Is the following function convex ?

[thisnot]

convex function

how would i determine if the following function is convex?

f(x)=-[det(A0+x1A1+...+xnAn)]^(1/m) on {x | A0+x1A1+...+xnAn > 0}


~thanks

 

[me2please]

Re: convex function

I think you have \[A_{n} (m \times m)\]

Use
1. definition of convex function \[f(\alpha x+(1-\alpha)y)\leq \alpha f(x) + (1-\alpha ) f\]


2. \[det(cA)=c^m det(A)\]

3. Show 1. using Minkowski's inequality \[A\neq 0, B\neq 0 (m\times m)\] and Positive semidefinite
\[[[det(A+B)]]^{\frac{1}{m}}\geq (detA)^{\frac{1}{m}} + (detB)^{\frac{1}{m}}\]

Then, you will have
\[\alpha f(x) = -[det(\alpha A_{0} + \alpha x_{1}A_{1}+\cdots +\alpha x_{n}A_{n})]^{\frac{1}{m}}\]
and the same for \[(1-\alpha ) f\]

and
\[f(\alpha x + (1-\alpha )y) = -[det\{ (\alpha + (1-\alpha) )A_{0} + (\alpha x_{1}+(1-\alpha )y_{1} )A_{1}+\cdots +(\alpha x_{n} + (1-\alpha )y_{n})A_{n} \}]^{\frac{1}{m}}\]

 

[firephenix405]

Re: convex function

How did you do these equations on the post. it's amazing!

I think you have \[A_{n} (m \times m)\]

Use
1. definition of convex function \[f(\alpha x+(1-\alpha)y)\leq \alpha f(x) + (1-\alpha ) f\]


2. \[det(cA)=c^m det(A)\]

3. Show 1. using Minkowski's inequality \[A\neq 0, B\neq 0 (m\times m)\] and Positive semidefinite
\[[[det(A+B)]]^{\frac{1}{m}}\geq (detA)^{\frac{1}{m}} + (detB)^{\frac{1}{m}}\]

Then, you will have
\[\alpha f(x) = -[det(\alpha A_{0} + \alpha x_{1}A_{1}+\cdots +\alpha x_{n}A_{n})]^{\frac{1}{m}}\]
and the same for \[(1-\alpha ) f\]

and
\[f(\alpha x + (1-\alpha )y) = -[det\{ (\alpha + (1-\alpha) )A_{0} + (\alpha x_{1}+(1-\alpha )y_{1} )A_{1}+\cdots +(\alpha x_{n} + (1-\alpha )y_{n})A_{n} \}]^{\frac{1}{m}}\]

 

[M!k]

Re: convex function

Hi firephenix405,

when writing a post just klick on the LaTex-Button for an overview. In the text klick on TeX, write your formula as described in the overview, press again Tex* - that's it.
\[\alpha \sum \int \sqrt[n]{abc}\]


Mik