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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 ${1 \over 2}{\sin(\xi) \over \sin(2\xi)}$, which is generally not the same.

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.

This isn't necessarily true even when $\phi(x)$ and $\psi(x)$ are real-valued functions and $f(x)$ is nonnegative. For example, let $\phi(x) = x^2(1-x)^2$ and $\psi(x) = x^2(1-x)^3$ and suppose $f(x)$ has a local nondegenerate maximum at ${1 \over 2}$. Then one can integrate by parts twice to get $$\int_0^1 \phi''(x)f(x)^n dx = \int_0^1\phi(x)(n(n-1)(f'(x))^2+ nf''(x))f(x)^{n-2}dx$$ $$\int_0^1 \psi''(x)f(x)^n dx = \int_0^1\psi(x)(n(n-1)(f'(x))^2+ nf''(x))f(x)^{n-2}dx$$ 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 ${\sin(\xi) \over 2\sin(2 \xi)}$, which is generally not the same. (although I'm interpreting the ratio for the right-hand sides in a certain way).

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 ${1 \over 2}{\sin(\xi) \over \sin(2\xi)}$, which is generally not the same.

This isn't necessarily true even when $\phi(x)$ and $\psi(x)$ are real-valued functions and $f(x)$ is nonnegative. For example, let $\phi(x) = \cos(x)$ and $\psi(x) = \cos(2x)$. Then one can integrate by parts to get $$\int_0^{\pi} \cos(x)f(x)^n dx = \int_0^{\pi}n\sin(x)f'(x)f(x)^{n-1}dx$$ $$\int_0^{\pi} \cos(2x)f(x)^n dx = {1 \over 2}\int_0^{\pi}n\sin(2x)f'(x)f(x)^{n-1}dx$$ 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 ${\sin(\xi) \over 2\sin(2 \xi)}$, which is generally not the same. (although I'm interpreting the ratio for the right-hand sides in a certain way).

This isn't necessarily true even when $\phi(x)$ and $\psi(x)$ are real-valued functions and $f(x)$ is nonnegative. For example, let $\phi(x) = x^2(1-x)^2$ and $\psi(x) = x^2(1-x)^3$ and suppose $f(x)$ has a local nondegenerate maximum at ${1 \over 2}$. Then one can integrate by parts twice to get $$\int_0^1 \phi''(x)f(x)^n dx = \int_0^1\phi(x)(n(n-1)(f'(x))^2+ nf''(x))f(x)^{n-2}dx$$ $$\int_0^1 \psi''(x)f(x)^n dx = \int_0^1\psi(x)(n(n-1)(f'(x))^2+ nf''(x))f(x)^{n-2}dx$$ 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 ${\sin(\xi) \over 2\sin(2 \xi)}$, which is generally not the same. (although I'm interpreting the ratio for the right-hand sides in a certain way).