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!


1 Answer 1


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 $f:\Bbb R\to \Bbb R$.


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