Let $E$ be a measurable set with $m(E)>0$.
Let $f$, $\gamma$ be two measurable, real-valued function which are finite a.e. on $E$ with $f, \gamma$ and $f\cdot \gamma$ all integrable.
Assume $\gamma\geq 0$ and $\int_{E}\gamma>0$. Then if $\phi$ is a convex function on an open interval $(a,b)$ containing the range of $f$, then we have
$\phi\Big(\dfrac{\int_{E}f\cdot\gamma}{\int_{E}\gamma}\Big)\leq\Big(\dfrac{\int_{E}\phi(f)\cdot\gamma}{\int_{E}\gamma}\Big)$
I am trying to modify the proof of the simpler version (the one with only $f$) of Jensen's inequality to this more general version, but I don't really know how to deal with the new function $\gamma$.
Any hints or detailed explanations are really appreciated!!
${\bf Edit\ 1:}$
I got some but still limited development.
To make things simpler, I assume $\phi$ has $1^{st}$ derivative, and assume $\int_{E}\gamma=1$
As $\phi$ is convex, for each $a$, we have $\phi(y)>(y-a)\phi'(a)+\phi(a)$
so in particular, as $\gamma\geq 0$, we have
$\int \phi\big(f(x)\big)\cdot\gamma(x)dx\geq\int\big[(f(x)-a)\phi'(a)+\phi(a)\big]\gamma(x)dx=\int \big(f(x)-a\big)\gamma(x)dx\times \phi'(a)+\phi(a)$
Now, let $a=\int f(x)\cdot\gamma(x)dx$, we will have $\int \big(f(x)-a\big)\gamma(x)dx=0$ and we are done in this simple case.
In a general case, a convex function on the real line has only right derivative, and thus we have $\phi(y)>(y-a)\lim_{h\rightarrow 0^{+}}\dfrac{\phi(a+h)-\phi(a)}{h}+\phi(a)$, so with the complicated notation, our proof is still okay.
However, how do I expand this case to the case where $\int_{E}\gamma$ is not 1?
$\textbf{My Completed Proof:}$
Okay, I think I proved it.
To make things simpler, let $\int_{E}\gamma=b\geq 0$ and suppose $\phi(x)$ has the first derivative.
As $\phi$ is convex, for each $a$, we have $$\phi(y)>(y-a)\phi'(a)+\phi(a)$$
So in particular, as $\gamma\geq 0$, we have:
$\dfrac{\int_{E}\phi(f(x))\gamma(x)dx}{b}\geq\dfrac{\int_{E} [(f(x)-a)\phi'(a)+\phi(a)]\gamma(x)dx}{b}=\dfrac{\phi'(a)}{b}\int_{E}f(x)\gamma(x)dx-a\phi'(a)+\phi(a)$
Now, let $a=\dfrac{\int_{E}f(x)\gamma(x)dx}{b}$, the first two summands of the above equation will be killed, and we get what we want.
So in general case, convex function has only right derivatives, so we have $$\phi(y)>(y-a)\lim_{h\rightarrow 0^{+}}\dfrac{\phi(a+h)-\phi(a)}{h}+\phi(a)$$
By letting $F(a)=\lim_{h\rightarrow 0^{+}}\dfrac{\phi(a+h)-\phi(a)}{h}$ our proof is still valid, since $F(a)$ will be killed in the end as what happened to $\phi'(x)$, then we are done.
Is my proof okay here? Feel free to point out my mistake or better proof!
Thank you!
Best Answer
Most of the answer is in my post, $\textbf{ My Completed Proof}$ part. Only three minor points need to be added.
Firstly, in the third line of my proof, we need to comment that $a$ in is the domain of $\phi$.
Secondly, in the sixth line, we need to note that $\phi(f(x))$ makes sense since by hypothesis, the domain of $\phi$ contains the range of $f(x)$.
Finally, $a=\dfrac{\int_{E}f(x)\gamma(x)dx}{b}$ makes sense because $\gamma\geq 0$, so we can enlarge or reduce $f$ to its max or min so that $a$ is in the range of $f(x)$.
With all those details added, the proof in my post is complete.