# Convex functions - multi variables

I have $x_1,x_2,..,x_n\in\left[0,1\right]$.

Suppose $x=(x_1,x_2,..,x_n), F(x,i,j,\epsilon)=g(x_1,...,x_i+\epsilon,...,x_j-\epsilon,..,x_n)$ . We know that $F$ is convex with respect to $\epsilon$. How do you show that $F(x,i,j,\epsilon)\geq F(x,i,j,0)$ for all $\epsilon \in \left[-min\left\{ x_{i},1-x_{j}\right\} ,min\left\{ 1-x_{i},x_{j}\right\} \right]$ ?

Taken from section 2 at http://www2.warwick.ac.uk/fac/sci/dcs/people/maxim_sviridenko/dirmaxcut.pdf

Only that they said that it follows directly from $F$ being convex, but I can't figure out why.

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To abstract a little bit the situation, consider a convex function $u$ defined on some interval $[s,t]$ with $s\lt0\lt t$. Then the question is to show that $$u(0)\leqslant\max\{u(s),u(t)\}.$$ (Note that one does not prove that $u(0)\leqslant u(s)$ and $u(0)\leqslant u(t)$, and that these inequalities have no reason to be true.)
To prove this, note that $0=\theta s+(1-\theta )t$ for some $\theta$ in $(0,1)$ hence, by convexity, one has $u(0)\leqslant \theta u(s)+(1-\theta )u(t)$. Thus, $$u(0)\leqslant \theta \max\{u(s),u(t)\}+(1-\theta )\max\{u(s),u(t)\}=\max\{u(s),u(t)\}.$$ Finally, apply this to the function $u=F(x,i,j,\ )$.
They said that given the fact that it holds $\epsilon$-convexity, it holds that $F(x,i,j,\epsilon) \geq F(x,i,j,0)$ –  Shmoopy Feb 9 '13 at 14:22