Let $u:\mathbb{R}\rightarrow\mathbb{R}$ a convex and $C^2(\mathbb{R})$ function such that $\lim\limits_{|x|\rightarrow +\infty}u(x)=+\infty$ and $u(0)=0=\min_{\mathbb{R}}u$.
Now I know that I can consider the inverse functions of the restrictions of $u$ for positive and negative $x$:
if $u_+:[0,+\infty)\rightarrow[0,+\infty)$, $u_+(x)=u(x)$ for every $x\geq 0$ then exist the inverse function of $u_+$, i.e. $(u_+)^{-1}$;
if $u_-:(-\infty,0]\rightarrow[0,+\infty)$, $u_-(x)=u(x)$ for every $x\leq 0$ then exist the inverse function of $u_-$, i.e. $(u_-)^{-1}$.
I denote \begin{equation*} \phi(y)=\begin{cases}( u_+)^{-1}(y)\quad y\geq 0 \;\\ (u_-)^{-1}(y) \quad y\leq 0.\end{cases} \end{equation*}
I would like to prove that the function $y\rightarrow\phi(y)\;$ is $\;C^1(\mathbb{R})$.
Edit I have proved that $\phi$ is $C^0(\mathbb{R})$ and that it is derivable in $(0,+\infty)$ and $(-\infty,0)$. The only problem is the derivability in $y=0$.
How can I prove that $$\lim_{y\rightarrow 0^+}\phi'(y)=\lim_{y\rightarrow 0^-}\phi'(y)?$$
Thanks for the help!