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Let $f$ be continuous on $[a,b]$, let $f(x) \geq 0$ for all $x \in [a,b]$ and let $M_n = (\int_{a}^{b} f^n )^{\frac{1}{n}}$. Show that $\lim(M_n) = \sup\{f(x):x\in [a,b]\}$

Here is what I have so far: Let $M = \sup(f)$ and $p \in [a,b]$ such that $M = f(p)$. Given $\epsilon > 0$, there exists an interval $[c,d]$ containing $p$ such that

$$M - \epsilon \leq f(x) \leq M, ~~~x \in [c,d]$$

Note that $0 \leq (d-c) \leq (b-a)$ so I raise everything to the power of $n$ and then add $(d-c)$ and $(b-a)$ to the inequality. $$ (M-\epsilon)^n(d-c) \leq \int_{c}^{d}f^n \leq \int_{a}^{b}f^n \leq M^n(b-a)$$

Then I take the $nth$ root to get

$$(M-\epsilon)(d-c)^{1/n} \leq M_n \leq M(b-a)^{1/n}$$

Then since $\lim(\alpha)^{1/n} \to 0$, by the squeeze theorem, $\lim(M_n) = \sup(f(x)) = 0$?

Evan Kim
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  • Everything but the last line is correct. You can conclude that every limit value of $(M_n)$ lies between $M-\epsilon$ and $M$, and then you are almost done. – Martin R Apr 28 '20 at 18:14
  • I don't know what the measure $S_{/delta}$ is, we have not learned that so I am going to say no it doesn't answer my question – Evan Kim Apr 28 '20 at 18:25

1 Answers1

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Here's what you need to fix. From $$ (M-\epsilon)(d-c)^{1/n}\leq M_n\leq M(b-a)^{1/n}, $$ you cannot apply squeeze theorem since the left hand term converges to $M-\epsilon$. In fact, you don't know that the limit exists yet. What you can say is $$ M-\epsilon\leq \liminf M_n \leq M. $$ Since $\epsilon>0$ is arbitrary, $\implies \liminf M_n=M$. Apply the exact same argument but with $\limsup$. Then you have $\liminf M_n=\limsup M_n = M$. It's very similar to squeeze theorem, but a bit more delicate.

ProfOak
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