Can a convex/concave function have a saddle point?

**infernalmich** · May 31st 2012, 12:53 PM

My question is:
Can a convex/concave function have a saddle point?

My answer would be:
Convex and concave function do not have saddle points, because a saddle point is not a local extremum.

Is this answer correct? How could I explain it better?

**HallsofIvy** · June 1st 2012, 08:03 AM

Are you asserting that every point on the graph of a convex or concave function must be a local extremum?

If not, the fact that a saddle point is not a local extremum does not immediately imply that it cannot be a saddle point.

**infernalmich** · June 1st 2012, 08:33 AM

But the affermation that a convex function can`t have a saddle point is true, right? - So the problem is that my explaination of the why is not correct?

**jj323** · June 1st 2012, 10:42 AM

Neither convex or concave functions can have saddle points.

Let's show this for convex functions, then a similar argument can be used for concave. First, we need some definitions.

1. $N_\epsilon(x)$ denotes the $\epsilon$ -neighborhood of x.

2. Given a Frechet differentiable $f: X\rightarrow \Re$ where X is a Hilbert space (which includes $\Re^n$ ), a stationary point is a point x where $\nabla f(x)=0$ . I'm sure we could do this with Gatteaux (directional) derivatives, but it's simpler to assume we have a gradient.

3. A saddle point is a stationary point that is neither a local min or a local max.

4. Given $f: X\rightarrow \Re$ , a local min is a point x such that there exists an $\epsilon$ -neighborhood such that $f(x)\leq f(y)$ for all $y\in N_\epsilon(x)$ .

5. A convex function is a function $f: X\rightarrow \Re$ where for all $x,y\in X$ and $\lambda\in [0,1]$ , $f(\lambda x + (1-\lambda) y) \leq \lambda f(x) + (1-\lambda) f(y)$

Now, let us assume that f is Frechet differentiable and convex. Let us also assume that there exists a stationary point x that is a saddle point. Since x is a saddle point, x is not a local min. That means that for every $\epsilon$ -neighborhood of x there exists $y\in N_\epsilon(x)$ such that $f(y) < f(x)$ . Next, from convexity we have that

$f(\lambda y + (1-\lambda) x) \leq \lambda f(y) + (1-\lambda) f(x)$

for $\lambda \in (0,1)$ . Then, since $\lambda >0$ , we have that

$\frac{f(\lambda y + (1-\lambda) x)}{\lambda} \leq \frac{\lambda f(y) + (1-\lambda) f(x)}{\lambda}$

Rearranging terms, we have

$\frac{f(x-\lambda (y-x))-f(x)}{\lambda} + f(x) \leq f(y)$

Taking the limit as $\lambda\rightarrow 0$ , we have that

$\langle\nabla f(x),y-x\rangle + f(x) \leq f(y)$

where $\langle \cdot,\cdot\rangle$ denotes the inner product on X. In $\Re^n$ , we typically have that $\langle x,y\rangle=x^Ty$ . In any case, since x is stationary, $\nabla f(x)=0$ . Hene, we reduce the above inequality into

$f(x) \leq f(y)$

This contradicts the assumption that x is not a local min. Hence, if f is convex and x is stationary, then x is a local, and in fact global, min.

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