Concave implies subadditive

Question

Let $f: [0, \infty) \to [0,\infty)$ be concave, meaning; $$ f(tx + (1-t)y) \ge tf(x) + (1-t)f(y)$$

for $t \in [0,1]$. Also, assume $f(0) = 0$. I trying to show $f(x+y) \le f(x) + f(y)$ but I fail. Is there any advise you can give me for help? I appreciate.

I found proof here: http://en.wikipedia.org/wiki/Concave_function — tom b., Feb 05 '13 at 16:21
you can answer yourself... that way if someone has a similar question, they can read your answer — Albert, Feb 05 '13 at 16:23

Calvin Khor · Answer 1 · 2021-12-21T10:49:59.273

28

Notice by taking $y=0$ that $t f(x) ≤ f(tx)$. Then using $t = \frac{x}{x+y}\in[0,1]$, \begin{align} f(x) &= f(t(x+y))\phantom{(1-{})} \geq t f(x+y)\\ f(y) &= f((1-t)(x+y))\geq (1-t) f(x+y)\end{align} so $f(x)+f(y) \geq f(x+y)$.

As mentioned in the comments by Martin R: the same holds for all concave functions $f:[0,\infty)\to\mathbb R$ with $f(0)\ge 0$.

$\color{lightgrey} {\text{Remark: I quickly typed this proof because this question is a top google result for the title}}$

edited Dec 21 '21 at 10:49

answered Mar 04 '18 at 20:35

Calvin Khor

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One could mention that this holds for all concave functions $f: [0, \infty) \to \Bbb R$ with $f(0) \ge 0$. – Martin R Jun 04 '21 at 06:48
@MartinR that sounds like a very useful observation! – Calvin Khor Jun 04 '21 at 07:16

Concave implies subadditive

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