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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.

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    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.2012-03-02
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    @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 $\cosh(u)\geqslant\frac12\mathrm e^u$.2012-03-02
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    @DidierPiau that is true if $f(x)$ is nonnegative, but it's slightly different if $f(x)$ takes negative values.2012-03-02
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    @Zarrax: Right. Work with $|f|$ since $\cosh$ is even.2012-03-02
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    @Sasha Can you explain me how did you get to that intuitively? (I'm interested)2012-03-06

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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.

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    Mmmmh... you cheated. :-) Please solve the $\cosh$ case!2012-03-02
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    ok I edited it... happy now? :)2012-03-02
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    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).2012-03-02
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    @DidierPiau Could you [critique this](http://math.stackexchange.com/questions/114703/the-limit-of-f-nx/115037#115037) ? Or maybe give your ideas on how it can be proven.2012-03-02
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    @PeterT.off: OK, but on the other page.2012-03-02
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    @DidierPiau I added a hint for the last part.. it is homework so I figure I shouldn't add the whole argument2012-03-02
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    Hmmm... Kind of an odd choice since this is probably THE part of the argument the OP could feel uneasy with. Well, this is your post after all... :-)2012-03-02
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    @Zarrax Although I know little about measure thoery, I find your last insight very interesting.2012-03-06
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    @Zarrax , sorry after many years. Would you mind telling that why does it suffice to figure out $ \lim_{t \rightarrow \infty}{1 \over t}\log\int_0^1e^{t|f(x)|}\,dx $ and not $ \lim_{t \rightarrow \infty}{1 \over t}\log\int_0^1\frac{1}{2}e^{t|f(x)|}\,dx $???2017-08-26
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    I really like the measure theory argument! Really nice!2017-09-02
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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 $1 - \mu(E)$.