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.



Quotient of two Laplace integrals
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.


