# Real Analysis

• Jan 30th 2009, 01:29 PM
chiph588@
Real Analysis
Show that if $f$ is a continuous real-valued function on the interval $[a,b]$ and $f(x) \geq 0 \; \forall x \in [a,b]$, then
$\lim_{n \to \infty} \left( \int_a^b f^{n}(x) dx \right)^{\frac{1}{n}} = \max \{f(x) : x \in [a,b]\}.$
• Jan 30th 2009, 01:50 PM
Opalg
Quote:

Originally Posted by chiph588@
Show that if $f$ is a continuous real-valued function on the interval $[a,b]$ and $f(x) \geq 0 \; \forall x \in [a,b]$, then
$\lim_{n \to \infty} \left( \int_a^b f^{n}(x) dx \right)^{\frac{1}{n}} = \max \{f(x) : x \in [a,b]\}.$

Let $m = \max \{f(x) : x \in [a,b]\}$. It should be fairly clear that $\int_a^b f^{n}(x)\, dx \leqslant m^n$. Thus $\lim_{n \to \infty} \left( \int_a^b f^{n}(x) dx \right)^{\!\!1/n}$ (if it exists) is less than or equal to m.

To get the reverse inequality, let $\varepsilon>0$. The continuity of f ensures that there is some interval of length $\delta>0$ on which $f(x)\geqslant m-\varepsilon$. Then $\int_a^b f^{n}(x)\, dx \geqslant \delta(m-\varepsilon)^n$. Take the n'th root and let $n\to\infty$ to see that $\lim_{n \to \infty} \left( \int_a^b f^{n}(x) dx \right)^{\!\!1/n} \geqslant m-\varepsilon$. Finally, take the limit as $\varepsilon\to0$.
• Jan 31st 2009, 08:10 AM
chiph588@
Quote:

Originally Posted by Opalg
Let $m = \max \{f(x) : x \in [a,b]\}$. It should be fairly clear that $\int_a^b f^{n}(x)\, dx \leqslant m^n$. Thus $\lim_{n \to \infty} \left( \int_a^b f^{n}(x) dx \right)^{\!\!1/n}$ (if it exists) is less than or equal to m.

Shouldn't we say $\int_a^b f^{n}(x)\, dx \leq m^n(b-a)$, because what if the length of the interval could be really big. For example, $\int_0^2 m^n\, dx = 2m^n \geq m^n.$

So taking the $n$th root of both sides would yield $m (b-a)^{\frac{1}{n}}$ , and as $n \to \infty, \; (b-a)^{\frac{1}{n}} \to 1.$
• Jan 31st 2009, 08:41 AM
Opalg
Quote:

Originally Posted by chiph588@
Shouldn't we say $\int_a^b f^{n}(x)\, dx \leq m^n(b-a)$, because what if the length of the interval could be really big. For example, $\int_0^2 m^n\, dx = 2m^n \geq m^n.$

So taking the $n$th root of both sides would yield $m (b-a)^{\frac{1}{n}}$ , and as $n \to \infty, \; (b-a)^{\frac{1}{n}} \to 1.$

Absolutely correct! The length of the interval should be included, and its n'th root goes to 1 as n→∞. I overlooked that.
• Jan 31st 2009, 03:29 PM
chiph588@
This is one of my favorite theorems.
• Jan 31st 2009, 04:06 PM
NonCommAlg
it's not bad to know, if you haven't noticed yet, that the limit is some sort of "continuous version" of this: $\lim_{n\to\infty} \sqrt[n]{a_1^n + \cdots + a_k^n} = \max_{1 \leq j \leq k}{a_j},$ where $a_j$ are postive numbers.
• Jan 31st 2009, 04:10 PM
chiph588@
Ah yes, because an integral is just a Riemann sum.
• Jan 31st 2009, 05:15 PM
ThePerfectHacker
I do not want to argue against Opalg because I am a noob compared to him in analysis (Blush), but I offer a suggestion.
I think this problem is better solved using the concepts of limsup's and liminf's.

Let $\ell$ be the length of the interval. Let $I_n = \left( \smallint_a^b f^n \right)^{1/n}$.
Opalg hath shown us that $\delta^{1/n} \cdot \ell^{1/n} \cdot (m-\epsilon) \leq I_n \leq m \cdot \ell^{1/n}$.
Then he says $m-\epsilon \leq \lim I_n \leq m \implies I_n = m$ by $\epsilon \to 0^+$.
But we need to know the limit exists. (Thinking)

So a way around this problem is to perhaps say first take $\epsilon \to 0^+$ and then $n\to \infty$.
But the problem is $\delta$ depends on $\epsilon$ and so how we know that $\delta^{1/n} \not \to 0$ as $n\to \infty$? (Thinking)

Maybe I am missing something but I think that if we use limsup's and liminf's the solution is much smoother. Because $\delta^{1/n}\cdot \ell^{1/n}\cdot (m-\epsilon) \leq I_n$ and so $m-\epsilon \leq \liminf I_n \implies m \leq \liminf I_n$. However, $I_n \leq m \implies \limsup I_n \leq m$. Thus, we have shown $\liminf I_n \geq \limsup I_n$ but $\liminf I_n \leq \limsup I_n$. This means that $I_n$ is a convergent sequence with $\lim I_n = \liminf I_n = \limsup I_n = m$.
---
• Feb 1st 2009, 02:00 AM
Opalg
Quote:

Originally Posted by ThePerfectHacker
I do not want to argue against Opalg because I am a noob compared to him in analysis (Blush), but I offer a suggestion.
I think this problem is better solved using the concepts of limsup's and liminf's.

That is quite correct. The efficient way to explain the solution is to use limsup in connection with the upper estimate, and liminf for the lower estimate.

I didn't do that, because I didn't want to deter anyone who might be unfamiliar with limsup and liminf. So I just included a parenthetical comment to the effect that you need to establish that the limit exists.
• Feb 17th 2009, 03:13 PM
chiph588@
Quote:

Originally Posted by ThePerfectHacker
I do not want to argue against Opalg because I am a noob compared to him in analysis (Blush), but I offer a suggestion.
I think this problem is better solved using the concepts of limsup's and liminf's.

Let $\ell$ be the length of the interval. Let $I_n = \left( \smallint_a^b f^n \right)^{1/n}$.
Opalg hath shown us that $\delta^{1/n} \cdot \ell^{1/n} \cdot (m-\epsilon) \leq I_n \leq m \cdot \ell^{1/n}$.
Then he says $m-\epsilon \leq \lim I_n \leq m \implies I_n = m$ by $\epsilon \to 0^+$.
But we need to know the limit exists. (Thinking)

So a way around this problem is to perhaps say first take $\epsilon \to 0^+$ and then $n\to \infty$.
But the problem is $\delta$ depends on $\epsilon$ and so how we know that $\delta^{1/n} \not \to 0$ as $n\to \infty$? (Thinking)

Maybe I am missing something but I think that if we use limsup's and liminf's the solution is much smoother. Because $\delta^{1/n}\cdot \ell^{1/n}\cdot (m-\epsilon) \leq I_n$ and so $m-\epsilon \leq \liminf I_n \implies m \leq \liminf I_n$. However, $I_n \leq m \implies \limsup I_n \leq m$. Thus, we have shown $\liminf I_n \geq \limsup I_n$ but $\liminf I_n \leq \limsup I_n$. This means that $I_n$ is a convergent sequence with $\lim I_n = \liminf I_n = \limsup I_n = m$.
---

how do we know that $m-\epsilon \leq \liminf I_n$ and $\limsup I_n \leq m$?
• Feb 17th 2009, 09:06 PM
ThePerfectHacker
Quote:

Originally Posted by chiph588@
how do we know that $m-\epsilon \leq \liminf I_n$ and $\limsup I_n \leq m$?

It was shown that, $\delta^{1/n} \cdot \ell^{1/n} \cdot (m-\epsilon) \leq I_n \leq m \cdot \ell^{1/n}$.
Therefore, $\liminf \delta^{1/n} \cdot \ell^{1/n} \cdot (m-\epsilon) \leq \liminf I_n \implies m-\epsilon \leq I_n$.
Similary with $\limsup$. (Nod)
• Feb 18th 2009, 09:05 AM
heathrowjohnny
Quote:

Originally Posted by Opalg
Let $m = \max \{f(x) : x \in [a,b]\}$. It should be fairly clear that $\int_a^b f^{n}(x)\, dx \leqslant m^n$. Thus $\lim_{n \to \infty} \left( \int_a^b f^{n}(x) dx \right)^{\!\!1/n}$ (if it exists) is less than or equal to m.

To get the reverse inequality, let $\varepsilon>0$. The continuity of f ensures that there is some interval of length $\delta>0$ on which $f(x)\geqslant m-\varepsilon$. Then $\int_a^b f^{n}(x)\, dx \geqslant \delta(m-\varepsilon)^n$. Take the n'th root and let $n\to\infty$ to see that $\lim_{n \to \infty} \left( \int_a^b f^{n}(x) dx \right)^{\!\!1/n} \geqslant m-\varepsilon$. Finally, take the limit as $\varepsilon\to0$.

How did you conclude that $\int_{a}^{b} f^{n} \ dx \leq m^{n}$(I know there should be a $b-a$ also)? Are not you using what you are trying to prove?
• Feb 18th 2009, 06:32 PM
ThePerfectHacker
Quote:

Originally Posted by heathrowjohnny
How did you conclude that $\int_{a}^{b} f^{n} \ dx \leq m^{n}$(I know there should be a $b-a$ also)? Are not you using what you are trying to prove?

Because $0 \leq f \leq m \implies 0 \leq f^n \leq m^n \implies 0 \leq \int_a^b f^n \leq \int_a^b m^n$.
Thus, we get, $\int_a^b f^n \leq m^n(b-a)$.