Home » Uncategorized » Inequality of a concave function

Inequality of a concave function

Let f:[0, 1] \rightarrow [0, +\infty) be a concave function. Prove that

    \[\int_0^1 x^2 f(x) \, \mathrm{d}x \leq \frac{1}{2} \int_0^1 f(x) \,\mathrm{d}x\]

Solution

Since f is concave , it holds that

    \[f\left ( ax + \left ( 1-a \right )y \right ) \geq a f(x) + \left ( 1-a \right ) f(y) \geq a f(x) \quad \text{forall} \;\; a, x, y \in [0, 1]\]

By setting y=0 and x=a we get that x f(x) \leq f \left ( x^2 \right ). Thus,

    \begin{align*} \int_{0}^{1} x^2 f(x) \, \mathrm{d}x &= \frac{1}{2} \int_{0}^{1} x f(x) 2x \, \mathrm{d}x \\ &\leq \frac{1}{2}\int_{0}^{1} f \left ( x^2 \right ) 2x \, \mathrm{d}x \\ &=\frac{1}{2} \int_{0}^{1} f \left ( x^2 \right ) \, \mathrm{d} \left ( x^2 \right ) \\ &= \frac{1}{2} \int_{0}^{1} f(x) \, \mathrm{d}x \end{align*}

and the exercise is complete.

Read more

Leave a comment

Donate to Tolaso Network