Recall that a function $f: \mathbb{R}^n \rightarrow \mathbb{R}$ is quasiconvex if $\operatorname{dom}(f)$ is convex and the sublevel sets $$S_\alpha = \{x \in \operatorname{dom}(f)\,\vert\,f(x) \leq \alpha\}$$ are convex for all $\alpha$.

Equivalently, we can state $f$ is quasiconvex iff $\forall x,\,y \in \operatorname{dom}(f),\,\forall \theta \in [0,\,1]$, we have $$ f(z) \leq \max\{f(x),\,f(y)\},$$ where $z := \theta x + (1 - \theta) y$.


If $f$ also happens to be differentiable on its domain, the first order condition holds:

$$f\text{ is quasiconvex } \iff (f(x) \leq f(y) \implies \nabla f(x)^T(y - x) \leq 0).$$

There are two parts to this proof: in the "only if" ($\implies$) direction (which can be found here) and in the "if" ($\impliedby$) direction.

The latter direction is the focus of this question; I would like to know how to prove the statement $\forall x,\,y \in \operatorname{dom}(f)$ $$f\text{ is quasiconvex } \impliedby (f(x) \leq f(y) \implies \nabla f(x)^T(y - x) \leq 0).$$

Current atempt

Suppose $\forall x,\,y \in \operatorname{dom}(f)$ we know $f(x) \leq f(y) \implies \nabla f(x)^T(y - x) \leq 0$.

Without loss of generality, assume $x$ and $y$ are such that $f(x) \leq f(y)$. Then, we know

  1. $\nabla f(x)^T(y - x) \leq 0$.
  2. For $f$ to be quasiconvex, we need to have $f(z) \leq f(x)$.

This is not much in the way of an "attempt", but I have a feeling these two facts are related. I am just not sure if and how to use (1.) to show (2.). Does it have something to do with, maybe, the directional derivative of $f$ at $x$ in the direction $y - x$?


Your idea is right, but it is not sufficient to use the inequality in a single point, you have to use it "globally" and integrate. Here is what I mean.

Fix any $x,y \in \mathrm{dom}(f)$, and suppose without loss of generality that $f(x) \leq f(y)$. Let $\gamma : [0,1] \to \mathbb{R}^n$, $\gamma(t) = (1-t)y+tx$, and consider $z_0 = \gamma(t_0)$ the maximum point of $f(\gamma(t))$, i.e., the maximum point of $f$ restricted to the segment from $x$ to $y$. The point is well defined because $f(\gamma(t))$ is continuous on a compact set. Then, by the fundamental theorem of calculus, $$ f(z_0) = f(y) + \int_0^{t_0} \nabla f(\gamma(t)) \cdot (x-y)\ dx = f(y) + \int_0^{t_0} \frac{1}{t_0-t} \nabla f(\gamma(t)) \cdot ((t_0-t)x-(t_0-t)y)\ dx = f(y) + \int_0^{t_0} \frac{1}{t_0-t} \nabla f(\gamma(t)) \cdot (z_0 - \gamma(t))\ dx. $$

Since, for evey $t \in [0,1]$, $f(\gamma(t)) \leq f(\gamma(t_0))$ by definition of $t_0$, your properties implies $\nabla f(\gamma(t)) \cdot (z_0 - \gamma(t)) \leq 0$. Moreover the factor $\frac{1}{t_0-t}$ is positive, hence $$ f(z_0) = f(y) + \int_0^{t_0} \frac{1}{t_0-t} \nabla f(\gamma(t)) \cdot (z_0 - \gamma(t))\ dx \leq f(y), $$ as wanted.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.