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Smooth function that vanishes only on unit cube

Cantor Set and ternary expansions.The level set of a smooth functionProve that there is no function $f:BbbRtoBbbR$ with $f(0)>0$ such that $forall x,yinBbbR, f(x+y)geq f(x)+y f(f(x))$Deciding $displaystyle o,omega,Theta$ notationsProve that the function $f(n)=n^3+(fracn2^n)^5$ satisfies some property (where $ninBbbN$)Convergence when the derivative is uniformly continuousIdeas on how to find the number of unit cubes that intersect ball with radius RTrying to prove that this function is continuous for all real numbers.proving limits by ϵ−δ and ϵ−MQuestion on notation of 3rd degree Taylor expansion

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 $f:Bbb Rto Bbb R$, such that $f$ vanishes only on $[0,1]$ but failed in the process.

I would really appreciate any help,
Thanks in advance!

calculus differential-geometry

share|cite|improve this question

edited May 13 at 17:52

Adam Chalumeau

1,617420

asked May 13 at 17:25

Oriki77Oriki77

185

$endgroup$

add a comment | 

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 $f:Bbb Rto Bbb R$, such that $f$ vanishes only on $[0,1]$ but failed in the process.

I would really appreciate any help,
Thanks in advance!

calculus differential-geometry

share|cite|improve this question

edited May 13 at 17:52

Adam Chalumeau

1,617420

asked May 13 at 17:25

Oriki77Oriki77

185

$endgroup$

add a comment | 

$begingroup$

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 $f:Bbb Rto Bbb R$, such that $f$ vanishes only on $[0,1]$ but failed in the process.

I would really appreciate any help,
Thanks in advance!

calculus differential-geometry

share|cite|improve this question

edited May 13 at 17:52

Adam Chalumeau

1,617420

asked May 13 at 17:25

Oriki77Oriki77

185

$endgroup$

share|cite|improve this question

edited May 13 at 17:52

Adam Chalumeau

1,617420

asked May 13 at 17:25

Oriki77Oriki77

185

share|cite|improve this question

edited May 13 at 17:52

Adam Chalumeau

1,617420

asked May 13 at 17:25

Oriki77Oriki77

185

edited May 13 at 17:52

Adam Chalumeau

1,617420

edited May 13 at 17:52

Adam Chalumeau

1,617420

asked May 13 at 17:25

Oriki77Oriki77

185

asked May 13 at 17:25

Oriki77Oriki77

185

1 Answer
1

active

oldest

votes

10

$begingroup$

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

share|cite|improve this answer

edited May 14 at 10:25

answered May 13 at 17:32

Adam ChalumeauAdam Chalumeau

1,617420

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
  $endgroup$
  – Dan Uznanski
  May 13 at 21:47
  

  
  
  
  

add a comment | 

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10

$begingroup$

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

share|cite|improve this answer

edited May 14 at 10:25

answered May 13 at 17:32

Adam ChalumeauAdam Chalumeau

1,617420

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
  $endgroup$
  – Dan Uznanski
  May 13 at 21:47
  

  
  
  
  

add a comment | 

10

$begingroup$

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

share|cite|improve this answer

edited May 14 at 10:25

answered May 13 at 17:32

Adam ChalumeauAdam Chalumeau

1,617420

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can use the ideas from en.wikipedia.org/wiki/Non-analytic_smooth_function to generate $f$.
  $endgroup$
  – Dan Uznanski
  May 13 at 21:47
  

  
  
  
  

add a comment | 

$begingroup$

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

share|cite|improve this answer

edited May 14 at 10:25

answered May 13 at 17:32

Adam ChalumeauAdam Chalumeau

1,617420

$endgroup$

share|cite|improve this answer

edited May 14 at 10:25

answered May 13 at 17:32

Adam ChalumeauAdam Chalumeau

1,617420

answered May 13 at 17:32

Adam ChalumeauAdam Chalumeau

1,617420

answered May 13 at 17:32

Adam ChalumeauAdam Chalumeau

1,617420