0
$\begingroup$

Recently I've been introduced to the following definition of a concave function.

Let $X \subset \mathbb{R}^n$. A function $f:X \rightarrow \mathbb{R}$ is concave if for all $x,y \in X$ and $\lambda \in [0,1]$, \begin{equation} f(\lambda x+(1-\lambda)y) \ge \lambda f(x)+(1-\lambda)f(y). \end{equation}

My question is, how do we show that in the case of $X\in \mathbb{R}$, the above definition of concavity is equivalent to saying the second derivative $f''(x)\ge 0$ on $X$? I'm having trouble starting, because the derivatives have limits involved, while the more general definition does not. Thanks!

$\endgroup$

1 Answer 1

2
$\begingroup$

So first off, the relevant condition is actually $f''(x) \leq 0$ in this situation. After correcting this, the two are still not equivalent; the definition of concavity that you gave is more general in that it applies to functions that are not even once differentiable.

That said, when $f$ is twice continuously differentiable, we can derive this definition of concavity by Taylor expanding both terms on the right side to first order around $\lambda x + (1-\lambda y)$ and then examine the quadratic Lagrange remainder. The constant term is what is on the left side already (this is why we chose to expand at this point in particular), so those cancel out. The linear terms cancel one another as well. The quadratic remainder is then nonpositive, which gives the desired result.

Note that in this derivation we needed to know that $f''(z) \leq 0$ for all $z$ between $x$ and $y$, not just at $\lambda x + (1-\lambda) y$, because we had no further control than that over the points where $f''$ got evaluated in the Lagrange remainder.

Geometrically what you are seeing here is that a concave function lies above its secant lines, which is easily visualized in the case of just a downward opening parabola.

You can show the converse by assuming this more general definition and that $f$ is twice continuously differentiable, using the same expansion, and then just sending $y \to x$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .