Let $g: V \to \mathbb{R}^{m}$ be a $C^{2}$ function in the open $V \subset \mathbb{R}^{m}$. Given $b \in V$, suppose that $g'(b): \mathbb{R}^{m} \to \mathbb{R}^{m}$ be an isomorphism. Prove that there is an open ball $B$ of center, such that $\varphi: B \to \mathbb{R}$, defined by $\varphi(y) = |g(y) - g(b)|^{2}$, is convex.
I cannot gather the information I have to find a way to the solution. By hypothesis, the Inverse Theorem Function says that $g$ is a local diffeomorphism. I know many criteria to show that a function defined in $\mathbb{R}$ is convex, however, I don't know if they are true in $\mathbb{R}^{m}$. A result that I can use is: $\varphi((\alpha x + \beta y) \leq \alpha\varphi(x) + \beta\varphi(y)$ with $\alpha+\beta=1$. But, I dont know how to connect the convexity with diffeomorphism. CAn someone help me? Thanks for the advance.
I don't know if there is a solution without the Inverse Function Theorem, but I have to use it to solve.
Write $g = (g_1,\dots,g_m)$. Then $\varphi(y) = (g_1(y)-g_1(b))^2+\dots+(g_m(y)-g_m(b))^2$. Note $$\varphi(\alpha x+\beta y) = \sum_{j=1}^m (g_j(\alpha x+\beta y)-g_j(b))^2$$ $$\alpha \varphi(x)+\beta \varphi(y) = \sum_{j=1}^m \alpha(g_j(x)-g_j(b))^2+\beta (g_j(y)-g_j(b))^2.$$ So, to prove that $\varphi(\alpha x +\beta y) \le \alpha \varphi(x)+\beta\varphi(y)$, it suffices to show that for each $j$, $$(g_j(\alpha x+\beta y)-g_j(b))^2 \le \alpha (g_j(x)-g_j(b))^2 + \beta (g_j(y)-g_j(b))^2.$$ This is precisely showing that the function $\varphi_j : \mathbb{R} \to \mathbb{R}$ given by $\varphi_j(y) = (g_j(y)-g_j(b))^2$ is convex. The good news is that $\varphi_j$ is a function on $\mathbb{R}$! So, we know it suffices to show that $\varphi_j''(y) \ge 0$ for each $y$. [Note that since $g$ is a $C^2$ function, by definition each $g_j$ is $C^2$, and thus $\varphi_j$ is.] But $\varphi_j''(y) = 2$ for each $y$.