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If $\text E$ is an arbitrary closed set in $\mathbf R^n$, show that there is an $\text f\in \text C^\infty(\mathbf R^n)$ such that $\text f(x)=0~$for every $x \in \text E$ and $\text f(x)>0$ for every other $x\in \mathbf R^n$.

My approach is as follows :

Since $\text E$ is given to be closed hence $\text E^c=\text V$ will be an open set in $\mathbf R^n$ and hence $\text V$ can be written as countable union of open sets $\{\text V_i\}_{i=1}^{\infty}$ and hence one can have a partition of unity $\{\psi_i\}_{i=1}^{\infty}\subset \text C^\infty (\text V)$ such that $\text {supp}(\psi_i)\subset \text V_i$. Then if i take $f=\Sigma_{i=1}^{\infty} \psi_i$ then my this $\text f$ is in $\text C^\infty (\text V)$ and $f(x)>0$ in $\text E^c$ and $f(x)=0$ in $\text E$. Now i am just left with to show that $f\in \text C^\infty (\mathbf R^n)$ or how can i generalize my $f$ to get the desired result.

Any type of help will be appreciated. Thanks in advance.

bunny
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    This is called a bump function, the proof of existence is important, and the general idea is described in the wikipedia article https://en.wikipedia.org/wiki/Bump_function – Teddy Baker Aug 30 '17 at 17:04
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    This is a theorem of Whitney: https://mathoverflow.net/questions/179445/non-zero-smooth-functions-vanishing-on-a-cantor-set – Angina Seng Aug 30 '17 at 17:04
  • @TeddyBaker It seems the function $f$ in OP's post is different from bump functions. While bump functions have compact support, the set of zeros of $f$ have to be exactly $E$. It is not clear for me how the Wikipedia's idea (on the construction of a smooth function which is $1$ inside a compact and is zero outside another compact) can be applied for this case (construction of a smooth function which is zero inside a closed and positive outside the same closed). – Pedro Aug 30 '17 at 17:56
  • Good point, it is kind of the opposite of a bump function. I think you could do the same thing as to construct the bump function, just convolve the desired domain characteristic function (in this case not compact) with a bump function, should still work. – Teddy Baker Aug 30 '17 at 17:58
  • related – Pedro Aug 30 '17 at 18:06
  • @Teddy Baker can u please give some hint on how to solve this problem – bunny Aug 30 '17 at 19:17
  • It is basically what I said in the previous comment. Check out Lee's Introduction to Smooth Manifolds for a proof of the existence of bump functions. Your function is essentially a bump function just without compact support, but I think the basic idea is the same. – Teddy Baker Aug 30 '17 at 19:47
  • @TeddyBaker I don't think that construct is that simple. The support for that function will the pointwise sum of the set and the support for the function. If the function is symettrically supported it will be the points nearer than a fixed distance from the set, but such a set may not be that easy to construct as the size of the gaps in a closed set need not have a lower bound. – skyking Aug 31 '17 at 05:53
  • The title says 'Finding a function' while the problem is 'Show there is a function'. But proving the existence is not necessarily the same as showing an example. – CiaPan Aug 31 '17 at 06:50
  • You might be right, that is why I wrote it in the comments instead of an answer. It seems ok to me, but there might be a technicality I'm missing. Anyways it seemed it was a duplicate, so no worries. – Teddy Baker Sep 01 '17 at 18:14

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Following the lead from Teddy Baker, the function you look for is 1 - φ where φ is the bump function defined as below: (However, do pay attention to the "Edit 1" below, as $K$ needs not to be a compact set.)

If K is an arbitrary compact set in n dimensions and U is an open set containing K, there exists a bump function φ which is 1 on K and 0 outside of U. Since U can be taken to be a very small neighborhood of K, this amounts to being able to construct a function that is 1 on K and falls off rapidly to 0 outside of K, while still being smooth.

(https://en.wikipedia.org/wiki/Bump_function)

The construction of the bump function is described here:

The construction proceeds as follows. One considers a compact neighborhood V of K contained in U, so K ⊂ Vo ⊂ V ⊂ U. The characteristic function ${\chi} _{V}$ of V will be equal to 1 on V and 0 outside of V, so in particular, it will be 1 on K and 0 outside of U. This function is not smooth however. The key idea is to smooth $\chi _{V}$ a bit, by taking the convolution of $\chi _{V}$ with a mollifier. The latter is just a bump function with a very small support and whose integral is 1. Such a mollifier can be obtained, for example, by taking the bump function $\phi$ from the previous section and performing appropriate scalings.

The key idea for the construction is the use of convolution, and the fact that the convolution of a discontinuous function and a smooth function is a smooth function.

For example in 1D, let $E = [-1,1]$, simply choose sets $V$ and $U$ such that:

$E = [-1,1] \subset V = [-2,2] \subset U = (-3,3)$

Now let $\chi_V$ be the characteristic function of V. That is: $$ \chi_V(x) = \begin{cases} 1 \quad \text{if $x \in V$} \\ 0 \quad \text{otherwise.} \end{cases} $$

Convolve $\chi_V$ with a mollifier f (a summetric bump function: https://en.wikipedia.org/wiki/Mollifier) with a small enough support. In this case any mollifier f with support size less than or equal to 1 will work (since the distance between a boundary of V to that of E and U is 1). One example of such a mollifier is: $$ f(x) = \begin{cases} \frac{1}{C} e^{-\frac{1}{1-x^2}} \quad \text{if $-1 \leq x \leq 1$}\\ 0 \quad \text{otherwise,} \end{cases} $$ where $C$ is a (normalization) constant such that the integral over $[-1,1]$ equals 1.

Since $$ (\chi_V \star f)(x) = \begin{cases} 1 \quad &\text{if $x \in E$}\\ \text{a value between $0$ and $1$} \quad &\text{if $x$ is in $V$ but not in $E$ or $U$,} \\ 0 \quad&\text{if $x$ is in $U$ but not in $V$,} \end{cases} $$

the following function $g$ is the answer to your question (satisfying the condition that it equals 0 inside E and being non-zero outside E):

$$ g = 1-\phi = 1 - \chi_V \star f. $$

Edit 1 (Adding an example with unbounded sets as a response to comment from user Pedro): Interestingly, the mentioned convolution technique can be straightforwardly extended to unbounded sets (however with certain conditions on $E$--to be mentioned more later). As an example, one can slightly hijack the example above to have a new example with unbounded sets. For example, consider the following sets:

$E = (-\infty,1] \subset V = (-\infty,2] \subset U = (-\infty,3)$

Now, the answer to the original question (find a smooth function that equals 0 inside $E$ and non-zeros outside $E$) is the same as before: $g = 1 - \chi_V \star f$.

Conditions on $E$ so that the argument still works: I think the condition $E$ is closed is enough (will keep thinking a bit more).

MMM
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  • But $E$ is not bounded in the original question. – Pedro Sep 07 '17 at 03:41
  • @Pedro: it is a good point, I added a few lines regarding the case where $E$ is unbounded. – MMM Sep 08 '17 at 07:08
  • In $\mathbb{R}$ you can write $U = \bigcup_{i}^\infty (a_i,a_{i+1})$ and set $\varphi = \sum_i \phi_{a_i,a_{i+1}}$ where $\phi_{a_i,a_{i+1}}$ is $C^\infty$ and non-zero exactly on $(a_i,a_{i+1})$. In $\mathbb{R}^2$ it is a little different and you should look at Robert's answer – reuns Sep 08 '17 at 07:38