Q: Suppose $\varphi$ is a real valued function on $\mathbb{R}$ such that: \begin{align*} \varphi \left[ \int_0^1 f(x) \, dx \right] &\le \int_0^1 \varphi [f(x)] \, dx \\ \end{align*} for every real valued bounded measurable function $f$ on $[0,1]$. Prove that $\varphi$ is convex.
My attempt:
Consider for fixed $x_1, x_2 \in \mathbb{R}$:
\begin{align*} f(\theta) &= \theta x_1 + (1 - \theta) x_2 \\ \int_0^1 f(\theta) \, d\theta &= x_1 \int_0^1 \theta \, d\theta + x^2 \int_0^1 (1 - \theta) \, d\theta = \frac{x_1 + x_2}{2} \\ \end{align*}
Plugging this into the inequality given in the problem:
\begin{align*} \varphi \left[ \frac{x_1 + x_2}{2} \right] &\le \int_0^1 \varphi [\theta x_1 + (1 - \theta) x_2] \, d\theta \\ \end{align*}
From here, I'm stuck. If I can show the midpoint property:
\begin{align*} \varphi \left( \frac{x_1 + x_2}{2} \right) &\le \frac{\varphi(x_1) + \varphi(x_2)}{2} \\ \end{align*}
I know proofs from there that $\varphi$ is convex. But I don't see how to get that.