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A function is convex if and only if

  1. Nov 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Show that a differentiable function f is convex if and only if the following inequality holds for each fixed point x0 in Rn:
    f(x) ≥ f(x0) + ∇tf(x0)(x-x0) for all x in Rn, where ∇tf(x0) is the gradient vector of f at x0.


    2. Relevant equations



    3. The attempt at a solution

    by definition of a convex function we have:

    f(λx + (1-λ)x0) ≤ λf(x) + (1-λ)f(x0)

    f(λ(x-x0)+x0) ≤ λ(f(x) - f(x0)) + f(x0)

    f(λ(x-x0)+x0) - f(x0) ≤ λ(f(x) - f(x0))

    Setting Δx = λx-x0 we'll have:

    f(Δx + x0) - f(x0) ≤ λ(f(x) - f(x0))

    I'm very hesitant to use this step, because vector division is not defined, but if we were dealing with real numbers (i.e, vectors of dimension 1) I could've gone further to obtain:

    [tex] \frac{f(\Delta{x} + x_0) - f(x_0)}{\lambda} \leq f(x) - f(x_0)[/tex]
    [tex] \frac{f(\Delta{x} + x_0) - f(x_0)}{\lambda(x-x_0)}(x-x_0) \leq f(x) - f(x_0)[/tex]
    [tex] \frac{f(\Delta{x} + x_0) - f(x_0)}{\Delta{x}}(x-x_0) \leq f(x) - f(x_0)[/tex]

    taking limits from both sides as Δx goes to 0 gives us the desired result.

    I don't know how to show that the converse is true, and also I don't know how to generalize what I've written to functions of several variables since vector division is not defined. any helps would be appreciated.
     
  2. jcsd
  3. Nov 24, 2012 #2

    Ray Vickson

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    Most of theses types of results generalize almost immediately from the 1-dimensional case to the multivariate case, because if ##x_0, \, x \in \mathbb{R}^n## and ##p = x - x_0,## then
    [tex] f \text{ convex in }\mathbb{R}^n \Rightarrow \phi(t) = f(x_0 + t p)\text{ is convex in } t \in \mathbb{R},[/tex] and conversly. So, you need to show that ##\phi(t)## is convex if and only if
    [tex] \phi(t) \leq \phi(t_0) + \phi'(t_0) \, (t - t_0) \, \forall \, t, t_0.[/tex]

    RGV
     
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