# 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.

-
Speaking intuitively, the answer should be $M = \sup_{0 \leqslant x \leqslant 1} | f(x) |$, because contribution from around this point will dominate the integral. – Sasha Mar 2 '12 at 16:41
@Sasha's intuition is right, even when $f=M$ at more than one point. For the upper bound, use $f\leqslant M$ and $\cosh(u)\leqslant\mathrm e^u$ for every $u\geqslant0$. For the lower bound, use $f\geqslant M-\varepsilon$ on an interval $I_\varepsilon$ of positive length and $\cosh(u)\geqslant\frac12\mathrm e^u$. – Did Mar 2 '12 at 16:45
@DidierPiau that is true if $f(x)$ is nonnegative, but it's slightly different if $f(x)$ takes negative values. – Zarrax Mar 2 '12 at 16:56
@Zarrax: Right. Work with $|f|$ since $\cosh$ is even. – Did Mar 2 '12 at 17:02
@Sasha Can you explain me how did you get to that intuitively? (I'm interested) – Pedro Tamaroff Mar 6 '12 at 3:28

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.

-
Mmmmh... you cheated. :-) Please solve the $\cosh$ case! – Did Mar 2 '12 at 17:04
ok I edited it... happy now? :) – Zarrax Mar 2 '12 at 18:05
Not quite: you forgot (1) a logarithm and (2) an argument to explain why the second term converges to zero (as it stands, one could replace $M$ by any $M'\gt M$ and you would still get rid of the second term, wrongly obviously). – Did Mar 2 '12 at 18:17
@DidierPiau Could you critique this ? Or maybe give your ideas on how it can be proven. – Pedro Tamaroff Mar 2 '12 at 18:24
@PeterT.off: OK, but on the other page. – Did Mar 2 '12 at 18:27