Prove that if a function $f(x)$ is concave down on the interval $(a,b)$ then the second derivative cannot be greater than $0$ for $x \in (a, b)$

Question

I have tried approaching this by drawing out the graph and the second derivative but I’m thoroughly confused on where to start and what would constitute a proof of this. I believe I am supposed to use some form of MVT but I’m not sure where that would fit into this.

I recommend that instead of jumping from $f\to f''$, first describe the relationship between $f$ and $f'$, then $f'$ and $f''$, and finally $f\to f'\to f''$. — Andrew Chin, Dec 12 '19 at 21:56

score 2 · Answer 1 · answered Dec 12 '19 at 22:09

2

Here is an outline for the proof.

Note that for $x$ in a neighborhood of $a$, $$f(x)=f(a)+f'(a)(x-a)+\frac {f''(\zeta)}{2}(x-a)^2$$

The first part of the RHS is the equation of the tangent line.

For the function to be concave down you have to have your $f(x)$ to be below the tangent line so $$ \frac {f''(\zeta)}{2}(x-a)^2\le 0$$

Thus your second derivative should be less than or equal zero.

answered Dec 12 '19 at 22:09

Mohammad Riazi-Kermani

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How do you get the $$\frac {f''(\zeta)}{2}(x-a)^2$$ part and what does it mean? – Abhijeet Saha Dec 13 '19 at 03:44
@AbhijeetSaha The remainder theorem for Taylor polynomial. – Mohammad Riazi-Kermani Dec 13 '19 at 07:14
I think your solution needs the continuity of the second derivative to apply the Lagrange form of remainder. And even after that, it will only mean that the second derivative is negative on a dense set, so you will need the continuity again. – Botond Dec 13 '19 at 09:08
@Botond yes, we need the continuity of the second derivative in my answer. – Mohammad Riazi-Kermani Dec 13 '19 at 12:31

Botond · Answer 2 · 2019-12-13T09:12:20.870

We can't use Taylor's theorem with mean-value form of the remainder in the general case, because it would need the continuity of the second derivative. Instead, we can use Taylor's Theorem with Peano's Form of Remainder. Let $c \in (a,b)$ be given. We have that $$f(x)=f(c)+f'(c)(x-c)+\frac{1}{2}f''(c)(x-c)^2+o((x-c)^2)$$ Concavity implies that the tangent line is over the function, i.e $$f(x)\leqslant f(c)+f'(c)(x-c)$$ Which implies that $$\frac{1}{2}f''(c)(x-c)^2+o((x-c)^2)\leqslant 0$$ I.e. for $x\neq c$ we have that $$\frac{1}{2}f''(c)+o(1)\leqslant 0$$ And this implies that the derivative is negative, because if we take $x \to c$ we get that $$\frac{1}{2}f''(c)\leqslant 0$$

Prove that if a function $f(x)$ is concave down on the interval $(a,b)$ then the second derivative cannot be greater than $0$ for $x \in (a, b)$

2 Answers2