Let $E$ be a closed subset of $\mathbb R$. I proved that there is a real continuous function $f$ on $\mathbb R$ whose zero set is $E$. Is it possible, for each closed set $E$, to find such an $f$ which is differentiable on $\mathbb R$, or one which is $n$ times differentiable, or even one which has derivatives of all orders?

Thanks for your help.

  • $\begingroup$ The answer is yes! I've given an example below. $\endgroup$ – Fly by Night Aug 30 '12 at 20:01

Let's say your interval is $[p,q]$ where $p \le q$. Would the following function do?

$$ f(x) := \left\{ \begin{array}{ccc} e^{-1/(x-p)^2} & : & x < p \\ 0 & : & p \le x \le q \\ e^{-1/(x-q)^2} & : & x > q \end{array}\right. $$

We have $f(x) = 0$ for all $p \le x \le q$ and

$$\lim_{h\to 0^{\pm}} \frac{f(p+h)-f(p)}{h} = \lim_{h\to 0^{\pm}} \frac{f(q+h)-f(q)}{h} = 0 \, . $$

In fact this function is infinitely differentiable over all of $\mathbb{R}$.


Yes, we can find a function which is infinitely differentiable with $E$ as its zero set. Since $E$ is closed, we know that $E^c$ is open, so $E^c$ is an at most countable union of disjoint open intervals $I_j$, at most two of which are unbounded. Then for each interval $I_j$ we define the associated function $f_j$ which is infinitely differentiable and vanishes outside of $I_j$, while being positive on $I_j$. Then we choose our function to be $$ f(x) := \begin{cases} f_j(x) & x\in I_j, \\ 0 & x \in E, \end{cases} $$ Then we note that this is well defined since each $I_j$ are disjoint, and infinitely continuous at all points in $E^c$ by construction. It is also infinitely differentiable at all points in $E$ since for each $x\in E$ we assign the same value that any $f_i$ would assign. Moreover, this function is $0$ exactly on $E$.

Thus all we need to do is create these $f_i's$. So we begin with the function $$ g(x) := \begin{cases} e^{-\frac{1}{x^2}} & x > 0, \\ 0 & x \leq 0, \end{cases}; $$

Claim : $g(x) \in C^{\infty}$. Certainly the only problem point is $x = 0$, so we consider $g^{(n)}(0)$. \ To do this we first show that for $x > 0$, $g^{(n)}(x) = p(x)e^{-\frac{1}{x^2}}$ where $p(x)$ is a polynomial with negative powers. We see that $g'(x) = 2x^{-3}e^{-\frac{1}{x^2}}$ so certainly this holds here, and now we suppose inductively that $g^{(n)}(x) = p(x)e^{-\frac{1}{x^2}}$ and consider $g^{(n+1)}$:

by the product rule $$ g^{(n+1)}(x) = p'(x)e^{-\frac{1}{x^2}} + p(x)( 2x^{-3}e^{-\frac{1}{x^2}}) = e^{-\frac{1}{x^2}}(p'(x) + p(x)2x^{-3})$$ but since $p(x)$ was a polynomial with negative powers by assumption, we have $$ p'(x) + p(x)2x^{-3} $$ is a polynomial with negative powers, and hence the induction is complete.

So now we consider $$ g^{(n)}(0) = \lim_{x\rightarrow 0}\frac{g^{(n-1)}(x) - g^{(n-1)}(0)}{x-0} $$ but $$\lim_{x\rightarrow 0^{-}}\frac{g^{(n-1)}(x) - g^{(n-1)}(0)}{x - 0} = \lim_{x\rightarrow 0^{-}} 0 = 0$$ and $$\lim_{x\rightarrow 0^{+}}\frac{g^{(n-1)}(x) - g^{(n-1)}(0)}{x - 0} = \lim_{x\rightarrow 0^{+}}\frac{p(x)e^{-\frac{1}{x^2}} - 0}{x - 0} = \lim_{x\rightarrow 0^{+}}p(x)x^{-1}e^{-\frac{1}{x^2}} = 0 $$ Thus we have $$ \lim_{x\rightarrow 0^{-}}\frac{g^{(n-1)}(x) - g^{(n-1)}(0)}{x - 0} = \lim_{x\rightarrow 0^{+}}\frac{g^{(n-1)}(x) - g^{(n-1)}(0)}{x - 0} = 0 = g^{(n)}(0) $$ Therefore we have $g^{(n)}$ exists for every $n$ and hence, $g(x) \in C^{\infty}$ as desired.

Now that we have $g(x) \in C^{\infty}$ we can manipulate it to give us the functions we need on any interval.

If we have $I_k = (-\infty , a)$ we can define $$f_k := g(a - x)$$ and if $I_k = (a, \infty)$ we can define $$f_k := g(x-a)$$. These functions give us what we need for the two unbounded intervals, since if $x \geq a$, $g(a - x) = 0$ and if $x \leq a$, $g(x - a) = 0$ while both are positive otherwise.

Next let $I_k = (a,b)$ be an interval, then we will consider the function $$f_k := g(x - a)g(b - x)$$. Then clearly for $x \leq a$ we have $g(x - a) = 0$ so $f_k (x) = 0$ and for $x \geq b$ we have $g(b-x) = 0$ so $f_k(x) = 0$, and moreover for $x\in (a,b)$, we know $g(x - a) > 0$ and $g(b - x) > 0$ so $f_k > 0$. Finally this is infinitely differentiable since it is the product of two infinitely differentiable functions. Thus for every possible interval $I_k$ we have the associated $f_k$ we need, and hence we are done.


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.