Prove an integral inequation Prove that if a function is descending on [0; 1], then
$\forall\theta\in(0;1)\quad \theta\int\limits_0^1f(x)dx \leq \int\limits_0^\theta f(x)dx$.
If a see $\leq$, then there is an idea to use limits. I've tried to play with integral sums but all I've got is that $\int\limits_0^\theta f(x)dx \gt f(\theta)\theta$ (but it's obvious without integral sums).
 A: Hint Let $g(y):= \int_0^y f(x) dx$. Then $g(y)$ is concave down.
Apply Jensen's inequality with $x_1=0, x_2=1$ and $t=\theta$.
Added Let $0 < \theta <1$.
Then
$$
g(\theta)-g(0)=\int_{0}^\theta f(x)dx \geq \theta f(\theta) \\
g(1)-g(\theta)=\int_\theta^1f(x)dx \leq (1-\theta)f(\theta)
$$
Multiplying the first ineqaulity by $(1-\theta)$ and the second by $\theta$ gives your inequality.
If you replace $0,1$ by $x_0,x_1$ and $\theta$ by $t$ you get the Jensen inequality for $g$ under the assumption that $f$ is decreasing.
A: Without assuming that $f$ is differentiable, we can show that $F:[0,1] \to \mathbb{R}$ where $F(\theta) =\int_0^\theta f(x) \, dx$ is concave.
Assuming that $f$ is bounded we have that $F$ is continuous. Since $f$ is nonincreasing we have for $\theta_1 < \theta_2$
$$F(\theta_2) - F\left(\frac{\theta_1 + \theta_2}{2}\right) \leqslant F\left(\frac{\theta_1 + \theta_2}{2}\right) - F(\theta_1)$$
Hence,
$$F\left(\frac{\theta_1 + \theta_2}{2}\right) \geqslant \frac{1}{2} ( F(\theta_1) + F(\theta_2))$$
Since $F$ is continuous and midpoint concave it must be concave and for all $\theta \in (0,1)$,
$$\frac{F(\theta)}{\theta} =\frac{F(\theta)-F(0)}{\theta-0}\geqslant \frac{F(1)-F(0)}{1-0} = F(1)$$
Thus,
$$\frac{1}{\theta}\int_0^\theta f(x) \, dx \geqslant \int_0^1 f(x) \, dx$$
