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Let $\varphi(x)$ and $\psi(x)$ be two complex-valued continuous functions on $[a,b]$, and let $f(x)$ be a complex-valued continuously differentiable function on $[a,b]$. Suppose that $|f(x)|$ has an absolute maximum at an interior point, say $\xi$, of the interval. Prove or disprove \begin{equation}\label{eq3} \lim_{n\to\infty}\frac{\int_a^b\varphi(x)[f(x)]^ndx}{\int_a^b\psi(x)[f(x)]^ndx}=\frac{\varphi(\xi)}{\psi(\xi)}. \end{equation}

Remark 1: This is true for $f(x)\in C^2$, by Laplace's method.

Remark 2: Micheal has given a counter example, which makes the use of the fact $f'(\xi)\neq0$. This is a good example. Now, if we further assume that $f'(\xi)=0$, and also we assume that $\phi\neq0$ and $\psi(\xi)\neq0$. how about now? I believe this will be more difficult.

Remark 3: Sorry for my question style. Well, this is not just a home work. It's an open problem, when one try to prove a theorem of Chung and Erd\"os 1951. That theorem is essentially said that the ratio of two coefficients in Fourier series of $f(x)^n$ will tend to 1. Where the assumption on $f$ could be translated as $f\in C^1[-\pi,\pi]$, $f(0)=1$, $f'(0)=0$ and $|f(x)|<1$ for $x\neq0$. This theorem will be a corollary if the limit is true here. So I make a further assumption in Remark 2 that $f'(\xi)\neq0$.

Thank you.

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The way you have phrased this question makes it seem like an exercise or coursework. If this is not the case, could you please give some more background, e.g. where did you come across this problem? – Yemon Choi Dec 4 '10 at 19:49
Hum, I voted to close, but after looking closer at the question and noticing that it asks for $[f(x)]^n$ rather than $|f(x)|^n$ as I initially misread, I feel I may have been a bit too hasty with the vote: that $f(x)$ is required to be $C^1$ is a tiny bit subtle. Since I don't think I can undo my vote: can the next person who wants to vote to close instead leave a comment "cancelling" this request? Thanks. – Willie Wong Dec 4 '10 at 22:37
This problem comes from an attempt of proving a probability theorem (of K.L. Chung and P. Erd\"os, 1951) using analytic argument. – LI Yutian Dec 5 '10 at 0:18
I would like to echo Yemon's comment. I'm not a fan of "questions" without question marks: this one is written like a homework problem: "prove or disprove". Much better would be for you to provide some motivation --- why did you think about this question? how does it relate to your research? --- or at least some more background --- what other results are related to this? Remember that this site is for questions (and "research" questions at that), so demands like "prove or disprove" come across (usually unintentionally) as rude. – Theo Johnson-Freyd Dec 5 '10 at 4:38
Also, can someone explain to me why this problem is difficult? I can rescale $f$ by $f(\xi)$ to assume that $f(\xi)=1$ and $|f(x)|<1$ for $x\neq\xi$. Pick $\epsilon$, and find an interval around $\xi$ in which $\phi,\psi,f$ don't vary more than $\times(1\pm\epsilon)$ (I guess I'm assuming $\phi(\xi)\neq0\neq\psi(\xi)$). Moreover, outside this small interval $f(x)\leq c<1$. So in the limit only this interval contributes to the integrals. But then I approximate the integrals by something with error $\times(1\pm \epsilon)^4 \leq \times(1 \pm 5\epsilon)$. (continued) – Theo Johnson-Freyd Dec 5 '10 at 4:46
up vote 7 down vote accepted

This isn't necessarily true. Let $\phi(x) = \cos(x)$ and $\psi(x) = \cos(2x)$. Let $f(x)$ be a complex-valued function such that $|f(x)|$ has its absolute maximum at some $\xi$ for which the (complex-valued) derivative $f'(x)$ is nonzero at $\xi$. Then one can integrate by parts to get $$\int_0^{\pi} \cos(x)f(x)^n dx = n\int_0^{\pi}\sin(x)f'(x)f(x)^{n-1} dx$$ $$\int_0^{\pi} \cos(2x)f(x)^n dx = {n \over 2}\int_0^{\pi}\sin(2x)f'(x) f(x)^{n-1}dx$$ If your statement were true, then by looking at the ratio of the left-hand sides and taking limits as $n$ goes to infinity one should get ${\cos(\xi) \over \cos(2\xi)}$, while looking at the ratio of the right-hand sides and taking limits as $n$ goes to infinity one should get ${2 \sin(\xi) \over \sin(2\xi)}$, which is generally not the same.

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Ha. I took the OP's word for the fact that it is true in the $C^2$ case. Good example. This shows that even in the $C^2$ case the estimate requires stationary phase also. – Willie Wong Dec 5 '10 at 14:45
@Michael: Unless I am mistaken, the ratio $\frac12\frac{\sin(\xi)}{\sin(2\xi)}$ in your answer should read $\frac{2\sin(\xi)}{\sin(2\xi)}=\frac1{\cos(\xi)}$. This does not alter the rest of your argument. – Did Dec 5 '10 at 16:57
thanks, corrected it. – Michael Greenblatt Dec 5 '10 at 17:47
@ Michael: Thanks. Very good example. How about if we add another assumption that $f'(\xi)=0$? Is the limit correct in that case? – LI Yutian Dec 6 '10 at 21:37

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