I need help understanding the proof for Jensen's inequality in "Real and Complex Analysis" by Rudin.
3.3 Theorem (Jensen's Inequality) Let $\mu$ be a positive measure on a $\sigma$-algebra $\mathfrak M$ in a set $\Omega$, so that $\mu(\Omega)=1$. If $f$ is a real function in $L^1(\mu)$, if $a<f(x)<b$ for all $x\in\Omega$, and if $\varphi$ is convex on $(a,b)$, then $$\varphi(\int_\Omega f\,d\mu)\leq\int_\Omega(\varphi\circ f)\,d\mu$$
PROOF Put t=$f_{\Omega}fd\mu$. Then $a<t<b$. If $\beta$ is the supremum of the left side of $$\frac{\varphi(t)-\varphi(s)}{t-s}\leq\frac{\varphi(u)-\varphi(t)}{u-t}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(a<s<t<u<b)$$
It follows that $$\varphi(s)\geq\varphi(t)+\beta(s-t)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(a<s<b)$$ Hence $$\varphi(f(x))-\varphi(t)-\beta(f(x)-t)\geq 0\,\,\,\,\,\,\,\,\,\,\,\,(3)$$ for every $x\in\Omega$. Since $\varphi$ is continuous, $\varphi\circ f$ is measurable. If we integrate both sides of $(3)$ with respect to $\mu$ we get the inequality.
My first question is whether $t$ should be $t=\int_\Omega fd\mu$. It appears to be so, looking at this question Help with proof of Jensens inequality
. My second question is what does $\beta(s-t)$ mean? Does it mean $$\left(\sup \left(\frac{\varphi(t)-\varphi(s)}{t-s}\right)\right)\cdot(s-t)$$Or does it mean something like $$\sup \left(\frac{\varphi(t)-\varphi(s-t)}{t-s}\right)$$
My third question is why does $s$ change from being $a<s<t$ to $a<s<b$.My last question is how do we obtain Jensen's inequality by integrating both sides of $(3)$ with respect to $\mu$. Integrating both sides of $(3)$ yields
$$\int_\Omega\varphi(f(x))d\mu-\int_\Omega\varphi(t)d\mu-\int_\Omega\beta(f(x)-t)d\mu\geq 0$$
$$\Rightarrow \int_\Omega\varphi(f(x))d\mu-\int_\Omega\varphi(\int_\Omega f d\mu)d\mu-\int_\Omega\beta(f(x)-\int_\Omega f d\mu)d\mu\geq 0$$ I don't know how to proceed from here. Should I treat $\varphi(t)$ as a constant? so that $$\int_\Omega\varphi(\int_\Omega f d\mu)d\mu\Rightarrow \varphi(\int_\Omega f d\mu)\mu(\Omega)\Rightarrow\varphi(\int_\Omega f d\mu)$$
What about
$$\int_\Omega\beta(f(x)-\int_\Omega f d\mu)d\mu$$
Best Answer
t is the integral, as you surmise.
β is a number, not a function. Specifically, it must refer to the largest value taken by the left hand of the inequality which follows.
The desired inequality follows by integration, as the passage suggests, as β∫fdμ = βt so the final two terms cancel after integration.