# Showing that a real function is convex

How can I show that if $\varphi$ is a real function such that $$\varphi \left(\int_0^1 f\right)\leqslant \int_0^1 \varphi (f)$$ for any Borel-measurable real function $f$, then $\varphi$ is convex.

Ps: I realize homework questions have to be tagged as such. This isn't a homework problem. I came across this question in a text book, and I thought it was interesting.

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Could you give the name of the book? –  Martin Sleziak Nov 16 '11 at 20:11
I was using a random book at the school library and I didn't look at the title/author. Sadly, I'm home now. –  Colin Nov 16 '11 at 20:25
There is an amibiguity in the question. Does it mean "if there is any Borel-measurable function such that this inequality holds", or does it mean "If it is the case that for any Borel-measurable function, for which this inequality holds"? It can be read either way. If you mean what I think you mean, then merely changing "any" to "every" would disambiguate it. "Any" is often a potentially ambiguous word. Be careful with it. –  Michael Hardy Nov 16 '11 at 21:50

You can verify the definition of convexity directly by taking a suitable $f$. Given $x_1, x_2 \in \mathbb R$ and $t \in [0,1]$, we apply the hypothesis to the step function $f : [0,1] \to \mathbb R$ given by $$f(s) = \begin{cases} x_1, &0 \leqslant s \leqslant t, \\ x_2, &t \lt s \leqslant 1. \end{cases}$$ This gives us $$\varphi\left(\int_0^t x_1 ds + \int_t^1 x_2 ds \right) \leqslant \int_0^t \varphi(x_1) ds + \int_t^1 \varphi(x_2) ds$$ $$\implies \quad\varphi(tx_1 + (1-t)x_2) \leqslant t \ \varphi(x_1) + (1-t) \ \varphi(x_2),$$ which is what we want to prove.

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Thanks for your answer. So essentially, coming up with the function $f(s)$ is critical. –  Colin Nov 16 '11 at 20:15
@Colin Yes, that is crucial. But once we are told what to show and staring at the definition of a convex function, this is not that hard, I think. –  Srivatsan Nov 16 '11 at 20:17

I assume that $\varphi:[0,1]\to\mathbb R$.

Definition of convex is: For each $\alpha\in[0,1]$ and for each $x$, $y$ the inequality $$\varphi(\alpha x+(1-\alpha)y) \le \alpha \varphi(x) + (1-\alpha) \varphi(y).$$

Try to use the following measure: $$\mu(A) = \alpha \chi_{\{x\}}(A) + (1-\alpha) \chi_{\{y\}}(A)$$ and the function $f(x)=x$.

Here $\chi_B$ denotes the characteristic function of the set $B$ (a.k.a. indicator function). The measure $\mu$ is basically a convex combination of two Dirac measures $\delta_x$ and $\delta_y$, i.e. $$\mu(A)=\alpha \delta_x(A)+ (1-\alpha) \delta_y (A).$$

To prove that this implies convexity it suffices to show that for any function $g$ we have $$\int g \mathrm{d}\mu = \alpha g(x) + (1-\alpha) g(y).$$

EDIT: I understood the question as follows: Suppose that Jensen's inequality holds for arbitrary measure and arbitray $\varphi$, $f$. Then prove convexity. Srivatsan's answer is better, since he only uses the usual (Lebesgue) measure - which was probably the original intention of the question. (Although I have a simpler function $f$...)

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Is $A$ an arbitrary set? –  Colin Nov 16 '11 at 20:05
$A$ is arbitrary subset of $[0,1]$. (I am defining the measure on $[0,1]$.) I might have misunderstood the question - I am not sure whether we can choose arbitrary measure or we should use the usual measure on $[0,1]$ only. –  Martin Sleziak Nov 16 '11 at 20:17