A limit with integral

Solution 1:

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 $\max_x |f(x)|$.