# Using implicit function theorem without using the inverse function theorem.

Let $f:U\rightarrow \mathbb{R}$ defined on the open set $U \subset \mathbb{R}^m$. It the function $g(x):U\rightarrow \mathbb{R}$, given by the expression $$g(x)= \int_{0}^{f(x)} (t^2 + 1)dt,$$ of class $C^{\infty}$, then f is $C^{\infty}$.Explicitly using the implicit function theorem, knowing that the inverse function is equivalent.

Let $H:U\times\mathbb{R}^n\times\mathbb{R}^n\to \mathbb{R}$ the function $$H(x,y,z)= \int_{\langle x,x\rangle -{\langle z,z\rangle}}^{\mbox{sgn}\big(f(x)\big)\cdot\langle y,y\rangle} \big( t^2+1\big) \; \mathrm d t$$ By chain rule we have \begin{array}{rl} \left.\frac{\partial H(x,y,z)}{\partial y}\right|_{(x,y,z)=(x_0,y_0,x_0)} = & \left. \frac{d}{du}\left( \int_0^{u} \big( t^2+1 \big) \mathrm d t\right) \right|_{u=\mbox{sgn}\big(f(x_0)\big)\cdot\langle y_0,y_0\rangle} \cdot \left.\frac{\partial}{\partial x} \Big( \mbox{sgn}\big( f(x)\big)\langle y,y\rangle \Big)\right|_{y=y_0} \\ & \\ & \\ & \\ & \\ = & \left. \big( u^2+1 \big) \right|_{u=\mbox{sgn}\big(f(x_0)\big)\cdot\langle y_0,y_0\rangle} \cdot \left.2 \cdot\mbox{sgn}\big( f(x)\big)\cdot y\right|_{(x,y)=(x_0,y_0)} \\ & \\ & \\ & \\ = & 2 \cdot\mbox{sgn}\big( f(x_0)\big)\cdot\big( \langle y_0,y_0\rangle^2+1 \big) \cdot y_0 \\ \end{array} Then for $x_0\in U$ such that $f(x_0)\neq 0$ and $y_0\neq 0$ we have $$\left.\frac{\partial H(x,y,z)}{\partial y}\right|_{(x,y,z)=(x_0,y_0,x_0)}\neq 0$$ By Implicit Function Theorem there is:

• a open sets $V=V_{(x_0,z_0)}\subset U\times\mathbb{R}^n$ and $W=W_{y_0}\subset \mathbb{R}^n$ such that $V_{(x_0,z_0)}\times W_{y_0}\subset U\times\mathbb{R}^n\times\mathbb{R}^n$.

• a function $\varphi: V\to W$ such that $$H(x,\varphi(x,z),z)=H(x_0,y_0,z_0=x_0) \quad \forall (x,z)\in V \quad \forall y\in W$$

• The function $\varphi: V\to W$ is unique in the sense that if there is a function $\psi :V\to W$ with the previous property $$H(x,\psi (x,z),z)=H(x_0,y_0,z_0) \quad \forall (x,z)\in V \quad \forall y\in W$$ then $\varphi = \psi$.

• The function $\varphi$ has the same class of differentiability of $H$. In particular if $H$ is of class $C^\infty$ then $\varphi$ is class $C^\infty$.

Now to finalize the point suffices to observe that if $f(x)=\mbox{sgn}\big( f(x)\big)|f(x)|=\mbox{sgn}\big( f(x)\big)\langle y,y\rangle$ $$g(x)=H(x,y,x)= \int^{\mbox{sgn}\big( f(x)\big)\langle y,y\rangle}_0 (t^2+1) \mathrm d t =\int^{\mbox{sgn}\big( f(x)\big)f(x)}_0 (t^2+1) \mathrm d t$$ On the other hand, for all neighborhood $V_{(x_0,z_0)}\times W_{y_0}$ of a point $(x_0,y_0,z_0)$, with $x_0 = z_0$, that implicit function $\varphi: V_{(x_0,z_0)}\to W_{y_0}$ of Implicit Function Theorem ensures the equality $$g(x)=H(x,\varphi(x,x),x)= \int^{\varphi(x,x)}_0 (t^2+1) \mathrm d t =\int^{\mbox{sgn}\big( f(x)\big)\langle y,y\rangle}_0 (t^2+1) \mathrm d t$$ By uniqueness of $\varphi : V_{(x_0,z_0)}\to W_{y_0}$ for each point $(x_0,y_0,z_0)=(x_0,x_0,z_0)$ we have $$\varphi(x,x)= \mbox{sgn}\big( f(x) \big)\cdot |f(x)|=f(x)$$ Then we can conclude $f$ is of class $C^\infty$ by cause $\varphi$ is $C^\infty$. How is any point $(x_0,y_0,z_0)$ in $U\times\mathbb{R}^n\times\mathbb{R}^n$ we can extend $\varphi(x,x)$ (again because of the uniqueness) for all $U\times U = \bigcup_{x_0,z_0\in U} V_{x_0,z_0}$.

The function $$\phi:\quad {\mathbb R}\to{\mathbb R},\qquad y\mapsto y+{y^3\over3}$$ is $C^\infty$ and has first derivative $\geq1$; therefore it maps ${\mathbb R}$ diffeomorphically onto ${\mathbb R}$. It follows that $\phi$ has a $C^\infty$ inverse $\psi:\ {\mathbb R}\to{\mathbb R}$.

Now $$g(x)=\phi\bigl(f(x)\bigr)\qquad(x\in U)$$ and therefore $$\psi\bigl(g(x)\bigr)= f(x)\qquad(x\in U)\ .$$ This proves that the function $f$ is $C^\infty$ in $U$.