4

I got some useful information from this Question: Jensen's inequality in measure theory

Theorem 3.1 Jensen's Inequality

Let $(X,\mathcal{M},\mu)$ be a probability space (a measure space with $\mu(X) = 1$ ), $f: X \to \mathbb R \in L^1(X, \mu)$, and $\psi:\mathbb R \to \mathbb R $ be a convex function, then $$\psi\int_X f d\mu \le \int_X (\psi \circ f)d\mu$$

And that question asked whether Jensen's inequality still hold in general finite measure space ? A nice man d.k.o. answered:

Yes. In this case for convex $\varphi$ :$$\varphi\left(\frac{1}{\mu(X)}\int fd\mu\right)\le \frac{1}{\mu(X)}\int \varphi\circ fd\mu$$

However, this result is basically rescale $\mu$ to a probability measure.

So whether the following proposition hold?

Let $(X,\mathcal{M},\mu)$ be a general measure space, and $\mu(X) < \infty $,
$f: X \to \mathbb R \in L^1(X, \mu)$, and $\psi:\mathbb R \to \mathbb R $ be a convex function, then $$\psi\int_X f d\mu \le \int_X (\psi \circ f)d\mu$$

Tiger Zhao
  • 105

2 Answers2

4

No. Indeed, Jensen's inequality in its basic form only holds if $\mu$ is a probability measure. Setting $f=1$ shows that we have $\psi(\mu(X)) \le \psi(1) \mu(X)$ for every convex function $\psi$. If $\mu(X) \ne 1$ then we could take $\psi$ to be a linear function with $\psi(1) = 0$ and $\psi(\mu(X)) > 0$, yielding a contradiction.

Nate Eldredge
  • 97,710
4

I am very appreciated for the answer given by Nate Eldredge and Keba.
And I carefully reviewed proof process of Jensen's Inequality and found Why "Jensen's inequality Do Not hold in general finite measure space".
I write this thought down for anyone who has the same confusion.

In the original Proof:

Proof:

Since $\psi$ is convex, at each $x_0 \in \mathbb R$, there exist $a,b \in \mathbb R$ such that $\psi(x_0) = ax_0 + b$ and $\psi(x) \ge ax + b, \forall x \in \mathbb R$, (here, $y = ax + b$ defines a supporting plane of the epigraph of $\psi$ at $x_0$). Let $x_0 = \int_X fdµ$, then we have $$\psi(\int_Xf d\mu) = \psi(x_0) = ax_0+b=a\int_Xf\mu + b = \int(af+b)d\mu \le \int(\psi\circ f)d\mu$$, q. e. d.

When $\mu$ is a general finite measure, below equation do not hold:
$$a\int_Xf\mu + b = \int(af+b)d\mu $$ Specifically, $$b \neq \int b d\mu $$

In other words, below equations hold only when $\mu$ is a probability measure: $$\int_X c\ d\mu = c , \ (c\ is\ constant)$$ $$E[E(x)] = E(x)$$

Tiger Zhao
  • 105