Integral inequality: $\forall t>0, \frac{1}{t}\ln(\int_{0}^{1}e^{-tf(x)}dx)\leq-\min f(x)$

Question

Given a continuous function $f:[0,1]\to\mathbb R$, prove that $$\forall t>0, \frac{1}{t}\cdot\ln\left(\int_{0}^{1}e^{-tf(x)}dx\right)\leq-\min f(x).$$

I have no idea where to begin. Thought I could use the FTOC to come up with some form of antiderivative but I don't know if this is even the right intuition to approach this.

Answer 1

The inequality says $$ \int_0^1 e^{-tf(x)} \, dx \leq e^{-t\min_x f(x)}. $$

Since $t>0,$ the quantity $e^{-t\min_x f(x)}$ is the largest value of the function $e^{-tf(x)}$ within $0\le x\le 1.$ So this just says that

$$ \int_0^1 g(x) \,dx \le (1-0)\times \max\{ g(x) : 0\le x\le 1 \}. $$

Answer 2

A good first approach might be exploring what happens for constant functions $f$, or for something simplie like $f(x) = ax$.

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Note that

$$\int_0^1 e^{-t f(x)} dx \le \int_0^1 e^{-t \min f} dx = e^{-t \min f}$$

so that the left hand side of your inequality is

$$\frac 1 t \ln \left(\int_0^1 e^{-t f(x)} dx\right) \le \frac 1 t \ln e^{-t \min f} = - \min f. $$