Suppose $\mu(\Omega) = 1$, and $h: \Omega\to [0,\infty]$ is measurable. If $A = \int_\Omega h\, d\mu$, prove that $$\sqrt{1+A^2} \le \int_\Omega \sqrt{1+h^2}\, d\mu$$ and that equality holds iff $h = A$ a.e.
So I chose $\varphi(x) = \sqrt{1+x^2}$ and used Jensen's inequality to get the result. It is also clear that equality holds if $h = A$ almost everywhere. However, how does equality imply $h = A$ a.e.?
I guess what I'm really asking is, when does equality hold in Jensen's inequality? From what I understand, equality implies that the inequality (3) below in Rudin, is an equality a.e. Hence, $$ \varphi(f(x)) - \varphi(t) = \beta (f(x) - t) \quad \text{a.e.}$$ Would this mean that $\varphi$ is an affine function? If yes, it'd be of the form $\varphi(x) = Bx + C$. That's weird though, since $\sqrt{1+x^2} = Bx + C$ is impossible unless $x = 0$, and $C = 1$. What am I missing? Could someone please clarify this in detail?
Thank you!
For reference:
Jensen's inequality from Rudin:
Here's 3.1(2):
You were right, it was not so obvious.
$u(s) = \varphi(s) - \varphi(t) - \beta(s - t).$ Then $u \geq 0.$ Consider $\mathrm{A} = \{f = t\}$ and $\mathrm{B} = \{f \neq t\}.$ When $\varphi$ is strictly convex (as in your case), then $u \circ f = 0$ on $\mathrm{A}$ and $u \circ f > 0$ on $\mathrm{B}.$ However, when $\int\limits_\Omega \varphi \circ f d\mu = \varphi \left( \int\limits_\Omega f d \mu \right),$ the integral of $u \circ f$ is zero, this implies $$ 0 = \int\limits_\Omega u \circ f d\mu = \int\limits_\mathrm{B} u \circ f d\mu $$ and since the integrand in the last integral is positive while the integral is zero, we see that the set we are integrating against must be null relative to $\mu.$ Q.E.D.