I am a little bit confused with the following question from Elon Lima's Curso de Análise (volume 2):
Let $V$ be open in $\mathbb{R}^{m}$ and $g:V \to \mathbb{R}^{m}$ an application of class $C^{2}$. Given $b \in V$, suppose $g'(b)$ is an isomorphism. Show that there is a open ball $B$ with center in $b$ such that the function $\phi : B \to \mathbb{R}$, defined by $\phi(y) = |g(y)-g(b)|^{2}$ is a convex function.
How can I show this? Thanks.
The only difficulty here is that you are dealing with the second derivative of $g$.
Let $f(x) = g(x)-g(b)$ and $\phi(x) = {1 \over 2} \|f(x)\|^2$.
Then $D \phi(x)(h) = f^T(x) Df(x)(h)$ and $D^2\phi(x)(h,\eta) = (Df(x) (\eta))^T Df(x)(h) + f^T(x) D^2 f(x)(h,\eta)$, or $D^2\phi(x)(h,\eta) = \eta^T Df(x)^T Df(x)h + f^T(x) D^2 f(x)(h,\eta)$.
Since $g'(b)$ is injective, we see that $Df(b)$ is injective and hence so is $Df(b)^T Df(b)$. Furthermore, by continuity, there is some $\epsilon>0$ and $\delta >0$ such that for $x \in B(b,\delta) $ we have $Df(x)^T Df(x) \ge \epsilon I$.
Since $g$ is $C^2$, and $f(b) = 0$, there is some $\delta_1 \le \delta$ such that for $x \in B(b,\delta_1)$ we have $f^T(x) D^2 \phi(x)(h,h) \ge -{1 \over 2} \epsilon \|h\|^2$ and so $D^2 \phi(x) \ge {1 \over 2} \epsilon I$.
Hence $\phi$ is strictly convex in $B(b,\delta_1)$.