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Could anyone help me with the following problem? Because i have stuck.

Let $f:[a,b]\rightarrow [0,\infty)$ be continuous and not the zero function. Prove that $$\lim_{n\to \infty} \frac{\int\limits_a^b f^{n+1}(x)\, \mathrm{d}x}{\int\limits_a^b f^n(x)\, \mathrm{d}x}=\sup_{x\in[a,b]}f(x)$$

Some of my thoughts:

Let $\displaystyle M=\sup_{x\in[a,b]}f(x)$ and $\displaystyle m=\inf_{x\in[a,b]}f(x)$.

Moreover define the sequences $\{I_n\}_{n=0}^{\infty}$ and $\{a_n\}_{n=0}^{\infty}$,

So, $\displaystyle I_n=\int\limits_a^b f^{n}(x)\, \mathrm{d}x$

and $\displaystyle a_n=\frac{I_{n+1}}{I_n}$.

It easy to prove that $$m(b-a)\leq I_n \leq M(b-a)$$

and $$a_n \leq M\left(\frac{M}{m}\right)^n.$$

First I tried to use the above inequalities to prove it by definition or somehow with the squeeze theorem, but unfortunately $\displaystyle M\left(\frac{M}{m}\right)^n \xrightarrow[ n \to \infty]{} \infty$

Then I thought about the integral mean value theorem, but again nothing new came up.

So do you have any suggestions, thoughts, hints or\and solutions?

Another User
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karhas
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4 Answers4

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Thanks to Motyla Noga Tomka Mazura for part of what's below - this is easier than I've thought for years, because of the monotonicity that he pointed out. (I think this post is about half mine and half his. I'm posting it because I need the points a little more than he does - is there a way I can donate half of them to him?)

Define $M$, $I_n$ and $\alpha_n$ as you did. As @Motyla Noga Tomka Mazura points out, the Cauchy-Schwarz inequality shows that $$I_n\le\left(I_{n-1}I_{n+1}\right)^{1/2},$$ hence the sequence $\alpha_n$ is increasing. Hence $\lim\alpha_n$ exists, and hence it's enough to show that $$\lim_{n\to\infty}I_n^{1/n}=M.$$An this is easier than I thought.

First, it's clear that $f^n\le M^n$, hence $I_n\le(b-a)M^n$, so $$I_n^{1/n}\le(b-a)^{1/n}M\to M.$$ So $\limsup I_n^{1/n}\le M$.

Now let $\epsilon>0$. Since $M=\sup f$ and $f$ is continuous, there exists an open interval $E$ such that $$f(t)>(1-\epsilon)M\quad(t\in E).$$If $L$ is the length of $E$ then $$I_n\ge\int_Ef^n\ge L((1-\epsilon)M)^n,$$ so $$I_n^{1/n}\ge L^{1/n}(1-\epsilon)M\to(1-\epsilon)M.$$ Hence $\liminf I_n^{1/n}\ge(1-\epsilon)M$, and hence $\liminf I_n^{1/n}\ge M$.

David C. Ullrich
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  • maybe we should prove that $a_n$ is also bounded from above in order to the $\lim a_n$ exists? In this way we may construct a even shorter proof. – karhas Jun 25 '15 at 17:44
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    Valid point. Except it's trivial that $\alpha_n$ is bounded, just because $f^{n+1}\le Mf^n$. (Also we don't need to worry about it, because in fact the ratio test thing works just as well for infinite limits - if $\alpha_{n+1}/\alpha_n\to\infty$ then $\alpha_n^{1/n}\to\infty$.) (In any case I don't see how this will make the proof shorter...) – David C. Ullrich Jun 25 '15 at 17:49
  • you are right. I had in my mind the following result. If $a_n$ is bounded from above and increasing then $\lim a_n = \sup {a_n , n \in \mathbb{N}}$. – karhas Jun 25 '15 at 18:01
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Use the fact $$\lim_{n\to\infty} \frac{a_{n+1}}{a_n } =g \Longrightarrow \lim_{n\to\infty} \sqrt[n]{a_n } =g$$

  • sorry but it is not clear how to use the above fact. do you want to be more specific? – karhas Jun 25 '15 at 15:32
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    @karhas A more clear sketch: if $a_n=\int_a^b f(x)^n dx$, you want to show that $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \sup_{x \in [a,b]} f(x)=M$. So try to show that $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} a_n^{1/n}$ when both limits exist. Then what you want to prove becomes the Riemann version of this: http://math.stackexchange.com/questions/242779/limit-of-lp-norm – Ian Jun 25 '15 at 15:36
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    @Motyla You need the converse of the fact you mention to do this problem. The sequence 1,1,2,2,4,4,8,8,... shows that the converse is false. (If we knew somehow that the limit of the ratios exists then what you say would work. But we don't know that.) – David C. Ullrich Jun 25 '15 at 16:22
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    First observe that $$\int_I f^n (t) dt =\int_I f^{\frac{n-1}{2} } (t) f^{\frac{n+1}{2} } (t) dt \leq \sqrt{\int_I f^{n-1} (t) dt }\sqrt{\int_I f^{n+1} (t) dt }$$ by Cauchy = Schwarz hence the sequence is increasing , thus the limit $\lim_{n\to \infty }\frac{a_{n+1}}{a_n} $ exist and then apply the fact I wrote above. –  Jun 25 '15 at 16:50
  • @MotylaNogaTomkaMazura Ok, that shows the limit exists. Still seems like cheating, using that result about $L^p$ norms. There's a bigger problem: We're talking about the Riemannn intergral! Thinking about the problem, I have a slightly complicated proof. But the proof I have in mind only works for the Lebesgue integral, at least without some fixing up. The problem is that if $f$ is continuous and $E$ is an open set or a closed set then $f\chi_E$ need not be Riemann integrable. (This also shows that the proof you give a link to does not apply here, at least without some fixing up). – David C. Ullrich Jun 25 '15 at 17:04
  • @MotylaNogaTomkaMazura Wait! Actually to get the result about the $L^p$ norm we just need $\int_Ef$ for a single interval $E$, and there's no problem with that. Very nice - this is altogether a nicer proof than the one I knew. – David C. Ullrich Jun 25 '15 at 17:13
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WLOG, assume the maximum value of $f$ is $1.$ Then $f^{n+1} \le f^n,$ so the ratio of integrals is $\le 1$ for all $n.$ On the other hand, Holder shows

$$\int_a^b f^n \le (\int_a^b f^{n+1})^{n/(n+1)}\cdot(b-a)^{1/(n+1)}$$

by Holder. Use this and simplify to see

$$(1)\,\,\,\,\frac{(\int_a^b f^{n+1})^{1/(n+1)}}{(b-a)^{1/(n+1)}}\le \frac{\int_a^b f^{n+1}}{\int_a^b f^n} \le 1.$$

But as is well known, $\lim_{p\to \infty}(\int_a^b f^p)^{1/p} = \sup_{[a,b]}|f| = 1.$ It follows that the left side of $(1) \to 1,$ and we're done.

zhw.
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Let $M=\max_{x\in[a,b]}f(x)$. By assumption, $M>0$. Choose $x_{0}\in[a,b]$ such that $M=f(x_{0})$. By continuity, there exists $\delta>0$ such that $f(x)>\frac{M}{2}$ for whenever $x\in[x_{0}-\delta,x_{0}+\delta]$ (If $x_{0}=a$ or $x_{0}=b$, we modify the interval in an obvious way). For each $n\in\mathbb{N}$, Observe that $\int_{a}^{b}f^{n}(x)dx\geq\int_{x_{0}-\delta}^{x_{0}+\delta}(\frac{M}{2})^{2}dx=2\delta(\frac{M}{2})^{n}>0$. Therefore, $\frac{\int_{a}^{b}f^{n+1}(x)dx}{\int_{a}^{b}f^{n}(x)dx}$ is well-defined.

We go to show that $\frac{\int_{a}^{b}f^{n+1}(x)dx}{\int_{a}^{b}f^{n}(x)dx}\rightarrow M$ as $n\rightarrow\infty$. Firstly, we prove under the assumption that $M=1$. Clearly, $f^{n+1}(x)\leq f^{n}(x)$, so $\int_{a}^{b}f^{n+1}\leq\int_{a}^{b}f^{n}$. Therefore, $\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}}\leq1$. In particular, $\limsup_{n}\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}}\leq1$.

Let $\alpha\in(0,1)$ be arbitrary. Let $A=\{x\in[a,b]\mid f(x)>\alpha\}$, then $\mu(A)>0$ (i.e., the Lebesgue measure of $A$ is positive) because $f$ is continuous. We have that $\int_{a}^{b}f^{n+1}\geq\int_{A}f^{n+1}\geq\int_{A}\alpha f^{n}$. Therefore, \begin{eqnarray*} \frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}} & \geq & \frac{\alpha\int_{A}f^{n}}{\int_{A}f^{n}+\int_{A^{c}}f^{n}}\\ & \geq & \frac{\alpha\int_{A}f^{n}}{\int_{A}f^{n}+\alpha^{n}(b-a)}.\\ & = & \alpha\frac{1}{1+\frac{\alpha^{n}(b-a)}{\int_{A}f^{n}}}. \end{eqnarray*} We assert that $\frac{\alpha^{n}}{\int_{A}f^{n}}\rightarrow0$ as $n\rightarrow\infty$. Choose $\beta\in(\alpha,1)$. Let $B=\{x\in[a,b]\mid f(x)>\beta\}$, then $\mu(B)>0$. Observe that $B\subseteq A$, so $\int_{A}f^{n}\geq\int_{B}f^{n}\geq\mu(B)\beta^{n}$. It follows that $\frac{\alpha^{n}}{\int_{A}f^{n}}\leq\frac{\alpha^{n}}{\mu(B)\beta^{n}}\rightarrow0$ as $n\rightarrow\infty$. Now, it is clear that \begin{eqnarray*} \liminf_{n}\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}} & \geq & \liminf_{n}\alpha\frac{1}{1+\frac{\alpha^{n}(b-a)}{\int_{A}f^{n}}}\\ & = & \alpha. \end{eqnarray*} Since $\alpha\in(0,1)$ is arbitrary, we actually have $\liminf_{n}\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}}\geq1$. Combining, we have $\liminf_{n}\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}}=\limsup_{n}\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}}=1$. This shows that $\lim_{n}\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}}=1$.

Finally, we drop the assumption that $M=1$. Let $g(x)=\frac{1}{M}f(x)$, then $g$ satisfies all the conditions in above with $\max g=1$. Therefore $\lim_{n}\frac{\int_{a}^{b}g^{n+1}}{\int_{a}^{b}g^{n}}=1$ and it follows that $\lim_{n}\frac{\int_{a}^{b}f^{n+1}}{\int_{a}^{b}f^{n}}=M.$

Danny Pak-Keung Chan
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