Built on Facts

Sunday Function

One of the most important mathematical concepts in physics and pure mathematics is continuity. There’s a formal definition for it which for the moment isn’t too relevant, but for our purposes we can think of it in terms of smoothness. Put your finger at a point on the graph, and if the function smoothly approaches from all directions with no instantaneous jumps or gaps and it’s continuous at that point. But it can get kind of tricky. Here’s one that’s frequently presented in intro multivariable calculus classes (I first encountered it on page 904 of Stewart).


Its mug shot:


Ignore the rendering difficulties near the origin for the moment. The function is not technically continuous at the origin anyway, because the denominator is 0 where x and y are zero. But if we could find what value the function approaches we could make an appropriate redefinition and fill in the missing point. Let’s do it with some trickery. We can put our finger somewhere else on the graph and move it toward the origin and see what the value of the function is doing. And furthermore let’s do it from every direction at once! It’s easier than it sounds: just let y = mx and you’ve got every straight line approach at once for the infinite possible values of m. Thus:


Which is manifestly equal to 0 when x approaches 0. Case closed, let’s define f(0,0) = 0 and be done, with a function which is continuous everywhere.

Or is it? Every straight line approach to the origin is not the same as every approach. What about that parabolic-shaped ridge that seems to get pretty near the origin? Let’s approach the origin along that ridge by setting x = y^2.


Which is clearly not 0 at all! So we do have an instant jump at the origin, and thus the function is irretrievably discontinuous at that point.

Alas. But these things do happen, and it’s better to recognize them before they work their way into your theory undetected later and wreck everything.


  1. #1 abb3w
    January 18, 2009

    IIR, the formal definition of continuity for a function requires you be working with a continuous starting set such as the Reals (or some similar set), where for any subset of numbers {3, 3.1, 3.14, 3.141, 3.1415…} if there is an Upper Bound to the subset (EG: 4) there must exist a Least Upper Bound within the original set (EG: π).

    However, I’m not sure physics requires this per se, or merely requires this hold true for the all of Turing-Computable sequences of numbers from the original set (that is, the set of output tapes possible on halt from some Turing machine).

    Anyone care to gift me with enlightenment?

  2. #2 DMI
    January 18, 2009

    The definition of a function being continuous at a point x0 is that for every epsilon greater than 0, there exists delta greater than zero such that the distance between x and x0 being less than delta (i.e. d(x,x0)

  3. #3 Tyler DiPietro
    January 18, 2009

    “Matt also mentioned smoothness, which is a bit stronger than continuity: smoothness also implies that the function has infinite derivatives. Since differentiability implies continuity, we get continuity, but we miss out on some important guys, like absolute value of x, which are certainly not smooth but are definitely continuous.”

    I thought the definition of a smooth function was that it generated a smooth surface (i.e., the first, second and mixed partial derivatives exist and are continuous functions of both variables). I don’t recall anything about differentiability to arbitrary order, though I’m no expert and that may well be true.

  4. #4 Joshua Zelinsk
    January 18, 2009

    Abb3w, you can when you have a function just from the reals to the reals define continuity more or less the way you want to, but it doesn’t generalize that well (R^2 has no order on it even though it has a metric). Thus, one usually uses a definition using epsilons and deltas. And if one wants to get complete generality and want continuity to make sense even when one doesn’t have a metric then you define it in terms of the underlying topology. Thus, the most general definition of a continuous function is one where the pre-image of an open set is open. It takes a little work to show that this is actually equivalent to our intuition about continuity for well-behaved metric spaces.

    Tyler, it depends somewhat on the context but in most contexts smooth means differentiable to arbitrary order.

  5. #5 Matt Heath
    January 19, 2009

    I think the using the word “smoothness” for continuity isn’t very helpful, even for giving an intuitive picture. A continuous function can have “corners”, it just can’t have “jumps”.

    The example is damn cool though. Have you done/do you intend to do sin(1/x) (varying over the positive reals) in this series? I guess maybe that’s the sort of function that mathematicians like more than physicists.

  6. #6 abb3w
    January 19, 2009

    Joshua Zelinsk: you can when you have a function just from the reals to the reals define continuity more or less the way you want to

    Errr… actually, I’m not defining continuity of the functions, there. I’m defining what I understood was required for “continuity” within the set: informally what distinguishes the “blip blip blip” nature of the integers and rationals from the smooth “varooooomm” of the reals, and (looking around some more) to have a complete ordering.

    Joshua Zelinsk: the most general definition of a continuous function is one where the pre-image of an open set is open

    Aha! So, provided the function is from the Turing-Computables to the Turing-Computables, you can get a definition of continuous functions without requiring Dedekind-completeness of the set.

    That looks useful for my twisted plan….

  7. #7 Eric Lund
    January 19, 2009

    Tyler: While “smooth” generally means differentiable to arbitrary order, there do exist functions which have continuous Nth derivatives but not (N+1)st derivatives. For example, you can fit a cubic spline to your data set, which gives you something smoother than linear interpolation–the first derivative is continuous–but the third derivative is discontinuous (I’m not sure offhand about the second). Another example is the Daubechies wavelets: the first order (Haar) wavelet is discontinuous, the second and third are continuous but not differentiable, the fourth is differentiable once but not twice, and so on.

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