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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 $fcdot gamma$ all integrable.

Assume $gammageq 0$ 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

$phiBig(dfrac{int_{E}fcdotgamma}{int_{E}gamma}Big)leqBig(dfrac{int_{E}phi(f)cdotgamma}{int_{E}gamma}Big)$

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 $gammageq 0$, we have

$int phibig(f(x)big)cdotgamma(x)dxgeqintbig[(f(x)-a)phi'(a)+phi(a)big]gamma(x)dx=int big(f(x)-abig)gamma(x)dxtimes phi'(a)+phi(a)$

Now, let $a=int f(x)cdotgamma(x)dx$, we will have $int big(f(x)-abig)gamma(x)dx=0$ 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 $phi(y)>(y-a)lim_{hrightarrow 0^{+}}dfrac{phi(a+h)-phi(a)}{h}+phi(a)$, 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 $int_{E}gamma=bgeq 0$ 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 $gammageq 0$, we have:

$dfrac{int_{E}phi(f(x))gamma(x)dx}{b}geqdfrac{int_{E} [(f(x)-a)phi'(a)+phi(a)]gamma(x)dx}{b}=dfrac{phi'(a)}{b}int_{E}f(x)gamma(x)dx-aphi'(a)+phi(a)$

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 $$phi(y)>(y-a)lim_{hrightarrow 0^{+}}dfrac{phi(a+h)-phi(a)}{h}+phi(a)$$

By letting $F(a)=lim_{hrightarrow 0^{+}}dfrac{phi(a+h)-phi(a)}{h}$ 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!

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 $gammageq 0$, 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.