In a book on PDEs that I'm reading, I am trying to fill in the details of the proof of Jensen's Theorem, and am having a little trouble with the algebra.
Here is the statement of Jensen's Theorem in the book:
Assume $f:\mathbb{R}^{m}\to \mathbb{R}$ is convex and $U\subset \mathbb{R}^{n}$ is open, bounded. Let $\mathbf{u}:U\to \mathbb{R}^{m}$ be summable. Then, $f\left(\oint_{U}\mathbf{u}dx\right) \leq \oint_{U}f(\mathbf{u})dx$
(In this case, though, $\oint$ is not the regular contour integral, it actually means the "average integral", $\frac{1}{|U|}\int_{U}\mathbf{u}dx=$ average of $\mathbf{u}$ over $U$. It's normally written as a regular integral sign with a slash through it, but I don't know the tag for it.)
Here is their proof:
Since $f$ is convex, for each $p \in \mathbb{R}^{m}$, there exists $r \in \mathbb{R}^{m}$ such that
$f(q) \geq f(p)+r\cdot (q-p)$ for all $q \in \mathbb{R}^{m}$.
Let $p=\oint_{U}\mathbf{u}dy$, $q=\mathbf{u}(x)$:
$f(\mathbf{u}(x))\geq f\left(\oint_{U}\mathbf{u}dy\right)+r\cdot \left(\mathbf{u}(x)-\oint_{U}\mathbf{u}dy\right)$.
Integrate with respect to $x$ over $U$. $\square$
So, here's what I've done:
$f(\mathbf{u}(x))\geq f\left(\oint_{U}\mathbf{u}dy\right)+r\cdot \left(\mathbf{u}(x)-\oint_{U}\mathbf{u}dy\right) \to \\ f(\mathbf{u}(x)) \geq f\left(\oint_{U}\mathbf{u}dy\right)+r\mathbf{u}(x)-r\oint_{U}\mathbf{u}dy \to \\ \oint_{U}f(\mathbf{u}(x))dx \geq \oint_{U}\left[f\left(\oint_{U}\mathbf{u}dy\right)\right]dx + r\oint_{U}\mathbf{u}(x)dx - r\oint_{U}\left(\oint_{U}\mathbf{u}dy\right)dx$.
Now, could someone help me figure out how to make the right side turn into $f\left(\oint_{U}\mathbf{u}dx\right)$? Thank you!
You just need to note that
$$\oint_U \mathbf{u}\,dx$$
is a constant, and since you divide by the measure of $U$ for the average, you have
$$\oint_U c\,dx = c$$
for any constant $c$. So
$$\begin{align} \oint_U\underbrace{f\left(\oint_U \mathbf{u}\,dy\right)}_{c = f(p)}\,dx &= f\left(\oint_U \mathbf{u}\,dy\right) = f(p),\\ \oint_U \underbrace{\left(\oint_U \mathbf{u}\,dy\right)}_p\,dx &= p, \end{align}$$
and your right hand side simplifies to
$$f(p) + r\cdot p - r\cdot p = f(p) = f\left(\oint_U \mathbf{u}\,dx\right),$$
as desired.