Convex functions, prove another definition. I have the next problem: Suppose $f$ is continuos then $f:\mathbb{R}^n\to R$ is convex iff $\forall x,y \in \mathbb{R}^n \left( \int_0^1 f(x+\theta (y-x))d\theta \leq \dfrac{f(x)+f(y)}{2}\right)$.
The implication $\Rightarrow$ is really easy but I have problems with the other way any suggestions are welcomed.
 A: If $f$ is convex it is straightforward to prove the desired inequality. Conversely, we need to notice that the right-hand side of your inequality is
$$
\int_0^1 \left((1-\theta)f(x) + \theta f(y)\right)\mathrm{d}\theta = \frac{f(x)+f(y)}{2}\tag{1}
$$
So then, we have
$$
\begin{align}
&\int_0^1 f(x+\theta (y-x)) \mathrm{d}\theta \leq \int_0^1 \left((1-\theta)f(x) + \theta f(y)\right)\mathrm{d}\theta \\
\Leftrightarrow& \int_0^1 \left[ (1-\theta)f(x) + \theta f(y) - f(x+\theta (y-x))  \right] \mathrm{d}\theta \geq 0
\end{align}
$$
which means that $f$ is convex. Actually, it means that $(1-\theta)f(x) + \theta f(y) \geq f(x+\theta (y-x))$ for almost all $\theta \in [0,1]$, but you said that $f$ is continuous, so it holds for all $\theta\in[0,1]$.
There is another interesting inequality we may prove: You may use the integral form of Jensen's inequality according to which if $f$ is convex and $g$ and $f\circ g$ are integrable on $[0,1]$ then
$$
f\left(\int_0^1 g(\theta)\mathrm{d}\theta\right) \leq \int_0^1 f(g(\theta))\mathrm{d}\theta\tag{2}
$$
Here, choose $g(\theta) = x + \theta(y-x)$; we have
$$
\int_0^1 g(\theta)\mathrm{d}\theta = \int_0^1 (x + \theta(y-x))\mathrm{d}\theta = \frac{x+y}{2}\tag{3}
$$
Then, using (3) and recalling (1) we have the double-sided inequality
$$
f\left(\frac{x+y}{2}\right) \leq \int_0^1 f(\theta x + (1-\theta)y)\mathrm{d}\theta \leq \frac{f(x)+f(y)}{2}.\tag{4}
$$
Assuming that $f$ is continuous, then $f$ is convex if and only if at least one of these inequalities holds.
