Riemann Integral and limits

[dotzo]

Let \(\displaystyle $f$\) be a continuous, non-negative function on the interval \(\displaystyle $[0,1]$\), where \(\displaystyle $M = \sup_{[0,1]} f(x)$\). Prove that \(\displaystyle $\lim_{n\to\infty} \left[\int_0^1 f(x)^n dx\right]^{\frac{1}{n}}\right] = M$\).

I'm really lost as even where to begin. I tried the definition of the limit but I don't know if that even leads anywhere. I know that the limit is less than or equal to M, but I have no clue if that is even helpful. Any clue on where to start would be greatly appreciated.

 

[SlipEternal]

Whenever you have equality, you can break it down into two inequalities. The left side is less than or equal to the right. The right side is less than or equal to the left. Prove both, and you have proven equality.

 

[dotzo]

I thought of this originally but was over-thinking the \(\displaystyle M \le f(x)\), but now I realize that ONLY equality can hold in that case because of the definition of M and the extreme value theorem, but it doesn't have to be presented like that.

I worked it out and I think the proof holds. I essentially built from \(\displaystyle f(x) \le M\) to the limit and said that an analogous conclusion can be drawn from \(\displaystyle M \le f(x)\), guaranteed to be true for some \(\displaystyle x\) by the EVT. Therefore equality must hold.


Thanks for forcing me to rethink a proof that I had originally thrown out.