Does $\lim_{x\to0}f(x)=L$ imply convergence of $\lim_{h\to0}\frac{1}{h}\int_{-h}^{h}f(x)dx$?

Question

Let $f:[-1,1]\to \mathbb{R}$ with $\lim_{x\to0}f(x)=L<\infty $. Does this imply convergence of $\lim_{h\to0}\frac{1}{h}\int_{-h}^{h}f(x)dx$?

I thought we can apply L'Hospital to $\lim_{h\to0}\frac{\int_{-h}^{0}f(x)dx}{h}$ and $\lim_{h\to0}\frac{\int_{0}^{h}f(x)dx}{h}$ and with the fundamental theorem we'd have $$\lim_{h\to0}\frac{1}{h}\int_{-h}^{h}f(x)dx=2L$$

Is this correct?

Answer 1

Yes, you are correct. This is a direct approach without L'Hopital and just assuming that $f$ is integrable in a neighbourhood of $0$ (no need of continuity).

Since $\lim_{x\to0}f(x)=L\in\mathbb{R}$, by definition of limit it follows that for a given $\epsilon>0$ there is $h>0$ such that for $|x|\leq h$, $$L-\epsilon\leq f(x)\leq L+\epsilon.$$ Hence, $$2L-2\epsilon=\frac{1}{h}\int_{-h}^{h}(L-\epsilon)dx\leq \frac{1}{h}\int_{-h}^{h}f(x)dx\leq \frac{1}{h}\int_{-h}^{h}(L+\epsilon)dx=2L+2\epsilon$$ which implies that $$\left|\frac{1}{h}\int_{-h}^{h}f(x)dx-2L\right|\leq 2\epsilon.$$

Answer 2

Since you make a reference to “fundamental theorem” (of Calculus, I suppose), then my guess is that you are assuming that $f$ is continuous. Then, by the definition of derivative,$$\lim_{h\to0}\frac{\int_0^hf(x)\,\mathrm dx}h=f(0)=L$$and$$\lim_{h\to0}\frac{\int_{-h}^0f(x)\,\mathrm dx}h=-\lim_{h\to0}\frac{\int_0^{-h}f(x)\,\mathrm dx}h=f(0)=L.$$Therefore,$$\lim_{h\to0}\frac{\int_{-h}^hf(x)\,\mathrm dx}h=2L.$$