Question about convex property in Jensen's inequality

[psie]

TL;DR

I am reading a proof of Jensen's inequality. I am getting stuck on an "elementary property" of convex functions.

I am reading a proof of Jensen's inequality. The proof goes like this.

Theorem 4.3: Let $(\Omega, \mathcal A,\mu)$ be a probability space and let $\varphi:\mathbb R\to\mathbb R_+$ be a convex function. Then for every $f\in L^1(\Omega, \mathcal A,\mu)$, $$\int_\Omega \varphi\circ f\, d\mu\geq\varphi\left(\int_\Omega f\, d\mu\right).$$ Proof: Set $$\mathcal E_\varphi=\{(a,b)\in\mathbb R^2:\forall x\in\mathbb R,\varphi(x)\geq ax+b\}.$$ Then by elementary properties of convex functions, $$\varphi \left(x\right)=\sup_{\left(a{,}b\right)\in \mathcal E_{\varphi }}\left(ax+b\right).\tag1$$ ... ... ...

I do not know much about convex functions, but why does (1) hold?

The definition of convex I'm using is that $$\varphi(tx+(1-t)y)\leq t\varphi(x)+(1-t)\varphi(y)$$ holds for all $x,y\in\mathbb R$ and all $t\in[0,1]$.

 

[psie]

From here, I found the answer. However I still have some questions:

If $\phi$ is convex, for each point $(\alpha, \phi(\alpha))$, there exists an affine function $f_\alpha(x) = a_\alpha x + b_\alpha$ such that
- the line $L_\alpha$ corresponding to $f_\alpha$ passes through $(\alpha, \phi(\alpha))$;
- the graph $\phi$ lies above $L_\alpha$.
Let $A = \{f_\alpha: \alpha \in \mathbb{R}\}$ be the set of all such functions. We have
- $\sup_{f_\alpha \in A} f_\alpha(x) \geq f_x(x) = \phi(x)$ because $f_x$ passes through $(x, \phi(x))$;
- $\sup_{f_\alpha \in A} f_\alpha(x) \leq \phi(x)$ because all $f_\alpha$ lies below $\phi$.

How does one show that there exist such an affine function $f_\alpha(x) = a_\alpha x + b_\alpha$ with those properties given the definition I gave above?

 

[fresh_42]

A function is convex iff its epigraph is convex. In case of differentiability, I assume we can simply use the tangents at $(a, \varphi (a)).$ The difficulty is the points that do not have one. My next assumption is, that such points have at least one-sided tangents that will do. These ideas are based on the functions $x^2$ and $|x|$ and I think they are typical.

Eine auf einem offenen Intervall definierte, konvexe bzw. konkave Funktion ist lokal Lipschitz-stetig und somit nach dem Satz von Rademacher fast überall differenzierbar. Sie ist in jedem Punkt links- und rechtsseitig differenzierbar.

(A convex or concave function on an open interval is locally Lipschitz continuous and by Rademacher's theorem almost everywhere differentiable. It is at each point left- and right differentiable.)

Looks like my intuition is correct but the proof needs some consideration.

 

[psie]

Here is Durret's proof of the inequality (in his book Probability: Theory and Examples). He uses $\phi$ to denote the function $\varphi$ in the theorem in my original post:

Proof. Let $c=\int f \,d \mu$ and let $l(x)=ax+b$ be a linear function that has $l(c)= \phi(c)$ and $\phi(x) \geq l(x)$. To see that such a function exists, recall that convexity implies
$$\lim_{h \to 0^+} \frac{\phi(c)−\phi(c−h)}{h} \leq \lim_{h \to 0^+} \frac{\phi(c+h)−\phi(c)}{h}\tag2$$(The limits exist since the sequences are monotone.)

If we let $a$ be any number between the two limits and let $l(x) = a(x − c) + \phi(c)$, then $l$ has the desired properties. With the existence of $l$ established, the rest is easy. From the fact that if $g \leq f$ a.e., then $\int g\, d\mu \leq \int f \,d\mu$, we have
$$ \int \phi(f ) \,d\mu \geq \int (af + b) \,d\mu = a \int f \,d\mu + b = l\left(\int f \,d\mu \right)= \phi\left(\int f \,d\mu \right),$$ since $c = \int f \,d \mu$ and $l(c) = φ(c)$.

I can see why $(2)$ holds, but I do not see why the function $l(x) = a(x − c) + \phi(c)$ satisfies $\phi(x) \geq l(x)$. Is this clear to someone? Also, when Durret says "The limits exist since the sequences are monotone", which sequences does he mean?

 

[fresh_42]

Here is Durret's proof of the inequality (in his book Probability: Theory and Examples). He uses $\phi$ to denote the function $\varphi$ in the theorem in my original post:


I can see why $(2)$ holds, but I do not see why the function $l(x) = a(x − c) + \phi(c)$ satisfies $\phi(x) \geq l(x)$. Is this clear to someone? Also, when Durret says "The limits exist since the sequences are monotone", which sequences does he mean?

Isn't $l(x)=a(x − c) + \phi(c) =ax + \underbrace{(\phi(c)-c)}_{=b}\leq \phi(x)$ simply the definition of $\mathcal{E}_\varphi =\mathcal{E}_\phi\;$?

Durret probably means with sequence something like ...
$$
\lim_{h \to 0^+} f(h) = \lim_{n \to \infty} f(n^{-1})
$$