Calculate $\lim_{t\to\infty}\frac 1t\log\int_0^1 \cosh(tf(x))\mathrm d x$ Given that $f\in C([0,1])$, evaluate
$$\lim_{t\to\infty}\frac 1t\log\int_0^1 \cosh(tf(x))\mathrm d x.$$
I have thought about it for an entire day, but unfortunately I wasn't able to solve it. Now I'm asking you, more than a solution, which is in either case welcomed, an explanation on how to approach such a kind of problems. Thank you very much.
 A: Since ${1 \over 2}e^{|u|} \leq \cosh(u) \leq e^{|u|}$, it suffices to figure out 
$$\lim_{t \rightarrow \infty}{1 \over t}\log\int_0^1e^{t|f(x)|}\,dx$$
Let $M$ be the maximum of $|f(x)|$. Then you can write the above as
$$\lim_{t \rightarrow \infty}{1 \over t}\log\int_0^1e^{tM}e^{t(|f(x)|-M)}\,dx$$
$$=\lim_{t \rightarrow \infty}{1 \over t}\log [e^{tM}\int_0^1e^{t(|f(x)| - M)}dx]$$$$= M + \lim_{t \rightarrow \infty}{1 \over t}\log\int_0^1e^{t(|f(x)| - M)}dx$$
Show the limit of the right-hand term is zero by bounding below the portion of the integral where $|f(x)| - M > - \epsilon$ for arbitrary $\epsilon$. So the overall limit is $M$.

I thought of a better way of doing this, once it's been reduced to the exponential function. Note that 
$${1 \over t}\log\int_0^1e^{t|f(x)|}\,dx = \log[(\int_0^1 e^{t|f(x)|}\,dx)^{1 \over t}] $$
$$=\log||e^{|f(x)|}||_{L^t}$$
Since as $t$ goes to infinity the $L^t$ norm converges to the $L^{\infty}$ norm on a measure space of measure $1$ (a popular measure theory exercise), $||e^{|f(x)|}||_{L^t}$ converges to $e^M$, and thus the logarithm converges to $M$ since logarithms are continuous. Note you don't even need $f(x)$ to be continuous, just a bounded measurable function.
A: Here is another way of approaching the problem.
Notice, as already pointed out, that since $\cosh$ is even, it suffices to consider $f$ non-negative. Also, note that since $\cosh$ is increasing, if $f_1 \leq f_2$, then the limit for $f_1$ will be no more than the limit for $f_2$. Also, note that we do not necessarily have to consider continuous $f$ for this limit to make sense.
If you consider the step function $g(x) = M\chi_{[a,b]}(x)$, where $[a,b] \subset [0,1]$ and try to take the limit with this function, we get for each $t$:
$$\int_{0}^{1}\cosh(tf(x))dx =\int_{a}^{b}\cosh(Mt)dx + \int_{x\in [0,1]\setminus[a,b]}\cosh(0)dx $$
$$= (b-a)\cosh(Mt) + a + 1-b = \frac{b-a}{2}e^{Mt} + \frac{b-a}{2}e^{-Mt} + (a+1-b).$$
So, we get that:
$$\frac{1}{t}\log\int_{0}^{1}\cosh(tg(x))dx = \frac{1}{t}\log\left(e^{Mt}\left(\frac{b-a}{2} + \frac{b-a}{2}e^{-2Mt} + (a+1-b)e^{-Mt}\right)\right)$$
$$= \frac{1}{t}\log(e^{Mt}) + \frac{1}{t}\log\left(\frac{b-a}{2} + \frac{b-a}{2}e^{-2Mt} + (a+1-b)e^{-Mt}\right)$$
$$\to M + 0$$
since the argument for the logarithm is bounded for large $t$ between $b-a/2 > 0$ and $b-a$.
So, since $f$ is continuous (in fact, we only need $|f|$ to be lower semi-continuous) then if $x$ is the point that $|f|$ attains its maximum, $M$, then for every $\epsilon > 0$, there is an interval $[a,b]$ containing $x$ such that $(M-\epsilon)\chi_{[a,b]} \leq |f| \leq M\chi_{[0,1]},$ which tells us that the integral for large $t$ is between $M - \epsilon$ and $M$, which is exactly the definition of converging to $M$.
In fact, as the other responder has a proof for $f$ a bounded measurable function, my proof also generalizes to that case since instead of intervals $[a,b]$, you do the exact calculation for measurable subsets $E \subset [0,1]$ with non-zero measure, but the answer now is not $\max |f|$, but $\|f\|_{L^\infty}$ and in my proof instead of $(b-a)$ we have $\mu(E)$ and instead of $1-b + a$ we have $1 - \mu(E)$.
