I am trying to determine whether the limit of the following sequence exists and if so, find the limit. $f$ is a positive continuous function of $[a,b]$.

\begin{align*} \lim_{n\to\infty} \Big[\int_{a}^{b}f(x)^{n}dx\Big]^{\frac{1}{n}} \end{align*}

My thoughts so far on the problem: By definition, I know \begin{align*} \lim_{n\to\infty} \Big[\int_{a}^{b}|f(x)|^{n}dx\Big]^{\frac{1}{n}} = \sup\{|f(x)|: x \in [a,b] \} \end{align*}

because $f$ is positive I believe I can say the limits of these two sequences will be equivalent. We know because $f$ is continuous and on an compact interval $[a,b]$ it has it achieves it sup. I feel like I'm missing something here.

If anyone can provide any hints or insights that would be great. Thanks.

  • $\begingroup$ What part of that equality is "by definition"? $\endgroup$ – Thorgott Apr 18 '19 at 20:59
  • $\begingroup$ @Thorgott I am under the impression this is the definition of the $L^{p}$ norm as $p\to\infty$. $\endgroup$ – Matt Apr 18 '19 at 21:01
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    $\begingroup$ The LHS is the definition of the limit of the $L^p$ norm of $f$ as $p\rightarrow\infty$ (although you use the index $n$ which suggests that you're viewing it as a sequence). The RHS is the $L^{\infty}$ norm of $f$. The limit of something is already a notion with mathematical meaning, so you can't just define it to be whatever you want. The stated equality is true, but it certainly does not follow by definition; rather, it has to be demonstrated, which takes some potentially non-trivial effort. $\endgroup$ – Thorgott Apr 18 '19 at 21:50
  • $\begingroup$ @Thorgott Yes, thank you. $\endgroup$ – Matt Apr 19 '19 at 21:11

$\int_a^{b}[f(x)]^{n} dx \leq \int_a^{b} M^{n}=M^{n} (b-a)$ where $M=\sup \{f(x): a \leq x \leq b\}$. Hence $(\int_a^{b}[f(x)]^{n} dx)^{1/n} \leq M (b-a)^{1/n}$. Letting $n \to \infty$ we get LHS $\leq $ RHS.

Now there is a point $x_0$ such that $f(x_0)=M$. Suppose $a < x_0 <b$. Let $\epsilon >0$. By continuity there exists $\delta >0$ such that $f(x) >M-\epsilon$ for $x_0-\delta <x < x_0+\delta$. Hence $\int_a^{b}[f(x)]^{n} dx \geq \int_{x_0-\delta}^{x_0+\delta}[f(x)]^{n} dx \geq (M-\epsilon)^{n}(2\delta)$. This gives $(\int_a^{b}[f(x)]^{n} dx)^{1/n} \geq (M-\epsilon)(2\delta)^{1/n}$. Letting $n \to \infty$ we get LHS $\geq $RHS. I will leave the cases $x_0=a$ and $x_0=b$ to you.

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  • $\begingroup$ We are able to say M exists by the Extreme Value Theorem correct? $\endgroup$ – Matt Apr 19 '19 at 21:23
  • $\begingroup$ Yes, that's right. @Matthieu $\endgroup$ – Kavi Rama Murthy Apr 19 '19 at 23:04
  • $\begingroup$ Also, the cases $x_0 = a and b$ are analogous to the case you have shown above with the difference we would restrict our neighbourhood to the right side of a for $x_0 = a$ (or left side of b for $x_0 = b$). This is because continuity of $f$ is only within the interval $[a,b]$. Then perform similar calculations as above. $\endgroup$ – Matt Apr 19 '19 at 23:07
  • $\begingroup$ @Matthieu Exactly! You have got the complete proof now. $\endgroup$ – Kavi Rama Murthy Apr 19 '19 at 23:09
  • $\begingroup$ Awesome that was great thank you @KaviRamaMurthy $\endgroup$ – Matt Apr 19 '19 at 23:09

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