Question
Prove that $\int f \ln(f) d \mu = \sup \left \{ \int f \phi d \mu : \int e^{\phi} d\mu \leq 1 \right \}$
With $f$ verifying $ \int f d \mu = 1 $ and $ f \cdot \ln(f) $ is integrable, with $ \phi $ a real measurable function s.t. $ f \phi $ is $\mu$-integrable
Answer
1- $\int f \cdot \ln ( \frac{e^{\phi}}{f}) d \mu = \int f \cdot \ln(e^{\phi}) - f \cdot \ln(f) d\mu = \int f \phi - f \cdot \ln(f) d\mu = \int f \phi - \int f \cdot \ln(f) d\mu$ When the last equality came from the fact that by assumption $ f \cdot \ln(f) $ and $\int f \phi d \mu$ are integrable.
2- Because we have that: $d \nu = f d \mu $ we can write: $\ln (\int e^{\phi} d \mu) = \ln (\int e^{\phi} \frac{f }{f} d \mu) = \ln (\int e^{\phi} \frac{1 }{f} d \nu) $.
$\ln(.)$ is a concave function hence by Jensen inequality we have $ \ln (\int e^{\phi} d \mu) = \ln (\int e^{\phi} \frac{1 }{f} d \nu) \geq \int \ln(e^{\phi} \frac{1 }{f}) d \nu = \int \ln(e^{\phi} \frac{1 }{f}) f d \mu$
3- Let prove that $ \int f \ln(f) d \mu \geq \int f \phi d \mu \Leftrightarrow \int f \ln(f) - f \phi d \mu \leq 0$
(i)Using "2-" we have on one side : $\ln (\int e^{\phi} d \mu) \geq \int \ln(e^{\phi} \frac{1 }{f}) f d \mu = \cdots = \int f \phi d \mu - \int f \ln(f) d\mu $
(ii) While on the other side because $\ln(x) \geq 0$ for $x \geq 1 $ so for $ \int e^{\phi} d\mu \leq 1 \Rightarrow \ln ( \int e^{\phi} d\mu) \leq 0 $
Combining (i) and (ii) we have that $ 0 \geq \ln (\int e^{\phi} d \mu) \geq \int f \phi d \mu - \int f \ln(f) d\mu $
4- If $\phi = \ln(f) $ there is equality. So indeed it is a supremum.
Q.E.D.
Is it correct? Mostly my part "4 -" I have a doubt on is my justification good enough?
Thank you
According to https://math.stackexchange.com/users/915356/snoop (this remark is in the comment of my post)
You proved that $∫ϕfdμ≤∫fln(f)dμ$ for all real $∫ϕfdμ∈{∫ϕfdμ:∫eϕdμ≤1}=:F$
Therefore, $supF≤∫fln(f)dμ$
Since $∫fln(f)dμ∈R$ and $∫eln(f)dμ=1$ we have $∫fln(f)dμ∈F$ and so $∫fln(f)dμ≤supF$. It follows that $∫fln(f)dμ=supF$