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How do I show that if a function satisfies

$f(\alpha x + (1-\alpha)y) > \alpha f(x) + (1-\alpha)f(y)$ with $0 < \alpha < 1$,

it is true that (assuming it exists) the second derivative

$f''(x) < 0$.

I can see this is true visually in a graph but a more rigourous proof is elusive.

EDIT: I see a counterexample in the comment. I have now corrected the errors in my question (mixed up strictly concave and concave) and broadly ask about when I can use the second derivative to prove concavity (as is done to show $\log(x)$ is concave for instance) and how the equivalence is shown in these cases.

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1 Answer

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Yes if we assume that $f''(x) < 0$ esists then $f(x)$ is strictly concave but if $f(x)$ is strictly concave we can also have $f''(x) \le 0$ with $f''(x)=0$ is some points (not on an interval) as for example $f(x)=-x^4$.

For the proof the idea behind is to show that

• $f''(x)<0 \implies f'(x)$ strictly decreasing $\iff f(x)$ strictly concave

• $f(x)$ concave$ \iff f'(x)$ decreasing $\iff f''(x)\le0$

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