Well-known example: Consider the function $$f(x,y)=\left\{\begin{array}{c} \frac{x^2y}{x^4+y^2} & \text{if }(x,y)\neq(0,0)\\ 0 & \text{if }(x,y)=(0,0) \end{array}\right.$$ When restricted to any straight line through the origin, this function is continuous. However, if we approach the origin along the parabola $y=x^2$, we get a limit of $\frac 12$, so $f$ is actually discontinuous. The question is whether smooth curves can always ferret out discontinuity in this way.

Does there exist a function $f:\mathbb R^2\to \mathbb R$ which is discontinuous at a point $x$, but is continuous at $x$ when restricted to any smooth curve?

Discontinuity is witnessed by a sequence $\{x_i\}$ converging to $x$ so that $\{f(x_i)\}$ does not converge to $f(x)$. Since we could take $f$ to be the characteristic function on $\{x_i\}$, the above question is equivalent to the following.

Given a convergent sequence $\{x_i\}$ in $\mathbb R^2$, must there be a smooth curve passing through infinitely many of the $x_i$?

  • 1
    $\begingroup$ You know, it's weird. For some reason, just today, I was thinking about this exact example and asking myself this exact question. $\endgroup$
    – Ryan Reich
    Jun 8, 2012 at 1:08
  • 7
    $\begingroup$ Why can't you just choose a super-fast converging subsequence of the $x_i$ (say $\|x_i\|<2^{-i}$) and then put together curve segments that visit each of these. e.g. between $1-1/n$ and $1-1/(n+1)$ you pay a visit to $x_n$. Since $2^{-n}$ is smaller than any power of $1/n^2$, you can make this as smooth as you want. $\endgroup$ Jun 8, 2012 at 1:09
  • 1
    $\begingroup$ @Anthony: I don't follow. How do you make sure the curve is smooth at the limit point? What is the significance of $2^{-n}$ going to zero faster than any power of $1/n^2$? $\endgroup$ Jun 8, 2012 at 1:56
  • 3
    $\begingroup$ @Anton: Suppose you want a $C^1$ curve from $[0,1]$ going through the points that you chose. Suppose you take a sequence of excursions from the limit point to the points on your sequence, and that you will use a sub-interval $J_n$ of $[0,1]$ to visit $x_n$. Suppose further you want all derivs to vanish at the endpoints of $J_n$. Then a quick calc shows you need $|J_n|$ at least $\|x_n\|$ to achieve this. If you want $C^2$, you need $|J_n|\ge \|x_n\|^2$ etc. As long as you have all of this, then you can make each derivative converge to 0 as you approach the limit. $\endgroup$ Jun 8, 2012 at 2:02
  • 2
    $\begingroup$ In this context, does smooth mean $ C^k $ for as large a $ k $ as you want, or $ C^{\infty} $? $\endgroup$ Jun 8, 2012 at 10:51

3 Answers 3


Usually, by a "smooth curve through $0\in\mathbb R^2$", one means a $C^k$ mapping $f:[-a,a]\to\mathbb R^2$ where $k\ge 1$ is the desired degree of smoothness, $a>0$, $f(0)=0$, and $f'(x)\ne 0$ for all $x\in[-a,a]$ (without the last condition, you can get quite ugly images even of $C^\infty$ mappings). Equivalently, a smooth curve is a graph of a $C^k$-function near the origin after some rotation of coordinates. Now, the answer depends on $k$. If $k=1$, then there is a subsequence $x_n$ approaching $0$ from some limiting direction. Choosing this direction as the positive semiaxis, we see that we can rarefy the sequence a bit and put everything on a $C^1$ graph with $0$ derivative at the origin. On the other hand, to request any particular modulus of continuity for $f'$ is already impossible because that would require some estimate of the kind $|\mbox{arg}(x)|\le \omega(|x|)$ with some fixed function $\omega$ tending to $0$ at $0$ where $\mbox{arg}$ is measured from the limiting direction. However, we can construct a sequence of points whose absolute values tend to $0$ very fast while their arguments tend to $0$ very slowly.

  • $\begingroup$ Nice! I was secretly assuming $f'(x)\neq 0$, since as you said, the image can be quite bad otherwise. $\endgroup$ Jun 10, 2012 at 15:13

Yes you can do this. Suppose you have a sequence of points $c_n$ which converges very fast to zero, so that $n^k c_n$ is bounded for all $k$. Let $h$ be a smooth function that is equal to $0$ for $x\le -1$ and $1$ for $x\geq 0$. Let $t_n=\frac{1}{(n+1)^2}+2\sum_{k=1}^n \frac{1}{k^2}$ and $$g(x)=\sum_{n\geq 0}h(1+n^2(x-t_n))h(1-n^2(x-t_n))c_n$$ then $g(t_n)=c_n$ and g is smooth because the fast convergence of the $c_n$ implies that the derivatives of all orders of the terms in the summation are uniformly bounded. This proof works in greater generality and the result is called "the general curve lemma". See 12.2 in Kriegl-Michor, "The convenient setting of global analysis".

  • $\begingroup$ I see this was already described by Anthony Quas in the comments above. $\endgroup$ Jun 8, 2012 at 2:14
  • $\begingroup$ Nice. This answer plus fedja's answer resolve all variants of the question that I can think of. $\endgroup$ Jun 10, 2012 at 15:15

Just for the sake of exposition, take ${\Bbb C}$ as the model of ${\Bbb R}^2$. Given a sequence $(z_n)$ of non-zero complex numbers which converges to 0, the compactness of $S^1$ entails the existence of a subsequence such that Arg$(z_n)$ converges. Then, passing if necessary to an even thinner subsequence $(w_n)$, one can also get the convergence od the arguments monotonic and one-sided (in some appropriate sense). Connecting the dots, even just with line segments, gives a curve smooth at 0, and a little more care will make it smooth everywhere. (I assume "smooth" here means $C^1$...I don't think $C^n$ is much harder, but I haven't thought about $C^\infty$.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.