# Thread: Limit of definite integrals

1. ## Limit of definite integrals

Okay, I know I've seen this somewhere, but cannot prove it:
Given $\displaystyle f:[a,b]\rightarrow\mathbb{R}$ is continuous and nonnegative. Show that
$\displaystyle \lim_{n\rightarrow\infty}\left[\int_a^b (f(x))^n \; dx \right]^{1/n}=\mbox{max }\{f(x):x\in[a,b]\}$
All help would be greatly appreciated! Thanks!

2. How about an intuitive approach which hopefully might lead to a more rigorous proof?

By the Mean value Theorem:

$\displaystyle \int_a^b f^n(x)dx=(b-a)f^n(M)$

and surely as n goes to infinity, $\displaystyle M$ must necessarily migrate to the maximum value of $\displaystyle f(x)$. Suppose we could prove that part. Then if we take the n'th root of the expression, then $\displaystyle (b-a)^{1/n}\to 1$ in the limit and we're left with $\displaystyle f(M)$ where M tended to the maximum value of f.

I know, it's lame and unfit to be in an Analysis forum: "Fight and you may die, run and you will live . . . for a little while."

3. Hey thanks for the reply, but why M must tend toward the maximum of f I cannot see. I guess my inuition is failing me at the moment. I have gotten everything that you posted here before trying this problem, with the exception of convincing myself that M tends towards the x vaule giving a maximum. I've tried making the M's a function of n in an attempt to get a sequence and show it was at least convergent, and that didn't work either. So I guess, in summary, I really need that one part, but I need to get it rigorously, otherwise I won't be happy with myself... I have a qualifying exam in analysis tomorrow, and this is the one problem that has been giving me problems, and I've just got to solve it... So again, thanks for your help!

4. This is the best I can do for now:

As far as M migrating to the maximum of the function, as we begin raising f to higher and higher powers, it's maximum value will of course dominate all other values of f(x) and as n goes to infinity, this dominance will far exceed all other values of f(x) and thus f(M) will become the major contributing factor towards the value of the integral to the point where all other values of f(x) will have a negligible effect towards the integral and therefore for (b-a)f(M) to equal the integral, M must be at the maximum value. Weak, I know. Personally, if it were my problem, I'd look through some more Real Analysis books in the library. Surely there is a proof in one of them.

I don't think we're gonna' do well on this one ngramjets . . .

5. ## correction

I think there is a mistake in the proof. $\displaystyle f^{n}(x) \neq (f(x))^{n}$. Apply the mean value theorem for integrals and the fact that $\displaystyle [a,b]$ is compact and any continuous function on a compact metric space attains its bounds, ie there maximum and minimum values. This problem is there in the Introductory real analysis book by Bartle and Sherbert.

6. Originally Posted by Chandru1
I think there is a mistake in the proof. $\displaystyle f^{n}(x) \neq (f(x))^{n}$.
Yes. My mistake. Sorry and thanks for the reference. Mine's not a proof but rather a desperate attempt to try at least something: "we don't have to beat them, we just have to fight them".

7. This one too:

Art of Problem Solving Forum

Also, . . . hath?