(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

2. Relevant equations

Constrined optimzation

3. The attempt at a solution

("o" means dot product)

Let M={x|Ax=c} and f(x)=(1/2)x o Qx - b o x

Suppose x_{0}is a local min point.

Suppose, on the contrary, that x_{0}is NOT a global min point. Then there must exist a point x_{1}s.t. f(x_{1})<f(X_{0}).

Let x(s) = (1-s)x_{0}+ s x_{1}, s is any real number.

x(0)=x_{0}, x(1)=x_{1}

Let φ(s)=f(x(s)), s is any real number. Then φ has a local min at s=0.

=> φ'(0)=0 and φ''(0)≥0 (1st and 2nd order necessary conditions for a local min)

By chain rule, 0 = φ'(0) = x_{1}o Qx_{0}- x_{0}o Qx_{0}- b o (x_{1}-x_{0})

and 0 ≤ φ''(0) = x_{1}o Qx_{1}- x_{1}o Qx_{0}- x_{0}o Qx_{1}+ x_{0}o Qx_{0}

and I'm stuck here...how can I arrive at a contradiction from here? I tried adding them to get φ'(0)+φ''(0)≥0, but it doesn't seem to help.

Any help would be much appreciated!

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# Constrained Optimization Proof

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