College Math Teaching

February 24, 2014

A real valued function that is differentiable at an isolated point

A friend of mine is covering the Cauchy-Riemann equations in his complex variables class and wondered if there is a real variable function that is differentiable at precisely one point.

The answer is “yes”, of course, but the example I could whip up on the spot is rather pathological.

Here is one example:

Let f be defined as follows:

f(x) =\left\{ \begin{array}{c} 0, x = 0 \\ \frac{1}{q^2}, x = \frac{p}{q} \\ x^2, x \ne \frac{p}{q}  \end{array}\right.

That is, f(x) = x^2 if x is irrational or zero, and f(x) is \frac{1}{q^2} if x is rational and x = \frac{p}{q} where gcd(p,q) = 1 .

Now calculate lim_{x \rightarrow 0+} \frac{f(x) - f(0)}{x-0} = lim_{x \rightarrow 0+} \frac{f(x)}{x}

Let \epsilon > 0 be given and choose a positive integer M so that M > \frac{1}{\epsilon} . Let \delta < \frac{1}{M} . Now if 0 < x < \delta and x is irrational, then \frac{f(x)}{x} = \frac{x^2}{x} = x < \frac{1}{M} < \epsilon .

Now the fun starts: if x is rational, then x = \frac{p}{q} < \frac{1}{M} and \frac{f(x)}{x} = \frac{\frac{1}{q^2}}{\frac{p}{q}} = \frac{1}{qp} < \frac{1}{M} < \epsilon .

We looked at the right hand limit; the left hand limit works in the same manner.

Hence the derivative of f exists at x = 0 and is equal to zero. But zero is the only place where this function is even continuous because for any open interval I , inf \{|f(x)| x \in I \} = 0 .


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