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Does the following statement is true?
And if so, how can one prove it?
Given the function $f:R^n->R$
And it is given that for every $x_i\in \bar{x}$ setting $x_j$ $j\neq i$ to zero
The function $f(0,0,0,...,x_i,0,0,...0)$ is convex
The function $f$ is convex

DsCpp
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1 Answers1

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If $g$ is any non-convex function on $\mathbb R$ with $g(0)=0$ then $f(x)=g(x_1)g(x_2)...g(x_n)$ gives a counterexample.

Kavi Rama Murthy
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