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Here is a problem about proving concavity of a function

Problem 5. Assume $f$ and $g$ are concave. prove that $f+g$ is concave

Answer:

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How does this prove concavity? I am used to the idea that a function is concave iff $f''\leq 0$. I don't see how this relation proves concavity.

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user56834
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    Your definitions are equivalent. This one is the standard definition. You might want to draw the function to see it. –  Aug 25 '17 at 15:39
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    The definition of concave used is the standard one, which does not assume twice-differentiability. Of course, if you want to define concavity as $f'' \leq 0$, then the problem is very trivial: $f'' \leq 0$ and $g'' \leq 0$ implies $(f+g)'' = f'' + g'' \leq 0$. – MathematicsStudent1122 Aug 25 '17 at 15:40
  • @NiklasHebestreit Equivalent... for $\mathcal{C}^2$ functions. – anderstood Aug 25 '17 at 15:49

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A function is concave if lies above the chord between any two points on it: that is, for any $x,y$ and $0 \leq t \leq 1$, $$ f(tx+(1-t)y) \geq t f(x) + (1-t) f(y). $$ This is a much more general notion than $f'' \leq 0$, since $f$ may not be twice-differentiable. The former is also defined even if $f$ is not continuous, or in more than one dimension.

If $f$ is once-differentiable, this condition can be proved equivalent to $f'$ being decreasing. If $f$ is also twice-differentiable, you can also show that concave implies that $f'' \leq 0$.

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