Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $f:\mathbb{R}\to\mathbb{R}$ be Borel and bounded, then I was able to prove that the map $t\mapsto \int_{t-h}^tf(s)ds$ is Lipschitz continuous. Now if we assume in addition that $f$ is left-continuous then

$$\lim_{h\downarrow 0}\frac{1}{h}\int_{t-h}^tf(s)ds=f(t) $$

What I've done: let $r:=\frac{1}{h}(s-t)+1$ such that

$$\lim_{h\downarrow 0}\frac{1}{h}\int_{t-h}^tf(s)ds=\lim_{h\downarrow 0}\int_0^1f(h(r-1)+t)dr$$

Since $f$ is bounded I can use Dominated convergence to obtain

$$\int_0^1\lim_{h\downarrow 0}f(h(r-1)+t)dr=f(t)\int_0^1dr=f(t)$$

Is this correct?

However, I don't see the importance of left-continuity. Wouldn't be the same true for a right continuous $f$ and changing the integral boundaries:

$$\lim_{h\downarrow 0}\frac{1}{h}\int_{t}^{t+h}f(s)ds=f(t)$$

share|improve this question
    
It would be the same, with $h\downarrow 0$ in your last formula. –  Etienne Feb 28 at 10:55
    
@Etienne thanks for the comment. edited the last line. –  math Feb 28 at 16:12

1 Answer 1

up vote 1 down vote accepted

I think that using Dominated convergence theorem is an overkill; my preference is to use more rudimentary tools when possible. Fix $t$. By the definition of left-continuity, for every $\epsilon>0$ there is $\delta>0$ such that $|f(s)-f(t)|<\epsilon$ whenever $s\in (t-\delta,t)$. Therefore, for $0<h<\delta$ we have $$ \frac{1}{h}\int_{t-h}^t f(s)\,ds \le \frac{1}{h}\int_{t-h}^t (f(t)+\epsilon)\,ds = f(t)+\epsilon $$ and similarly $$ \frac{1}{h}\int_{t-h}^t f(s)\,ds \ge \frac{1}{h}\int_{t-h}^t (f(t)-\epsilon)\,ds = f(t)-\epsilon $$ By the definition of a limit, $\lim_{h\downarrow 0 }\frac{1}{h}\int_{t-h}^t f(s)\,ds =f(t)$.

share|improve this answer
    
Thanks for your answer. very nice solution –  math Mar 1 at 9:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.