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Let $f\in C[a,b]$. prove that $$\lim_{n \to \infty} \left(\int ^b_af(x)^n dx\right)^\frac{1}{n}=\sup _{x \in [a,b]} |f(x)|$$

i really have no idea where to start

Arnaud Mortier
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Inverse Problem
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    Wouldn't you have to write there some absolute value(s)? – DonAntonio Mar 06 '18 at 09:54
  • @DonAntonio..I did't get your point sir – Inverse Problem Mar 06 '18 at 09:57
  • There is a result that looks similiar and is about $L^p$-norms. But I think your equality might hold as well. – Yaddle Mar 06 '18 at 09:59
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    Suresh: if for example $;f(x);$ is negative ,that integral is going to get infinitely many times a negative value...and the same is true for positive values. – DonAntonio Mar 06 '18 at 10:02
  • Possible duplicate of Limit of $L^p$ norm – Arnaud Mortier Mar 06 '18 at 10:13
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    @ArnaudMortier This question is much more elementary than the one you link to, which belongs more to advanced calculus or functional analysis, and it would probably confuse the OP. This question can be given in a first undergraduate year course, the other one belongs more towards the end of undergraduate studies. – DonAntonio Mar 06 '18 at 10:16
  • @DonAntonio it is not necessary to understand the terms of this other question to understand the given answer and see that it answers exactly what the OP asks here. – Arnaud Mortier Mar 06 '18 at 10:19
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    @ArnaudMortier I don't agree with you at all: there are terms that for a first yearer could be like chinese and even in the case the OP was curious enough to read about them, that'd take him way too astray from a basic solution as the one given in the answer below, say. – DonAntonio Mar 06 '18 at 10:28

1 Answers1

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I assume $f$ is positive (as stated there is a problem with the left term).

Let $C := \sup_{[a,b]} f(x)$

$\left(\int ^b_af(x)^n dx\right)^\frac{1}{n} ≤ \left(\int ^b_aC^n dx\right)^\frac{1}{n} = C(b-a)^{1/n}$

Now, fix $\epsilon$ and consider $[c,d], d >c$ on which $f(x) > C - \epsilon$

$\left(\int ^b_af(x)^n dx\right)^\frac{1}{n} ≥ \left(\int ^d_cf(x)^n dx\right)^\frac{1}{n} ≥ (C- \epsilon)(d-c)^{1/n}$

To conclude, use $y^{1/n} \to 1$, for all $y > 0$.