# Variation of Jensen-Inequality

I just read a variation of Jensen's Inequality which states:

If $f: \mathbb{R} \rightarrow \mathbb{R}$ is a convex function, $\phi \in \mathcal{L}^1(\mathbb{R}^n)$ with $\phi \geq 0$ and $\int \phi dx=1$ then:

$$f \Big( \int_{\mathbb{R}^n} u(x)\phi(x)dx \Big) \leq \int_{\mathbb{R}^n} f(u(x))\phi(x)dx$$

for any $u: \mathbb{R}^n \rightarrow \mathbb{R}$ for which the integral makes sense. 

They don't offer a proof and I was not able to solve this. Does anyone have proof which is based on the "original" Jensen-Inequality?

It is the original Jensen's inequality applied to the real line endowed with the probability measure $\mu(A):=\int_A\phi(x)\mathrm dx$ (the assumptions guarantee that $\mu$ is a probability measure).

Assume, to make things simpler, that $f$ has a derivative.

As $f$ is convex, for each $a$:

$$f(y) \ge (y-a)f'(a) + f(a)$$

so in particular, as $\phi\ge 0$: $$\int f(u(x))\phi(x) dx \ge \int \left[ (u(x)-a)f'(a) + f(a)\right] \phi(x) dx \\= \int (u(x)-a) \phi(x) dx \times f'(a) + f(a)$$

Now with $a = \int u(x)\phi(x) dx$, $\int (u(x)-a) \phi(x) dx = 0$ and you get your result.

In the general case, a convex function on the real line has a right derivative, and you also have the inequality $$f(y) \ge (y-a) \lim_{h\downarrow 0} \frac{f(a+h)-f(a)}h + f(a)$$ and so the proof remains valid.