# Not continuous at a point

The simplest type of pathological function is probably a function which is well-behaved, except for having a jump discontinuity. An example of such a function is given by \[ f(x) = \left\{ \begin{array}{ccc} x &\text{if} & x \le 0 \\ x+1 & \text{if} & x\gt 0 \end{array} \right. \] What makes this function misbehave is essentially that it has two different definitions, applicable on different parts of the domain, and which somehow disagree where the different parts meet. This suggests a strategy for constructing 'monsters' by appropriate choices of how to split up the domain and how to define the function on the different parts.

# Nowhere continuous

We can take this notion of making a function discontinuous at one point and push it to the limit by splitting up the domain into two sets, each of which is dense. One familiar way of making such a split is to consider the rational numbers (\(\mathbb{Q})\) and the irrational numbers (\(\mathbb{R} \setminus \mathbb{Q}\)). Then if \(x \in \mathbb{R}\), no matter how small a number \(\varepsilon \ge 0\) is chosen, the interval \((x-\varepsilon,x+\varepsilon)\) contains both rational and irrational numbers.So let's now look at the function \[ f(x) = \left\{ \begin{array}{rcc} 0 &\text{if} & x \in \mathbb{Q} \\ 1 & \text{if} & x \in \mathbb{R} \setminus \mathbb{Q} \end{array} \right. \] If we choose some real number \(x\), then if \(x\) is rational, so \(f(x)=0\), we can find irrational numbers as close as we like to \(x\), so that arbitrarily close to \(x\) are values on which \(f\) takes on the value \(1\); likewise if \(x\) is irrational, so \(f(x)=1\), we can find rational numebrs as close as we like to \(x\), so that arbitrarily close to \(x\) are values on which \(f\) takes on the value \(0\).

It follows then that no matter what value of \(x\) we choose, \(f\) is not continuous.

We can sort of visualize the graph of this \(f\) as a pair of dotted lines, one at height \(0\) (for the rational values of \(x\)) and one at height \(1\) (for the irrational values). It's then pretty plausible than changing the value of \(x\) by any amount at all involves jumping up and down between the two lines, so the function is not continuous anywhere.

This rather went from the sublime to the ridiculous in one jump. But I think that the next case is still stranger.

# Continuous at just one point

This time we have \[ f(x) = \left\{ \begin{array}{rcl} 0 &\text{if} & x \in \mathbb{Q} \\ x & \text{if} & x \in \mathbb{R} \setminus \mathbb{Q} \end{array} \right. \] By the same argument as above, this function is not continuous at any value of \(x\) other than \(0\).But at \(0\), the story is different.

This time, if we chose a value of \(x\) close to zero, the value of \(f(x)\) is either \(0\), if \(x\) is rational, or \(x\) which is small, if \(x\) is irrational. In either case, we can make \(f(x)\) as close as we want to \(0\) by taking \(x\) sufficiently small: so \(f\) is continuous at just this one point.

In much the same way as before, we can try to visualize the graph as a pair of lines again, this time at height \(0\) for rational values, and at height \(x\) for irrational ones. Away from zero we have the same situation as before, but at zero, where the graphs 'intersect', the two definitions (almost) agree, and so the function is continuous.

So we can have a function that is continuous everywhere except at one point, or just at one point. It's not hard to see we can have as many discontinuities as we want. But now it's time for a really weird function.

# Continuous on just the irrationals

This one is hard to believe. First, though, agree that if \(x\) is a rational number, then we express it as the ratio of two integers \(m/n\) where \(n>0\) and \(\gcd(m,n)=1\). We then define \[ f(x) = \left\{ \begin{array}{rcl} 1/n &\text{if} & x \in \mathbb{Q} \\ 0 & \text{if} & x \in \mathbb{R} \setminus \mathbb{Q} \end{array} \right. \] So if \(x\) is a rational number, then \(f(x)\) is non-zero, but arbitrarily close to \(x\) are irrational numbers, on which \(f\) takes the value \(0\). So \(f\) is not continuous at \(x\).On the other hand, if \(x\) is not rational, then no matter how large an integer \(N\) we choose, there is an interval about \(x\) so small that no rational number with denominator less than \(N\) lies inside that interval. So any number inside the interval is either irrational, so \(f\) takes on the value \(0\), or is rational, so the denominator must exceed \(N\), and the value is less than \(1/N\).

So \(f\) is continuous at this \(x\).

This is very hard to comprehend. I can try to think of a graph analogous to the other two cases, but when I think about it carefully I realize I'm kidding myself.

# Continuous on just the rationals

You might expect (I certainly expected) that it would be possible to do this by a suitable clever adaptation of the previous example. It turns out that that doesn't work. In fact, nothing works. It isn't possible for a function to be continuous on just the rationals.More than anything else, this makes me realize that I don't really understand just how the previous example works.

This pretty much wraps it up for the continuity monsters. But there are other monsters to admire.

# Differentiable, but the derivative is not continuous

This is subtler than you might expect. You can't get it by integrating a function with a discontinuity, because the result isn't differentiable. So consider \[ f(x) = \left\{ \begin{array}{ccl} x^2\sin(1/x) &\text{if} & x \neq 0 \\ 0 & \text{if} & x =0 \end{array} \right. \] Then as long as \(x \neq 0\), the usual rules of differentiation give us \[ f'(x) = 2x \sin(1/x) - \cos(1/x) \] But as \(x\) approaches \(0\), this is not at all well-behaved. It just oscillates faster and faster between (approximately) \(\pm 1\).On the other hand, if \(h \neq 0\) then \[ \begin{split} \frac{f(h+0)-f(0)}{h} &= \frac{h^2\sin(1/h)}{h}\\ &= h \sin(1/h) \end{split} \] so as \(h \to 0\), since \(\sin(1/h\) is bounded above by \(1\) and below by \(-1\), this also tends to \(0\), so that \(f'(0)=0\).

So it is possible for a derivative to be discontinuous, but not because of a jump discontinuity.

We can fairly obviously make this work for higher and higher derivatives by using higher powers of \(x\). That isn't very interesting. Let's look at something a bit weirder.

# Differentiable at just one point

In fact, this will be even stranger than it sounds. We can have a function that is differentiable at just one point, but not even continuous anywhere else. \[ f(x) = \left\{ \begin{array}{rcl} x^2 &\text{if} & x \in \mathbb{Q} \\ 0 & \text{if} & x \in \mathbb{R} \setminus \mathbb{Q} \end{array} \right. \] Again, by just the same argument as before, this function is not continuous at any non-zero value of \(x\).But (rather like the above example) if \(h \neq 0\), we have \[ \frac{f(0+h)-f(0)}{h} = \left\{ \begin{array}{rcl} h &\text{if} & h \in \mathbb{Q} \\ 0 & \text{if} & h \in \mathbb{R} \setminus \mathbb{Q} \end{array} \right. \] so again we see that as \(h \to 0\), the limit is \(0\), so \(f'(0)=0\).

And again, we can make this as differentiable as we want at \(0\) by using higher powers of \(x\).

It's probably worth saying out loud that we can't do this by working out derivatives of derivatives, since the derivative only exists at \(0\). Instead, we say that \(f\) is differentiable \(n\) times at \(x\) if there exist numbers \(f^{(i)}(x)\) for \(i=1,\ldots n\) with \[ f(x+h) = f(x) + f'(x)h + \ldots + \frac{1}{n!}f^{(n)}(x) + e \] where \(e/h^n \to 0\) as \(h \to 0\).

Now, these are all interesting in their own right, just as examples of what can happen. Let's finish with an example which is interesting and useful.

# Infinitely differentiable but not analytic

We finish off with \[ f(x) = \left\{ \begin{array}{ccl} \exp(-1/x) &\text{if} & x \gt 0 \\ 0 & \text{if} & x \le 0 \end{array} \right. \] A straightforward calculation shows that this function has derivatives of all orders at \(0\), and the derivatives are all \(0\). So this function is infinitely differentiable, but since its power series is just \(0\), it is not analytic.Using this, we can build a new function: \[ g(x) = \frac{f(x)}{f(x)+f(1-x)} \] This function is \(0\) for negative values of \(x\), \(1\) for values of \(x > 1\), and smoothly interpolates between them. Functions built out of these are important in differential geometry and differential topology, to patch together smoothly objects which are defined locally. It is an argument using such functions that proves, for example, that any smooth manifold admits a Riemannian metric.

# You missed a bit!

There is one very well-known type pathological function which I haven't mentioned: the functions which are continuous but nowhere differentiable, first described by Karl Weierstrass. The reason is that I wanted to consider functions where there is an explicit (and simple) formula for the value of \(f(x)\), so that the properties could be seen directly. I've neglected the continuous but not differentiable functions although they are also interesting, because they are defined by means of limits rather than a simple formula.

# Enough is enough

Most of these functions are, basically, a freak show. We look at them to see objects with bizarre or surprising properties; and some actually have practical uses. But even those without obvious application are important. They help to find weaknesses in our understanding of basic concepts, and thereby to sharpen our understanding of them.Also, they're fun.

## Comment

It has been pointed out in the comments below that technically these functions are not freaks. In fact, amongst functions, the continuous functions are the freaks (i.e. the rarities), and amongst continuous functions, the differentiable ones are the freaks. But in terms of the kind of functions that students meet in practice, where (at worst piecewise) analytic is standard, these functions are still the monsters.

## Acknowledgement

- I probably wouldn't have written this if @panlepan hadn't tweeted about continuous but non-differentiable functions.
- Thanks to Manley Perkel for typo spotting.
- thanks to delio for his comments on what is (and is not) a freak.