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Problem of the Week #198 - March 15, 2016

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MHB Global Moderator
Staff member
Jun 20, 2014
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.


Remember to read the POTW submission guidelines to find out how to submit your answers!


Staff member
Jan 26, 2012
Congratulations to Opalg for his correct solution! You can find it below.

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.
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