# Convex Functions on 2 variables over an interval

It is required to show that $f(x) = x_1x_2$ is a convex function on $[a,ma]^T$ where $a\ge 0$ and $m\ge1$.To show convexity we need to show that for $\lambda \in [0,1]$:

$f(\lambda x + (1-\lambda )x')\le\lambda f(x)+(1-\lambda)f(x')$

What does the interval $[a,ma]^T$ mean in this case and where does it fit in the proof? Also how can we prove convexity in the case of many variables as is here?

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I think you should provide the source of this. Upon reading the first sentence of the question, I was about to ask in a comment what you mean by the $T$. Note that it is often the case that books have a list of symbols at the end. The definition of convexity was given to you: $f$ is convex if it satisfies the inequality you've written. I suspect that $[a,ma]^T\subseteq {\bf R}^2$, so maybe $[a,ma]^T=[a,ma]\times [a,ma]$? –  tomasz Nov 3 '12 at 13:09
This is from a coursework. The T means transpose, we have always used that notation. Yes I added the definition myself to the question. The thing I am not sure how the interval $[a,ma]^T$ fits in. –  MD1 Nov 3 '12 at 13:33
I'd say that it doesn't at all, because $f$ is a function of two variables... Not to mention that I'm not sure at all what would a transpose of an interval be. –  tomasz Nov 3 '12 at 18:06