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?

  • 1
    $\begingroup$ This property follows from the concavity of $s\to\ln s$. Just google "Jensen's inequality".. $\endgroup$ – GReyes Mar 4 at 1:10
  • $\begingroup$ See en.wikipedia.org/wiki/… $\endgroup$ – FDP Mar 4 at 1:14
  • $\begingroup$ @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? $\endgroup$ – Liv Mar 4 at 1:18
  • $\begingroup$ 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 $1/n$. $\endgroup$ – 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 $f$.


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