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Math Help - Can a convex/concave function have a saddle point?

  1. #1
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    Can a convex/concave function have a saddle point?

    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?
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  2. #2
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    Re: Can a convex/concave function have a saddle point?

    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.
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  3. #3
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    Re: Can a convex/concave function have a saddle point?

    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?
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  4. #4
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    Re: Can a convex/concave function have a saddle point?

    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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