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i'd like very much your help with this one :

given a positive function meaning $$ f(x) \geq 0 $$ and $f$ is continuous.

Let $$ M = \sup(f(x)) $$ where $x$ belongs to $[a,b]$.

How can i prove that $$\lim_{n\rightarrow\infty} \left[\int_a^b (f(x))^n \, dx\right]^\frac{1}{n} = M$$

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marked as duplicate by Guy Fsone, Raskolnikov, Rolf Hoyer, Lee Mosher, Peter Taylor Nov 11 '17 at 14:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ If you are luckier than me, you will find this answered here. Or you can observe that $\leq $ is trivial. For the other direction, show that the lhs is greater that $M(1-\epsilon)$ for every $\epsilon>0$, by use of continuity at a point achieving the sup=max. $\endgroup$ – Julien Apr 26 '13 at 16:43
  • $\begingroup$ To be more precise, you can show that $\limsup\leq M$ and $\liminf \geq M(1-\epsilon)$. $\endgroup$ – Julien Apr 26 '13 at 16:52
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Hints:

1) Recall that for any positive number $\alpha$, one has $$ \lim\limits_{n\rightarrow\infty} \alpha^{1/n}=1.$$ 2) One has the estimate: $$\biggl(\int_a^b (f(x))^n \,dx\biggr)^{1/n}\le \biggl(\int_a^b M^n \,dx\biggr)^{1/n} = (b-a)^{1/n}\cdot M .$$

3) Let $\epsilon>0$. By continuity of $f$, choose a non-degenerate interval $[c,d]$ such that $f(x)\ge M-\epsilon$ for all $x\in [c,d]$. Note $$ \biggl(\int_a^b (f(x))^n\, dx\biggr)^{1/n}\ge \biggl(\int_c^d (M-\epsilon)^n\,dx\biggr)^{1/n} =(M-\epsilon) (d-c)^{1/n}. $$


Be careful not to imply that the limit exists before proving that it indeed does exist. (From 2), you can show the $\limsup$ is at most $M$; and from $3$, you can show that the $\liminf$ is at least $M$.)

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  • $\begingroup$ the first part i understood and know what and why to do. but i dont understand how can i calim that that expression is greater from $$M -\epsilon$$) ? can you please give me more details? $\endgroup$ – Simba Apr 27 '13 at 9:05
  • $\begingroup$ @eyal $(f(x))^n$ is nonnegative, so integrating over a smaller domain gives a smaller quantity. Thus $\int_a^b (f(x))^n\,dx\ge\int_c^d (f(x))^n\,dx$. Since $f(x)\ge M-\epsilon$ for $x\in [c,d]$, you have $\int_c^d (f(x))^n\,dx \ge \int_c^d (M-\epsilon)^n\,dx$. $\endgroup$ – David Mitra Apr 27 '13 at 10:23
  • $\begingroup$ great, thats what i suspected. thank you very much! $\endgroup$ – Simba Apr 27 '13 at 10:38

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