I am trying to prove the following but even if it seems easy I am not sure how to start:

Let $R$ be a rectangle in the plane and $f\colon R\rightarrow {\mathbb{R}}$ a continuous non-negative function. Then the following holds:

$$\left(\iint_R f^n\right)^{\dfrac{1}{n}} \rightarrow \sup \left(f(x)\right)\text{ as }n\longrightarrow \infty.$$

Any hints will help. Thank you


Can you prove that the sequence $n\to (\int_{R} f^n)^{\frac{1}{n}}$ is bounded by $\sup_{x\in R} f(x)$ for all natural numbers $\mathbb{N}$?

The other direction is to show that for each $\epsilon > 0$, there exists $N$ sufficiently large such that $(\int_{R} f^n)^{\frac{1}{n}} > \sup_{x\in R} f(x) - \epsilon$ for all $n\geq N$.

Hint: Define $A_{\epsilon}=\{x\in R:f(x)>\sup_{x\in R} f(x) - \epsilon\}$. What can you say about $(\int_{A_{\epsilon}} f^n)^{\frac{1}{n}}$ for sufficiently large $n$?

Hope this helps!

  • $\begingroup$ The second part is unclear how to prove. $\endgroup$ – Whats My Name Apr 26 '14 at 10:19
  • $\begingroup$ Dear @WhatsMyName (it is funny to address you this way!), I have added a hint. Does that help? I would be very happy to elaborate further if you are still stuck after thinking about the hint. $\endgroup$ – Amitesh Datta Apr 26 '14 at 10:20
  • $\begingroup$ But if f has values less than 1 can we conclude the second part? Also since f is continuous I guess the sup f(x)= Max? $\endgroup$ – Whats My Name Apr 26 '14 at 10:25
  • $\begingroup$ Dear @WhatsMyName, yes the key point (as you may see if you follow my hint) is that for any positive number $c$ (whether less than $1$ or not), $\lim_{n\to\infty} c^\frac{1}{n}=1$. I believe this limit value is the main reason the result is true. Also, if the rectangle $R$ is a closed (hence, compact) rectangle, then $\sup_{x\in R} f(x)=\max_{x\in R} f(x)$; this would not necessarily be true if $R$ were an open rectangle. (However, the result is independent of whether or not $R$ is an open or closed rectangle and is also, in fact, independent of $R$ being a rectangle - $R$ could be anything.) $\endgroup$ – Amitesh Datta Apr 26 '14 at 10:29
  • $\begingroup$ Now it is clear indeed. thank you $\endgroup$ – Whats My Name Apr 26 '14 at 10:35

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