Limit of integral converges to esssup

Question

Given a finite measure space $(X,\mathcal{B},\mu)$ with $\mu(X)=1$. Let $f$ be a function with finite ess-sup. Can we prove that \begin{equation} \lim\limits_{t\to\infty} \frac{1}{t} \log \int_X e^{t\,f(x)}\mathrm{d}\mu(x)= \textrm{ess-sup}\, f(x). \end{equation}

Answer 1

Yes. One possible proof is this.

First case: $\textrm{ess-sup}f > 0$. Let $\phi(t)=\int_Xe^{tf(x)}d\mu(x)$. It's easy to show $\lim_{t\to\infty}\frac{\log(\phi(t))}{t}\le \textrm{ess-sup}f$.

To show the converse, begin by noting that for every $t\ge 0$, for $|h|\in(0,1)$ and all $x$ s.t. $f(x)$ is defined,

$$ \left|\frac{e^{(t+h)f(x)}-e^{tf(x)}}{h}\right|\le 2e^{(t+1)\textrm{ess-sup}f} $$

which is integrable over $X$. Hence, by dominated convergence theorem,

\begin{eqnarray} \frac{\phi(t+h)-\phi(t)}{h} & = & \int_X\frac{e^{(t+h)f(x)}-e^{tf(x)}}{h}d\mu(x)\\ & \to & \int_Xf(x)e^{tf(x)}d\mu(x) \end{eqnarray}

as $h\to 0$.

Consider now a sequence $a_n$ s.t. $a_n\nearrow \textrm{ess-sup}f$. For each $n$, we know

$$ \phi(t)=\int_{f< a_n}e^{tf(x)}d\mu(x) + \int_{f\ge a_n}e^{tf(x)}d\mu(x)\ge \int_{f< a_n}e^{tf(x)}d\mu(x) + e^{ta_n}\mu(f\ge a_n):=\phi_{a_n}(t) $$

By L'Hôpital we can show

\begin{eqnarray} \lim_{t\to\infty}\frac{\log(\phi_{a_n}(t))}{t} & = & \lim_{t\to\infty}\frac{\phi_{a_n}'(t)}{\phi_{a_n}(t)}\\ & = & \lim_{t\to\infty}\frac{\int_{f< a_n}f(x)e^{tf(x)}d\mu(x) + a_ne^{ta_n}\mu(f\ge a_n)}{\int_{f< a_n}e^{tf(x)}d\mu(x) + e^{ta_n}\mu(f\ge a_n)}\\ & = & \lim_{t\to\infty}\frac{\int_{f< a_n}f(x)e^{t(f(x)-a_n)}d\mu(x) + a_n\mu(f\ge a_n)}{\int_{f< a_n}e^{t(f(x)-a_n)}d\mu(x) + \mu(f\ge a_n)} \end{eqnarray}

Note, for all $x$ s.t. $f(x)<a_n$, $|e^{t(f(x)-a_n)}| \le 1$, hence by dominated convergence theorem, $\int_{f< a_n}e^{t(f(x)-a_n)}d\mu(x) \to 0$ as $t\to\infty$. Hence

$$ \int_{f< a_n}f(x)e^{t(f(x)-a_n)}d\mu(x)=\int_{f< a_n}(f(x)-a_n)e^{t(f(x)-a_n)}d\mu(x) + a_n \int_{f< a_n}e^{t(f(x)-a_n)}d\mu(x) $$

goes to 0 as $t\to\infty$ (again by dominated convergence theorem). Hence

$$ \lim_{t\to\infty}\frac{\log(\phi_{a_n}(t))}{t} = \lim_{t\to\infty}\frac{\int_{f< a_n}f(x)e^{t(f(x)-a_n)}d\mu(x) + a_n\mu(f\ge a_n)}{\int_{f< a_n}e^{t(f(x)-a_n)}d\mu(x) + \mu(f\ge a_n)} = a_n $$

which means, for all $n$, $\lim_{t\to\infty}\frac{\log(\phi(t))}{t}\ge a_n$, hence $\lim_{t\to\infty}\frac{\log(\phi(t))}{t}\ge \textrm{ess-sup}f$.

Second case: $\textrm{ess-sup}f \le 0$. We can use the previous case like this:

$$ \phi(t) = e^{t(\textrm{ess-sup}f-1)}\int_Xe^{t(f(x)-\textrm{ess-sup}f + 1)}d\mu(x) = e^{t(\textrm{ess-sup}f-1)}\int_Xe^{t\tilde{f}(x)}d\mu(x) $$

where the esential supremum of $\tilde{f}=f-\textrm{ess-sup}f+1$ is 1. Hence

\begin{eqnarray} \lim_{t\to\infty}\frac{\log(\phi(t))}{t} & = & \lim_{t\to\infty}\frac{t(\textrm{ess-sup}f-1) + \log(\int_Xe^{t\tilde{f}(x)}d\mu(x))}{t}\\ & = & \textrm{ess-sup}f-1 + \lim_{t\to\infty}\frac{\log(\int_Xe^{t\tilde{f}(x)}d\mu(x))}{t}\\ & = & \textrm{ess-sup}f-1 + 1\\ & = & \textrm{ess-sup}f \end{eqnarray}