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We are given a continuous function $f \colon [0,1] \to \mathbb{R}$. What is the value, if it exists, of the limit $$\lim_{t \to +\infty} \frac{1}{t} \log \int_0^1 \cosh (t f(x))\, \mathrm{d}x \ ?$$

PS: this is not homework. It's a question contained in a Ph.D test, and I am unable to make progress toward the solution :-(

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may be try L'Hôpital's rule –  Babak Miraftab May 3 '12 at 9:45
    
My first guess: consider $f(x)=mx+q$ and compute the limit in this particular case. Then remark that any continuous $f$ on $[0,1]$ can be uniformly approximated by affine functions. But also the affine case is not really easy... –  Siminore May 3 '12 at 9:55

1 Answer 1

up vote 9 down vote accepted

Without loss of generality, assume $f(x)\geq 0$ (as $\cosh$ is even). Let $m = \max f(x)$ and let $\delta$, $\epsilon$ be such that $|x-x_0|<\delta$ impies $m\geq f(x)>m-\epsilon$.

First, $$ I_t = \frac1t \log\int_0^1 \cosh(tf(x))\,dx \leq \frac1t \log \cosh(tm), $$ and since $\log\cosh(tm)\sim \log\frac12 + tm$ ($t\to\infty$), this implies that $$ \limsup_{t\to\infty} I_t \leq m. $$

Second, $$I_t \geq \frac1t \log \int_{x_0-\delta}^{x_0+\delta} \cosh(tf(x))\,dx \geq \frac1t \log \big(2\delta \cosh(t(m-\epsilon)),$$ which implies that $$ \liminf_{t\to\infty} I_t \geq m-\epsilon. $$

Taking the limit as $\epsilon\to0$, it follows that $\lim_{t\to\infty}I_t$ exists and equals $\max_x |f(x)|$.

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Very nice! Just a correction: $\cosh (tm) \sim \frac{1}{2}\mathrm{e}^{tm}$, so that $\log \cosh (tm) \sim \log \frac{1}{2}+tm$ as $t \to +\infty$. –  Siminore May 3 '12 at 11:31
    
Thanks. Fixed that. –  Kirill May 3 '12 at 11:34

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