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More generally, can the zero set $V(f)$ of a continuous function $f : \mathbb{R} \to \mathbb{R}$ be nowhere dense and uncountable? What if $f$ is smooth?

About two days ago I discovered that in this proof I am working on, I have implicitly assumed that $V(f)$ has to be countable if it is nowhere dense - hence this question.

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8 Answers

up vote 18 down vote accepted

Here's a semi-explicit construction for a smooth function f that is zero precisely on the classical Cantor set. By this set I mean the one that is obtained from I_0 = [0,1] by repeatedly removing the middle third of any ensuing interval. So let's denote by I_n the n-th set in this process.

Now let's make a smooth function f_n on [0,1] such that its zero set is exactly I_n. Starting with f_0 = 0 we obtain f_{n+1} from f_n as follows:

set f_{n+1} = f_n on I_{n+1} and on an interval that is removed from I_n make f_{n+1} equal to a bump function that is 0 only at the boundary of the interval. We can choose the bump function to be of height 2^{-2^n}.

This choice of heights of the bump functions will ensure that the derivatives of f all converge uniformly to their pointwise limits. Hence the limit function f_n is again smooth. By construction its zero set is exactly the Cantor set.

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If you work through the details, I suspect you may find you need the bumps at step $n$ to have heights decreasing faster than $2^{-n}$. The reason is that the bump has to fit into an interval of length $3^{-n}$, and the resulting squeeze makes the derivatives big (relatively speaking). –  Harald Hanche-Olsen May 9 '10 at 19:03
    
@Harald: Thanks you're absolutely right. I meant to say 2^{-2^n}. That should work. –  Roland van der Veen May 9 '10 at 19:21
    
Each bump in the bump function has width 1/3^n. If the height is h(n), then the largest value of the kth derivative of of the stage n bumps will be 3^{kn}h(n) times the largest value of the kth derivative of the original bump function. If you want these to converge uniformly, the exact condition you need is that 3^{kn}h(n) converges to 0. So the necessary and sufficient condition on h to make the construction work is that for all a>0, h(n) is eventually less than a^n. So yes, 2^{-2^n} will work. –  David Diamondstone May 16 '11 at 18:24
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It is a standard result that each closed subset of $\mathbb{R}^n$ is a zero set of some smooth function on $\mathbb{R}^n$. One proves this using smooth bump functions and partitions of unity.

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I admit I have never worked through the proof of this result. So I wonder, why partitions of unity? It seems to me that the local result is no easier than the global result. –  Harald Hanche-Olsen May 9 '10 at 19:06
    
It's true on any (paracompact) smooth manifold - you certainly need partitions of unity for that. –  Robin Chapman May 9 '10 at 19:10
    
Yes, that makes sense. –  Harald Hanche-Olsen May 9 '10 at 19:21
    
One can avoid using the partition of unity for that, just summation of "1-bump" functions with suitable coefficients.. –  Petya May 10 '10 at 3:22
    
Petya, I'm sure you're correct, but it's a bit more fiddly to ensure the uniform convergence of the sum of the derivatives of the bump functions without a convenient global coordinate system. –  Robin Chapman May 10 '10 at 9:07
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I can't resist trying my hand at sketching a proof of the general result given in Robin Chapman's answer. Let $F\subset\mathbb{R}^n$ be any closed set. Let $E_0=\{x\colon\operatorname{dist}(x,F)\ge1\}$, and for $k=1,2,\ldots$ let $E_k=\{x\colon2^{-k}\le\operatorname{dist}(x,F)\le2^{1-k}\}$. Let $\omega$ be a standard mollifier, and put $$f=\sum_{k=0}^\infty \alpha_k\chi_{E_k}*\omega_k,\qquad\omega_k(y)=2^{nk}\omega(2^ky),$$ where $\alpha_k>0$ decays fast enough so all derivatives converge uniformly ($\alpha_k=2^{-k^2}$ ought to be sufficient).

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The continuous function is very easy to construct: it's the distance to the closed set.

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Damn! My stupidity is now all the more evident. I wish it was possible to choose two answers! –  auniket May 9 '10 at 23:34
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In a normal topological space, the zero-sets of continuous functions are precisely the closed $G_{\delta}$ sets. Hence in any metric space all closed sets are, including the Cantor set.

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To mention a further point not covered in existing answers: while any closed subset of $\mathbb{R}$ can be the zero set of a smooth function, the zero set of an analytic function either consists entirely of isolated points, or is all of $\mathbb{R}$. To see this we note that if the zero set of an analytic function $f$ contains an accumulation point, then by taking a power series expansion of $f$ at the accumulation point we may extend $f$ locally to a small complex disc around that point, and apply the Identity Theorem from complex analysis to show that $f$ is everywhere zero within that disc. In particular the zero set contains an open neighbourhood in $\mathbb{R}$ of the accumulation point, and using connectedness we can repeat this argument to show that the zero set must be all of $\mathbb{R}$.

http://en.wikipedia.org/wiki/Identity_theorem

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Here's an answer from probability: a Brownian motion $B_t$ is a random, continuous function whose zero set is closed, nowhere dense, and has no isolated points. That is, $\{t : B_t = 0 \}$ is almost-surely a topological Cantor set (see, for example, Section 8 of Lalley's lecture notes).

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André Henriques answer for a closed set $C$ can easily be improved to $C^\infty$ by considering $e^{1/(\alpha-x)+1/(x-\beta)}$ if $x\not\in C$ where $\alpha$ is the supremum of all elements $< x$ in $C$ (and $\alpha=-\infty$ if $C$ contains no elements which are $< x$) and where similarly $\beta$ is the infimum of all elements $> x$ in $C$ (respectively $\beta=\infty$ if $C$ contains no elements $> x$).

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This is not C^inf. Example: take the closed set to be {-1,1}, then your function is not C^1 at 0. –  André Henriques May 10 '10 at 16:47
    
Of course. I have fixed this stupid error. –  Roland Bacher May 10 '10 at 17:54
    
Of course, a closed set need not have an upper bouned Also what is $\beta$? –  Robin Chapman May 10 '10 at 18:30
    
Sorry, part of the correction got lost du to a missing space between $<$ and $a$. –  Roland Bacher May 11 '10 at 7:13
    
So we back in one dimension again. –  Robin Chapman May 11 '10 at 17:36
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