6
$\begingroup$

Suppose $f:U \subseteq \mathbb{R}^2 \rightarrow \mathbb{R}$ is continous and $$(x^2+y^4)f(x,y)+(f(x,y))^3=1 \: \text{for all} \: (x,y) \in U. $$ Prove $f$ is $C^\infty$.

This kind of exercise is new to me and I don't really have any idea how to derive that the derivative exist infinitely and it's continous.

$\endgroup$
  • 3
    $\begingroup$ Not sure if this helps, but that implies $f$ satisfies the functional equations $$ \begin{split} \frac{1}{f(x,y)} &= x^2+y^4 + f(x,y)^2\\ f(x,y) &= \frac{1}{x^2+y^4 + f(x,y)^2} \end{split} $$ $\endgroup$ – gt6989b Apr 17 at 15:01
  • $\begingroup$ Yeah, we can see that $f(x,y)\ne 0$ otherwise $f$ would not satisfy the equation. $\endgroup$ – ipreferpi Apr 17 at 15:04
  • 3
    $\begingroup$ Idea: Implicit Function Theorem. $\endgroup$ – Martín-Blas Pérez Pinilla Apr 17 at 15:11
  • 2
    $\begingroup$ Idea: Cardano's formula. $\endgroup$ – Michael Hoppe Apr 17 at 15:14
  • 2
    $\begingroup$ More legible: $$ f(x,y)= \frac{\left(\sqrt3 \sqrt{4 a^3 + 27} + 9\right)^{1/3}}{2^{1/3} 3^{2/3}}-\frac{(2/3)^{1/3} a}{\left(\sqrt3 \sqrt{4 a^3 + 27} +9\right)^{1/3}}$$ where $a=x^2+y^4$. $\endgroup$ – Michael Hoppe Apr 17 at 15:30
3
$\begingroup$

The answers really are in the comments to the main question, and are due to Martin Blas Perez and Michael Hoppe. Hence this post is CW.

The function $z=f(x, y)$ satisfies the equation $$ (x^2+y^4)z+z^3=1, $$ so it can never vanish. In particular, by the implicit function theorem, $f$ must be $C^1$ and $$ \begin{array}{cc} \frac{\partial z}{\partial x} = -\frac{2xz}{x^2+y^4+3z^2}, & \frac{\partial z}{\partial y} =-\frac{4y^3z}{x^2+y^4+3z^2}, \end{array} $$ and we remark that the denominators can never vanish. By iterating this process we see that $f$ is infinitely differentiable.

Actually, Micheal even computed an explicit expression for $z$. See his comment to the main question, which I hope he turns into an answer. From that expression it is manifest that $z$ is smooth.

$\endgroup$
  • $\begingroup$ I'll accept this answer since it addresses Martin and Michael comments. Thanks. $\endgroup$ – ipreferpi Apr 17 at 16:31

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.