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Show that $\sup_{x\in [a,b]} f(x) = \lim_{n\rightarrow\infty} (\int_a^b (f(x))^n \;dx)^{\frac{1}{n}}$, for f continuous and positive on [a,b]. I can show that LHS is greater than or equal to RHS but I can't show the other direction.

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up vote 5 down vote accepted

Let $F = \sup_{\xi\in[a,b]} f(\xi)$. If $F=0$, the equality is trivial. So we may suppose suppose $F>0$.

Choose $\epsilon \in (0,F)$, and let $A_\epsilon = \{x | f(x) \geq F-\epsilon\}$. Since $f$ is continuous, we have $mA_\epsilon > 0$ ($m$ is the Lebesgue measure). Then the following is true (work from the middle towards the left or right to get the relevant bound):

$$mA_\epsilon (F-\epsilon)^n \leq \int_{A_\epsilon} (F-\epsilon)^n dx \leq I^n_n = \int_a^b (f(x))^n dx \leq \int_a^b F^n dx \leq (b-a) F^n$$

This gives $\sqrt[n]{mA_\epsilon} (F-\epsilon) \leq I_n \leq \sqrt[n]{b-a}F$.

We have $\lim_{n \to \infty}\sqrt[n]{mA_\epsilon} = 1$ and $\lim_{n \to \infty}\sqrt[n]{b-a} = 1$, hence $\limsup_{n \to \infty} I_n \leq F$ and $\liminf_{n \to \infty} I_n \geq F-\epsilon$. Since this is true for $\epsilon$ arbitrarily close to $0$, we have $\liminf_{n \to \infty} I_n \geq F$, from which the desired result follows.

(Note: If you wish to avoid the Lebesgue measure, note that $A_\epsilon$ could be taken to be any non-empty interval such that $f(x) \geq F-\epsilon$ for $x$ in this interval. Then $mA_\epsilon$ would be replaced by the length of this interval. The same reasoning still applies.)

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