ and all $y\in[a,b]$ (note that in contrast to "normal" pointwise convergence the constant $N$ may not depend on $y$).
Nice, for some reason I didn't think about reformulating it as a problem in uniform convergence of a sequence of functions. I was able to show that from the fact $g$ is continuously differentiable the necessary convergence holds. Thanks! – Serpahimz Mar 14 '13 at 15:58