3
$\begingroup$

I am having hard time defining a smooth function $f:\Bbb R^3 \to \Bbb R$ such that :

$f(x,y,z) = 0$ if and only if $(x,y,z)$ belongs to the unit cube $[0,1]^3$.

I tried generalizing the case of $f:\Bbb R\to \Bbb R$, such that $f$ vanishes only on $[0,1]$ but failed in the process.

I would really appreciate any help, Thanks in advance!

$\endgroup$

1 Answer 1

10
$\begingroup$

If $f:\Bbb R\to \Bbb R$ is a smooth function such that $f(x)=0\iff x\in[0,1]$ then the map $g:\Bbb R^3\to \Bbb R$ defined by $$g(x,y,z)=f(x)^2+f(y)^2+f(z)^2$$ is smooth and has the property that $$g(x,y,z)=0\iff f(x)=f(y)=f(z)=0\iff 0\leq x,y,z\leq 1,$$ so you just have to do the case $f:\Bbb R\to \Bbb R$.

$\endgroup$
1

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.