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Convexity in n-dimensions

Let Ω⊂Rn\Omega \subset \mathbb{R}^n be open and convex. A function f:Ω→Rf: \Omega \to \mathbb{R} is called convex if:


f(λx+(1−λ)y)≤λf(x)+(1−λ)f(y)∀x,y∈Ω, λ∈(0,1)f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda)f(y) \quad \forall x, y \in \Omega, \: \lambda \in (0,1)


Suppose f∈C1(Ω) f \in C^1(\Omega). Show that f is convex if and only if:


f(x)≥f(z)+⟨∇f(z),x−z⟩∀x,z∈Ωf(x) \geq f(z) + \langle \nabla f(z), x-z \rangle \quad \forall x,z \in \Omega


The forward argument is fairly simple; use the chain rule in n-dimensions to differentiate the entire first equation with respect to \lambda, and set \lambda equal to zero to give the second equation. I am finding the reverse argument harder... any helpful tips?
Reply 1
Avatar for nuodai
nuodai
17
By assumption, we have:
f(x)≥f(z)+∇f(z)⋅(z−x) ∀x,z∈Ωf(x) \ge f(z) + \nabla f(z) \cdot (z-x)\ \forall x,z \in \Omega
f(y)≥f(z)+∇f(z)⋅(z−y) ∀y,z∈Ωf(y) \ge f(z) + \nabla f(z) \cdot (z-y)\ \forall y,z \in \Omega

Take λ\lambda times the top equation and add it to (1−λ)(1- \lambda ) times the bottom equation, and rearrange... then find a suitable value of zz in terms of xx and yy.
Reply 2
Avatar for Dream Eater
Dream Eater
OP
nuodai
By assumption, we have:
f(x)≥f(z)+∇f(z)⋅(z−x) ∀x,z∈Ωf(x) \ge f(z) + \nabla f(z) \cdot (z-x)\ \forall x,z \in \Omega
f(y)≥f(z)+∇f(z)⋅(z−y) ∀y,z∈Ωf(y) \ge f(z) + \nabla f(z) \cdot (z-y)\ \forall y,z \in \Omega

Take λ\lambda times the top equation and add it to (1−λ)(1- \lambda ) times the bottom equation, and rearrange... then find a suitable value of zz in terms of xx and yy.


Lovely, thanks a lot :biggrin:.

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