1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $U\times U = \bigcup_{x_0,z_0\in U} V_{x_0,z_0}$.

$\endgroup$
1
$\begingroup$

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 $U$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.