Drew Williamson

2022-09-06

Consider a convex function f which is differentiable on closed convex set Y. Then it holds x is minimizer of f if $<\mathrm{\nabla}f(x),x-y>\le 0,\mathrm{\forall}y\in Y$

Proof:

Assume $<\mathrm{\nabla}f(x),x-y>>0$ and consider $h(t)=f(x+t(y-x))\Rightarrow {h}^{\prime}(t)<0$

Why is this a contradiction?

Proof:

Assume $<\mathrm{\nabla}f(x),x-y>>0$ and consider $h(t)=f(x+t(y-x))\Rightarrow {h}^{\prime}(t)<0$

Why is this a contradiction?

Shane Middleton

Beginner2022-09-07Added 7 answers

If ${h}^{\prime}(0)<0$ then you can find a t>0 such that $h(t)<h(0)$, which translates to

$$f(x+t(y-x))<f(x)$$

Therefore x can't be the minimizer, which contradicts your assumption.

$$f(x+t(y-x))<f(x)$$

Therefore x can't be the minimizer, which contradicts your assumption.

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