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If $f: [0, \infty) \to [0,\infty)$ is concave, show that $f(x+y) \leq f(x) + f(y)$ $\forall \ x,y$ in the domain.

We can assume $f$ to be strictly increasing on the interval $I = [0, x+y]$ since if $f(x+y) \leq \min(f(x), f(y))$, we are immediately done.

WLOG (due to symmetry), assume $x \leq y$. Then $f'(\theta_1) = \frac{f(x+y) - f(y)}{x} \leq \frac{f(x) - f(0)}{x} = f'(\theta_2) \leq \frac{f(x)}{x}$ where $\theta_1 \in (y,x+y)$ and $\theta_2 \in (0,x)$. $\theta_i$s exist due to the mean value theorem and then by concavity, $f'(\theta_1) \leq f'(\theta_2)$ holds. The rest follows.

But I am supposed to do this without assuming differentiability. I felt that the tangent argument may be extended for the non-differentiable case too and hence thought of writing down the above argument.

supremacy
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  • @IzaakvanDongen Yes, that's exactly what I was looking for! Thanks a lot. It's a bit tricky to find though. – supremacy Mar 27 '23 at 00:36

1 Answers1

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I will give a proof using some standard facts about concave funcions. We can write $f(x+y)-f(x)=\int_x^{x+y} g(t)dt$ for some decreasing function $g$. Also, $f(y)-f(0)=\int_0^{y} g(t)dt$. By a change of variable we get $f(y)-f(0)=\int_x^{x+y} g(s-x)ds$. Since $g$ is decreasing we get $f(y)-f(0)\geq\int_x^{x+y} g(s)ds=f(x+y)-f(x)$. We have proved that $f(x+y)\leq f(x)+f(y)-f(0)$ a stronger inequality than the one asked for.

Kavi Rama Murthy
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  • Thanks for the answer. (1) Where, in your answer, do we use the fact that $f$ is concave? (2) Is there a way we can do this just by using the inequality $\theta f(x) + (1-\theta)f(y) \leq f(\theta x + (1-\theta)y) \ \forall \ \theta \in [0,1]$ and $\forall \ x,y$ in the domain? [Note: I am not accepting your answer as I'm waiting for a non-calculus sort of a proof; in case you or anyone else isn't able to come up with one in the next few hours, I'll accept your answer.] – supremacy Mar 26 '23 at 07:14
  • @supremacy As far as I can see this cannot be proved just from the definition of a concave function. But I will be happy if someone comes up with such a proof. – Kavi Rama Murthy Mar 26 '23 at 07:20