Convex functions

Suppose \(x∈Ω\) satisfies

\(\hspace{4cm}∇f(x)^{T}(y-x)≥0,∀y∈Ω\)........................\((8)\).

By the first order characterization of convexity, we have

\(\hspace{3,5cm}f(y)≥f(x)+∇f^{T}(x)(y-x),∀y∈Ω.\).................\((9)\)

Then \((8)\)+\((9)\) \(⇒ f(y)≥f(x),∀y∈Ω⇒ x \) is optimal.

Suppose x is optimal but for some \( y∈Ω\) we had \(∇f^{T}(x)(y-x)<0\)

Consider \(g(α):=f(x+(α(y-x)))\). Because \(Ω\) is convex, \(∀α∈[0,1],x+α(y-x)∈Ω.\) Observe that

\(\hspace{1cm}g′(α) = (y-x)^{T}∇f(x+α(y-x))⇒ g′(0)=(y-x)^{T}∇f(x)<0.\)

This implies that

\(\hspace{1cm}∃δ > 0,s.t.g(α)<g(0),∀α∈(0,δ)⇒ f(x+α(y-x))<f(x),∀α∈(0,δ)\)

But this contradicts the optimality of \(x\).