$\int (\ln f) \, dx$ and $\ln(\int f \, dx)$

Question

Let $f \in C^0([0,1], \mathbb{R}_+^*)$

Compare $\int_0 ^1 \ln f \, dx$ and $\ln(\int_0^1 f \, dx)$

I think I have to use the Riemann sum so that's what I've tried:

$\frac{1}{n}\sum_{k=1}^n \ln f(\frac{k}{n}) \rightarrow \int_0^1\ln f(x) dx$

$\ln (\frac{1}{n}\sum_{k=1}^n f(\frac{k}{n})) \rightarrow \ln \int_0^1 f(x) dx$

So now I should be able to compare both sums and then conclude. I'm now sure how to compare them though. Any idea?

Answer 1

Consider any grid $\{a_1,\ldots,a_n\}$ over $[0,1]$. Then

$$ \frac{1}{n} \sum \log a_i = \log (\sqrt[n]{\Pi a_i}) \le \log (\frac{1}{n}\sum a_i) $$

Where the inequality follows from the AM-GM and $\log$ being a monotonically increasing function.

Answer 2

The function $g(t)=\log t$ is a concave function since $g''(t)=-\frac1{t^2}<0$, so for every convex combination $\sum_{k=1}^n\theta_k=1$, where $\theta_k\ge 0$, and for all $t_k>0$, $k=1,\dots, n$, you have $$\log\left(\sum_{k=1}^n\theta_kt_k\right)\ge \sum_{k=1}^n\theta_k\log t_k.$$ Use this inequality and Riemann sums.