# Real Analysis How to Prove Properties of Natural Logs and Integrals

Here is the question I am trying to answer:

Let $$f:[0,1] \to\Bbb R$$ be a Riemann integrable function with $$f \ge c>0$$. Prove that $$\int_0^1\ln(f(x))\ dx\le \ln\left(\int_0^1 f(x)\ dx\right).$$

I understand how to prove two integrals are equal by showing that their upper and lower Darboux sums are equal and that they converge to the same definite integral. But, I don't understand how to prove the less than or equal to part.

Ideas I've thought about include: integration by parts, improper integrals, partitions.

Does anyone know how to prove this?

• This property follows from the concavity of $s\to\ln s$. Just google "Jensen's inequality".. – GReyes Mar 4 at 1:10
• – FDP Mar 4 at 1:14
• @GReyes thank you for responding. I find the wikipedia page a bit hard to follow, as this is only my second class in real analysis. Do you think you could elaborate more? – Liv Mar 4 at 1:18
• Rafael Bailo gave a detailed proof. Just a comment: Jensen's inequality is just a "continuous" version of the usual inequality that defines convexity, using convex combinations of points. The combinations in the continuous cases are, essentially, the Riemann sums with weights $1/n$. – GReyes Mar 4 at 2:05

Jensen's inequality holds for an interval $$[a,b]$$, an integrable nonnegative function $$f$$ from $$[a,b]$$ to the real line and a convex function $$\varphi$$, and it states: $$\varphi\left(\int_a^bf(x) dx\right)\leq \int_a^b\varphi\left(f(x)\right)dx.$$ This does not apply to the $$\log(x)$$ function because it is concave away from zero, but this means the function $$(-\log(x))$$ is convex away from zero. Applying Jensen's: $$-\log\left(\int_a^bf(x) dx\right)\leq -\int_a^b \log\left(f(x)\right)dx,$$ which yields $$\int_a^b \log\left(f(x)\right)dx \leq \log\left(\int_a^bf(x) dx\right).$$ Zero is not a problem point because of your assumptions on $$f$$.