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I try to prove that $g(x)= K |x|^2/2 + z(x)$ is strictly convex, given that $z(x) \geq - m(1 + |x|^p)$ with $m \geq 0$, $0 \leq p \leq 2$, forall $x \in \mathbb{R}^n$, provided $K$ is sufficiently large. How to show that?

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What did you try? –  Did Nov 12 '12 at 10:50
    
I try to start with g(tx + (1-t)y), but the problem is that i need a $<$ and not a $\geq$. –  AlexisZorbas Nov 12 '12 at 10:52
    
i need a < and not a ≥... What do you mean by that? –  Did Nov 12 '12 at 10:53
    
for strict convexity i have to show that g(tx + (1-t)y) < tg(x) + (1-t)g(y). But i can only use $z(x)≥−m(1+|x|^p)$ –  AlexisZorbas Nov 12 '12 at 10:58

1 Answer 1

up vote 1 down vote accepted

The conclusion is false. For example, let $n=1$ and $z(x)=x^2(1+\sin x)\ge 0$. Then no matter how large $K$ is, $f''((2k+\frac{1}{2})\pi)<0$ when $k\in\mathbb{N}$ is large.

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