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Given a concave function $f(x)$ over a convex set $S$, so $f(a)+f(b) \leq 2 \times f (\frac{a+b}{2})$. Here, $a$ and $b$ lies in the same convex set $S$.

I know that the maximum value of $f(a)+f(b)$ is possible when $a=b$. But I don't know how to argue for it. Though I have some intuition, I need formal arguments to justify it.

For eg. the $\sin$ function is concave in the interval $[0, \pi]$, so $\sin(x)+ \sin (y) \leq 2 \times \sin\left(\frac{x+y}{2}\right)$. So, can we say that maximum of $\sin(x)+ \sin (y)$ is possible if $x=y$ ?

Patricio
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maths123456
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1 Answers1

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Here is a hint.

Notice that the line connecting two points $(x_{1},f(x_{1}))$ and $(x_{2},f(x_{2}))$ lies completely below the curve $y=f(x)$. The midpoint of this line is $\left(\frac{x_{1}+x_{2}}{2},\frac{f(x_{1})+f(x_{2})}{2}\right)$

This image will help

maths123456
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acat3
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  • This is my idea : We know $f(x_1)+f(x_2) \leq 2* f(\frac{x_1+x_2}{2})$. The equality holds only when x_1=x_2. So apart from this there is no other condition when equality will hold. :) – maths123456 Aug 06 '21 at 06:12
  • @maths123456 that is correct. In my hint, the two points and their midpoint are all the same point that lie on $y=f(x)$ – acat3 Aug 06 '21 at 06:14