I'm trying to show the following inequality. Let $f,g:X\to (0,\infty)$ be measurable functions with $\int f\; d\mu=\int g \;d\mu=1$. Show that if $f \log g \in L^1(\mu)$, then
$$\int f \log g \;d\mu\leq \int f \log f \;d\mu$$
This seems like an application of Jensen's inequality but I'm having difficulty applying it correctly. I defined a measure $$d\nu=f\;d\mu$$ and so I need to show that
$$\int \log g \;d\nu\leq \int \log f \;d\nu$$
Note that $(X,\nu$) is a probability space. I need help recognizing the convex function $\phi$ and the $\nu-$integrable function $F$ onto which I can apply Jensen's inequality:
$$\phi(\int F\;d\nu)\leq \int \phi(F)\;d\nu$$
Any help with this is appreciated. Thanks for your time.
Best Answer
Special Case. The claim easily follows from Jensen's inequality if we also assume that $f\log f$ is in $L^1(\mu)$:
$$ \int \log\left(\frac{g}{f}\right) \, f \, \mathrm{d}\mu \overset{\text{(Jensen)}}{\leq} \log \left( \int \frac{g}{f} \, f \, \mathrm{d}\mu \right) = \log \left( \int g \, \mathrm{d}\mu \right) = 0. $$
General Case. Now we turn to the general case. If we are only given that $\int f \, \mathrm{d}\mu = 1$, then it is possible to have both $\int \left(f\log f\right)_+ \, \mathrm{d}\mu = \infty$ and $\int \left(f\log f\right)_- \, \mathrm{d}\mu = \infty$:
So, OP's assumption $f \log g \in L^1(\mu)$ will play a crucial role in ensuring that $\int f \log f \, \mathrm{d}\mu$ is well-defined. More generally, we will prove:
Proof. To begin with, we fix a compact subset $K$ of $(0, \infty)$ so that $\mu(\{f, g \in K\}) > 0$. Then, for any compact intervals $I = [a, b]$ and $J = [c, d]$ of $(0, \infty)$ containing $K$, we define the measure $\nu_{I, J}$ on $X$ by
$$ \nu_{I, J}(\mathrm{d}x) = \frac{f(x) \mathbf{1}_{\{ f \in I, g \in J \}}(x) \, \mu(\mathrm{d}x)}{\int f \mathbf{1}_{\{ f \in I, g \in J \}} \, \mathrm{d}\mu}. $$
Since $ \int f \mathbf{1}_{\{ f \in I, g \in J \}} \, \mathrm{d}\mu \geq \int f \mathbf{1}_{\{ f, g \in K \}} \, \mathrm{d}\mu > 0 $, this is indeed a well-defined probability measure. Moreover, the choice of truncation $\mathbf{1}_{\{ f \in I, g \in J \}}$ guarantees that $\log f$ and $\log g$ are in $L^1(\nu_{I, J})$. So by the Jensen's inequality,
\begin{align*} \int \log\left(\frac{g}{f}\right) \, \mathrm{d}\nu_{I, J} \leq \log\left( \int \frac{g}{f} \, \mathrm{d}\nu_{I, J} \right) = \log\left( \frac{\int g \mathbf{1}_{\{ f \in I, g \in J \}} \, \mathrm{d}\mu}{\int f \mathbf{1}_{\{ f \in I, g \in J \}} \, \mathrm{d}\mu} \right). \tag{1} \end{align*}
Rearranging $\text{(1)}$, we get
\begin{align*} \int \left( f \log g \right) \mathbf{1}_{\{ f \in I, g \in J \}} \, \mathrm{d}\mu &\leq \int \left( f \log f \right) \mathbf{1}_{\{ f \in I, g \in J \}} \, \mathrm{d}\mu + E_{I, J}, \tag{2} \end{align*}
where $E_{I,J}$ is defined by
$$ E_{I,J} = \left( \int f \mathbf{1}_{\{ f \in I, g \in J \}} \, \mathrm{d}\mu \right) \log\left( \frac{\int g \mathbf{1}_{\{ f \in I, g \in J \}} \, \mathrm{d}\mu}{\int f \mathbf{1}_{\{ f \in I, g \in J \}} \, \mathrm{d}\mu} \right). $$
By our choice of $K$, we can check that $E_{I, J}$ is bounded. Let $M$ be a bound of $E_{I, J}$. Now we example two cases:
Case 1. Suppose $\int \left(f \log g\right)_- \, \mathrm{d}\mu < \infty$. Then rearranging $\text{(2)}$, we get \begin{align*} \int \left( - f \log f \right) \mathbf{1}_{\{ f \in I, g \in J \}} \, \mathrm{d}\mu &\leq \int \left( - f \log g \right) \mathbf{1}_{\{ f \in I, g \in J \}} \, \mathrm{d}\mu + E_{I, J} \\ &\leq \int (f \log g)_- \, \mathrm{d}\mu + M, \end{align*} and so, \begin{align*} \int \left(f \log f \right)_- \mathbf{1}_{\{ f \in I, g \in J \}} \, \mathrm{d}\mu &\leq \int \left(f \log f \right)_+ \mathbf{1}_{\{ f \in I \}} \, \mathrm{d}\mu + \int (f \log g)_- \, \mathrm{d}\mu + M. \end{align*} Recall that $I = [a, b]$ and $J = [c, d]$. Assuming WLOG $b > 1$ and then letting $a, c \to 0^+$ and $d \to \infty$, MCT shows that \begin{align*} \int \left(f \log f \right)_- \, \mathrm{d}\mu &\leq \int \left(f \log f \right)_+ \mathbf{1}_{\{ f \leq b \}} \, \mathrm{d}\mu + \int (f \log g)_- \, \mathrm{d}\mu + M < \infty. \end{align*}
Case 2. Suppose $\int \left(f \log f\right)_+ \, \mathrm{d}\mu < \infty$. Then arguing similarly as in Case 1, we can show that $\int \left(f \log g\right)_+ \, \mathrm{d}\mu < \infty$.
Hence, the assumption implies that both $\int \left(f \log f\right) \, \mathrm{d}\mu $ and $\int \left(f \log g\right) \, \mathrm{d}\mu $ exist. Finally, by letting $a, c \to 0^+$ and $b, d \to \infty$ in $\text{(2)}$,
\begin{align*} \int f \log g \, \mathrm{d}\mu \leq \int f \log f \, \mathrm{d}\mu + \underbrace{ \left( \int f \, \mathrm{d}\mu \right) \log\left( \frac{\int g \, \mathrm{d}\mu}{\int f \, \mathrm{d}\mu} \right) }_{=0} \end{align*}
and therefore the desired claim follows.