$f$ Concave function such as $f(0)=0$

Asked Sep 30 '19 at 13:02

Active Sep 30 '19 at 13:02

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Let $f$ a concave function such as $f(0)=0$.

$$f :\mathbb{R^+}\rightarrow \mathbb{R}$$

I want to prove that, for all $x,y\in \mathbb{R^+}$: $f(a+b)\leq f(a)+f(b)$

My idea: Since $f$ is concave then for all $t\in [0,1]$, $f(t.a+(1-t)b \geq tf(a)+(1-t)f(b).$

And by taking $t=1/2$ then we'll have : $f((a+b)/2)\leq 1/2(f(a)+f(b)).$

and if we take $b=0$, then we'll have $f(ta)\geq tf(a)$

But I could not get to resault.

asked Sep 30 '19 at 13:02

BrianTag

1

Use the inequality applied to $2a$ and $2b$ and $t=\frac{1}{2}$. – Frieder Jäckel Sep 30 '19 at 13:05
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Possible duplicate of If $f$ is a concave function and $f(0) \geq 0$ then $f(a+b) \leq f(a) + f(b)$ for all $a,b >0$ – Martin R Sep 30 '19 at 13:05
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Also: Concave implies subadditive. – Martin R Sep 30 '19 at 13:06
But $0$ is not in $\Bbb R^+$. Do you mean $\Bbb R_{\ge 0}$, that is, $[0,\infty)$? (In fact, the proof doesn't use that $a$ or $b$ are not negative. – ajotatxe Sep 30 '19 at 13:12
YES, I meant $\mathbb{R_{\geq 0}}$ – BrianTag Sep 30 '19 at 13:19

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