Prove: $\lim_{t \to \infty} \frac{1}{t}\ln\big(\int_0^1 e^{-tf(x)}dx\big) = -\min \ f(x)$ Let $f$ be a continuous function $[0,1] \rightarrow \mathbb{R}$.
Prove:

*

*$\forall \ t> 0;\frac{1}{t}\ln\Big(\int_0^1 e^{-tf(x)}dx\big) \le -\min \ f(x)$

*$\lim_{t \to \infty} \frac{1}{t}\ln\big(\int_0^1 e^{-tf(x)}dx\big) = -\min \ f(x)$
I solved 1) Let $L =\min \ f(x)$ so :
$\frac{1}{t}\ln(\int_0^1 e^{-tf(x)}dx) \le \frac{1}{t}\ln(\int_0^1 e^{-tL}dx) = \frac{1}{t}\ln(e^{-tL}) = \frac{-tL}{t} = -L$
But I'm not sure how to bound it from below or how to use another method to show the equality
 A: HINT:
Use the continuity of $f(x)$ on $[0,1]$.
Then, $f(x)$ attains a minimum, $f_{\min}$, at some point $x_0\in [0,1]$.  Then for all $\varepsilon>0$, there exists a neighborhood of $x_0$ for which
$$f_{\min}\le f(x)<f_{\min}+\varepsilon$$
Hence, for that $\varepsilon$ and neighborhood of $x_0$
$$ e^{-t\left(f_{\min}+\varepsilon\right)} \le e^{-tf(x)}\le  e^{-t\left(f_{\min}\right)}$$
Be careful to note that $f$ can attain its minimum at more than one point on $[0,1]$.
A: You have already established that the expression is less than or equal to $-L$ for any $t>0$, so the same must be true in the limit as well.
Now fix $\varepsilon>0$ and let $$I_{\varepsilon}\equiv\{x\in[0,1]\,|\,f(x)<L+\varepsilon\}.$$ Note that the set $I_{\varepsilon}$ has positive Lebesgue measure. This is because $f$ is continuous, so whenever it attains its minimum value $L$, there will be a small open interval around the minimum-attaining point that is contained in $I_{\varepsilon}$.
Given that the exponential function returns positive values, integrating on a smaller domain (weakly) decreases the value of the integral. Therefore, one can establish the following:
\begin{align*}
\frac{1}{t}\ln\left(\int_0^1e^{-tf(x)}\,\mathrm dx\right)&\geq\frac{1}{t}\ln\left(\int_{I_{\varepsilon}}e^{-tf(x)}\,\mathrm dx\right)\\&\geq\frac{1}{t}\ln\left(\int_{I_{\varepsilon}}e^{-t(L+\varepsilon)}\,\mathrm dx\right)=\frac{\ln\left(\int_{I_{\varepsilon}}\,\mathrm dx\right)-t(L+\varepsilon)}{t}.
\end{align*}
According to the observation made above, $\int_{I_{\varepsilon}}\,\mathrm dx>0$, so taking its logarithm is meaningful. Taking limit as $t\to\infty$, one can conclude that $$\liminf_{t\to\infty}\left\{\frac{1}{t}\ln\left(\int_0^1e^{-tf(x)}\,\mathrm dx\right)\right\}\geq-L-\varepsilon.\tag{$\star$}$$ Take $\varepsilon$ as small as you like and you’re done.

Technically, we took $\liminf$ in $(\star)$ instead of $\lim$, because we hadn’t yet known whether the limit existed to begin with. We can then put together the pieces of the puzzle to conclude:
\begin{align*}
-L\geq\limsup_{t\to\infty}\,(\cdots)\geq\liminf_{t\to\infty}\,(\cdots)\geq-L.
\end{align*}
Therefore, the limits superior and inferior share the common value $-L$, which is then the limit of the expression. Proceeding this way, not only have we found the value of the limit but we have also established that it exists in the first place.
