Problem of the Week #198 - March 15, 2016

Status
Not open for further replies.

Euge

MHB Global Moderator
Staff member
Here is this week's POTW:

-----
Let $\Lambda :\Bbb R \to \Bbb R$ be a mapping such that for all bounded measurable mappings $f : [0,1]\to \Bbb R$,

$$\Lambda\left(\int_0^1 f(x)\, dx\right) \le \int_0^1 \Lambda(f(x))\, dx.$$

Show that $\Lambda$ is a convex mapping.

-----

Jameson

Given $a < b$ in $\mathbb{R}$ and $0 < \lambda < 1$, apply the inequality $$\Lambda\left(\int_0^1 f(x)\, dx\right) \leqslant \int_0^1 \Lambda(f(x))\, dx$$ to the function $$f(x) = \begin{cases}a &(0\leqslant x \leqslant \lambda), \\b &(\lambda < x \leqslant 1).\end{cases}$$ On the left side, $$\displaystyle \int_0^1f(x)\,dx = \int_0^\lambda a\,dx + \int_\lambda^1 b\,dx = \lambda a + (1-\lambda)b.$$
On the right side, $$\displaystyle \int_0^1 \Lambda(f(x))\, dx = \int_0^\lambda \Lambda(f(x))\, dx + \int_\lambda^1 \Lambda(f(x))\, dx = \int_0^\lambda \Lambda(a)\, dx + \int_\lambda^1 \Lambda(b)\, dx = \lambda\Lambda(a) + (1-\lambda)\Lambda(b).$$
Putting those values into the given inequality gives $$\displaystyle \Lambda\bigl( \lambda a + (1-\lambda)b\bigr) \leqslant \lambda\Lambda(a) + (1-\lambda)\Lambda(b).$$ Since that holds whenever $a < b$ and $0 < \lambda < 1$, it follows that $\Lambda$ is convex.