Is Lambda a Convex Mapping?

[Euge]

Here is this week's POTW:

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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.

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[Jameson]

Congratulations to Opalg for his correct solution! You can find it below.

Spoiler

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.