$f:[a,b] \to [0, \infty)$ continuous , then $\lim_{n \to \infty} \Bigg(\int_a^b \big(f(x)\big)^ndx \Bigg)^{1/n}=\sup \{f(x):x \in [a,b]\}$ ?

Asked Aug 31 '15 at 13:54

Active Aug 31 '15 at 13:54

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Let $f:[a,b] \to [0, \infty)$ be continuous , then is it true that $\lim_{n \to \infty} \Bigg(\int_a^b \big(f(x)\big)^ndx \Bigg)^{1/n}=\sup \{f(x):x \in [a,b]\}$ ?

asked Aug 31 '15 at 13:54

Yes. The $L_p$ norms tend to the $L_{\infty}$ norm, if the function belongs to all of them. – uniquesolution Aug 31 '15 at 13:56
This is a duplicate of a question already solved by Did, I just need to find it. – Jack D'Aurizio Aug 31 '15 at 14:00

$f:[a,b] \to [0, \infty)$ continuous , then $\lim_{n \to \infty} \Bigg(\int_a^b \big(f(x)\big)^ndx \Bigg)^{1/n}=\sup \{f(x):x \in [a,b]\}$ ?

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