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Arian.D
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Homework Statement
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.
Homework Equations
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.