Prove that for any continuous function there exists a set of points satisfying a given criteria

Question

Prove that for any continuous function $f : [0, 1] \to (0, \infty)$ and for any natural number $n \geq 1$ there exist $x_1, x_2, ..., x_n \in [0, 1]$ with $0 \leq x_i - x_{i-1} \leq 2/n$ such that: $$e^{\int_0^1 \ln f(x) dx} \leq \frac{1}{n} \sum_{k=1}^n f(x_k)$$

I've noticed that the right-hand side of the inequality is a Riemann sum, which equals $\int_0^1 f(x) dx$. So, the problem would reduce to proving that: $$\int_0^1 \ln f(x) dx \leq \ln \int_0^1 f(x) dx$$

But I can't figure out how to prove this. Thank you!

Answer 1

$f$ is continuous on the compact $[0,1]$, such that $f$ is Riemann integrable. Therefore $S_n = \frac{1}{n} \sum_{k=1}^n f(x_k)$ converges to $\int_{0}^1 f(x)dx$ as $n \rightarrow \infty$, where $x_k = \frac{k}{n}$. Suppose that $f$ is non-constant a.e., then Jensen's inequality yields.

\begin{align}\int_0^1 \ln f(x) dx < \ln \int_0^1 f(x) dx\end{align}

Therefore it's quite obvious that given $N$ big enough, one has $S_N > \exp{\int_0^1 \ln f(x) dx}$

If $f = k$ is constant a.e., then the solution is trivial. Here, we must be careful to separate both cases, as the strict inequality was vital so that $S_n$ necessarily gets above the left-hand side as $n$ gets big enough.