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I came across a problem and I really don't know where to start from. It states that:

$$\lim_{n\rightarrow\infty} \left(\int_a^b f(x)^n dx\right)^{1/n} = \sup \{ f(x) :a \leq x \leq b \}$$

with $f : [a,b] \rightarrow \mathbb R $ being a continuous , nonnegative function.

I tried using the median value theorem, saying that:

$\int_a^b f(x)^n dx = f(\xi)^n(b-a)$, for some $\xi \in [a,b]$, and then concluded that the limit was $f(\xi)$.

However, I couldn?t find any other relationship between it and the other values of $f$.

I also tried using the definition of the supremum of a set , but I can't even prove that it is an upper bound.

Any help would be helpful

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  • $\begingroup$ If you know some measure theory take a look at this math.stackexchange.com/questions/242779/limit-of-lp-norm .You do not need continuity here, but there might be a simpler proof if you impose it. $\endgroup$ – Dayton Aug 23 '19 at 21:55
  • $\begingroup$ No, unfortunately I don't know much about it $\endgroup$ – Adolfo Garcia Aug 23 '19 at 22:18
  • $\begingroup$ This thing has appeared on MSE a zillion times. $\endgroup$ – zhw. Aug 24 '19 at 1:48
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Let $M=\sup \{ f(x) :a \leq x \leq b \}$. Then clearly, $$\left(\int_a^b f(x)^n\,dx\right)^{1/n}\leq M(b-a)^{\frac1n}\to M.$$ Since $f$ is continuous, there exists $x_0\in[a,b]$ such that $f(x_0)=M$ and for any $0<\epsilon<M$ there exists $(c,d)\subset[a,b]$ such that $f(x)>M-\epsilon$ for $x\in(c,d)$. Hence $$\left(\int_a^b f(x)^n\,dx\right)^{1/n}\geq \left(\int_c^d f(x)^n\,dx\right)^{1/n}\geq (M-\epsilon)(d-c)^{\frac1n}\to M-\epsilon.$$ Therefore $$\lim_{n\rightarrow\infty} \left(\int_a^b f(x)^n dx\right)^{1/n} = M=\sup \{ f(x) :a \leq x \leq b \}.$$

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  • $\begingroup$ It should be $M(b-a)^{1/n}$ in your first inequality. The proof however, still works. $\endgroup$ – uniquesolution Aug 23 '19 at 22:41
  • $\begingroup$ @uniquesolution Thanks! Fixed now. $\endgroup$ – Feng Shao Aug 23 '19 at 22:51

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