Convex Function on Many variables on an interval

[curiousguy23]

Homework Statement

Show that f(x) = x₁x₂ is a convex function on [a,ma]^T where a \geq 0
and m \geq 1.

Homework Equations

By definition f is convex iff

\forall x,y\in \Re \quad \wedge \quad \forall \lambda :\quad 0\le \lambda \le 1\quad \Rightarrow \quad f\left( \lambda x+(1-\lambda )y \right) \le \lambda f\left( x \right) +(1-\lambda )f\left( y \right)

The Attempt at a Solution

I am not really sure how to go about this. Firstly I was thinking on how to apply the above relation to multiple variables. I would assume that it applies in general and in this case x and y would be a vector of variables, but I still don't know how the proof follows. Also I am not sure is how the interval [a,ma]^T plays into the equation.

 

[Ray Vickson]

Homework Statement

Show that f(x) = x₁x₂ is a convex function on [a,ma]^T where a \geq 0
and m \geq 1.

Homework Equations

By definition f is convex iff

\forall x,y\in \Re \quad \wedge \quad \forall \lambda :\quad 0\le \lambda \le 1\quad \Rightarrow \quad f\left( \lambda x+(1-\lambda )y \right) \le \lambda f\left( x \right) +(1-\lambda )f\left( y \right)

The Attempt at a Solution

I am not really sure how to go about this. Firstly I was thinking on how to apply the above relation to multiple variables. I would assume that it applies in general and in this case x and y would be a vector of variables, but I still don't know how the proof follows. Also I am not sure is how the interval [a,ma]^T plays into the equation.

What is meant by [a,ma]^T? I doubt that [a,ma] is a 1-dimensional interval, because what in the world would possibly be meant by the transpose of an interval? My guess would be that [a,ma] means a point in R² of the form x₁ = a, x₂ = ma. However, that is just a guess.

RGV

 

[curiousguy23]

What is meant by [a,ma]^T? I doubt that [a,ma] is a 1-dimensional interval, because what in the world would possibly be meant by the transpose of an interval? My guess would be that [a,ma] means a point in R² of the form x₁ = a, x₂ = ma. However, that is just a guess.

RGV

Yes you are right in this, sorry I did not stipulate this, x1= a and x2=ma=mx1