If the graphs of two continuous real functions $$f$$ and $$g$$ “cross” each other, then there must be at least an intersection point for which $$f(x) = g(x)$$. This is a consequence of the Intermediate Value Theorem. But intersections are no longer guaranteed if one of the two functions is not continuous.

We draw again inspiration from a function we learned to know in an older post. We briefly recall here that $$f: [0, +\infty) \to [0, + \infty)$$ maps the point $$x$$ with decimal expansion  $x = \sum_k a_k 10^k,$ into $f(x) = \sum_{k} a_{2k}10^k,$ keeping thus only the enven positioned digits of the input number $$x$$. In case of ambiguity in the notation, the finite version of the number is adopted (so we will use $$1$$, rather than $$0.999\dots$$). We previously showed that $$f$$ is discontinuous on a dense subset of its domain, despite being continuous almost everywhere (in particular at points with infinite decimal representation).

Suppose a continuous function $$g$$ crosses one of the straight lines that approximate the behaviour of $$f$$ (see the Figure below). Can we say that there is a point $$a$$ such that $$f(a) = g(a)$$? In order to make sure this happens, we actually need – from $$g$$’s side – something stronger than continuity.

What we need is Lipschitz continuity, in an interval in which the two graphs cross. We say that $$g$$ is Lipschitz continuous in an interval $$[a, b]$$ if there is an $$M$$ such that, for all $$x$$ and $$y$$ in $$[a,b]$$, $|g(x)-g(y)| \leq M |x-y|\tag{1}\label{eq3375:1}.$ Of course if $$g \in C^1$$, that is if $$g$$ is continuously differentiable, then condition \eqref{eq3375:1} certainly holds.

We will demonstrate the following.

Theorem. Given the function $$f$$ defined above, and a function $$g$$ that is Lipschitz continuous in $$[0, b]$$, with $$g(0) = 0$$, then there exists a point $$a \in (0,b]$$ such that $f(a) = g(a).$

In order to prove the Theorem we will devise a procedure that either stops when $$a$$ is found or generates a sequence $$(a_n)$$ converging to the desired point.

• Firstly, we need to choose an appropriate “scale”. Thus we will choose $$k$$ in such a way that $10^k< \min\left\{\frac{20}M,\ \ \sqrt b\right\},$where $$M$$ is the number that guarantees \eqref{eq3375:1}. Show, then, that $$g$$ is Lipschitz continuous in $$\left[0, 10^{2k}\right]$$, and that the graph of $$g$$ must cross the second straight line from the left in Figure 1 (the one marked in dark blue).
• Our first “approximation” to $$a$$ is given by $$a_0$$, the intersection point between this line and $$g(x)$$, whose existence is guaranteed by the Intermediate Value Theorem. Show that $$a_0$$ is a solution to the equation $g(x) = 10^{-k+1}\left(x-10^{2k-1}\right).$ We can now zoom on the line and see what is shown below.
• Recall that $$f$$ is confined within the dashed parallelograms (which, once magnified, are scaled replicas of Figure 1). Show that either the point $$\left(a_0, g(a_0)\right)$$ is the vertex of one of such parallelograms, or $$g(x)$$ intersects its left side or its right side (here, again, Lipschitz continuity comes into play!). In the first case, the least significant digit of $$a_0$$ is the $$(k-1)$$th one, and we can set $$a=a_0$$, thus terminating the procedure. In the second case, however, we have $|f(a_0) – g(a_0)| \leq 10^{k-1},$ and we can refine our approximation by finding the intersection between the graph of $$g$$ and one of the “vertical” sides of the parallelogram.
• Find the equation of the left and right sides of the parallelogram containing $$\left(a_0, g(a_0)\right)$$. Conclude that our refinement $$a_1$$ is a solution to the equation $g(x) =10^{-k+2}\left(x – \frac{\left[10^{-2k+2}a_0\right]}{10^{-2k+2}}\right) + f\left(\frac{\left[10^{-2k+2}a_0\right]}{10^{-2k+2}}\right),$ such that $$\frac{\left[10^{-2k+2}a_0\right]}{10^{-2k+2}}<a_1<\frac{\left[10^{-2k+2}a_0\right]}{10^{-2k+2}}+10^{2k-3}$$, or to the equation \begin{eqnarray}g(x) &=&10^{-k+2}\left(x – \frac{\left[10^{-2k+2}a_0\right]}{10^{-2k+2}}+9\cdot 10^{2k-3}\right) + \\ & &+f\left(\frac{\left[10^{-2k+2}a_0\right]}{10^{-2k+2}}+9\cdot 10^{2k-3}\right),\end{eqnarray} such that $$\frac{\left[10^{-2k+2}a_0\right]}{10^{-2k+2}}+9\cdot 10^{2k-3}<a_1<\frac{\left[10^{-2k+2}a_0\right]}{10^{-2k+2}}+10^{2k-2}$$, where $$[\cdot]$$ denotes the integral part of the number in brakets.
• Demonstrate that if the procedure does not stop to a terminating decimal number, then, at each step $$n=2,3,\dots$$ you need to find the solution $$a_n$$ to the equation $g(x) = 10^{-k+n+1}\left(x – \frac{\left[10^{-2k+2n}a_{n-1}\right]}{10^{-2k+2n}}\right) + f\left(\frac{\left[10^{-2k+2n}a_{n-1}\right]}{10^{-2k+2n}}\right)$ or to the equation\begin{eqnarray}g(x) &=& 10^{-k+n+1}\left(x – \frac{\left[10^{-2k+2n}a_{n-1}\right]}{10^{-2k+2n}}+9\cdot 10^{2k-2n-1}\right) +\\ & &+ f\left(\frac{\left[10^{-2k+2n}a_{n-1}\right]}{10^{-2k+2n}}+9\cdot 10^{2k-2n-1}\right).\end{eqnarray}
• Show that $|f(a_n) – g(a_n)| \leq 10^{k-n},\tag{2}\label{eq3375:2}$ and that, for $$m,\ n \geq 0$$,$|a_{n+m}-a_n|\leq 10^{k-2n-1}\tag{3}\label{eq3375:3}.$
• Can you state, using \eqref{eq3375:3}, that $$(a_n)$$ is a Cauchy sequence? This implies that it converges to a non terminating decimal $$a$$.
• Use continuity of $$f$$ and $$g$$ in $$a$$ to conclude that the sequences $$\left(f(a_n)\right)$$ and $$\left(g(a_n)\right)$$ converge to $$f(a)$$ and $$g(a)$$ respectively.
• Finally, from \eqref{eq3375:2} deduce that $f(a) = g(a),$as desired.

If $$g$$ is continuous but not Lipschitz, no conclusion can be drawn on the existence of strictly positive intersections with $$f$$.

• Consider the function $g(x) = \sqrt x,$ for example. Observe that $g\left(25\cdot 10^{2k}\right) = 5 \cdot 10^k = f\left(25\cdot 10^{2k}\right), \forall \ k \in \Bbb Z.$
• Show that the smallest point $$x_k$$ for which $$f(x_k) = 10^k$$ is $x_k = \frac{10^{2k}}{11} = \sum_{h=1}^{+\infty} 9\cdot 10^{2k- 2h}.$Consider than the points $$A_k(x_k,f(x_k))$$ and conclude that the polygonal passing through the $$A_k$$’s, i.e. $g(x) = \begin{cases} 0 & (x=0)\\ 10^{-k}\left(x-\frac{10^{2k}}{11}\right) + 10^k & \frac{10^{2k}}{11}<x\leq \frac{10^{2k+2}}{11}, \ \forall k\in \Bbb Z\end{cases}$ is such that $$g(x)\geq f(x)$$, with equality only for $$x= 0$$ and for $$x=x_k$$, $$k\in\Bbb Z$$.
• Consider the concave function $g(x) = \sqrt{11 x}.$ Observe that $$g$$ passes through the points $$A_k$$. Conclude that any function $h(x) = K\sqrt x,$ with $$K>\sqrt{11}$$, has no intersections with $$f$$ other than $$x=0$$, despite being continuous in $$[0, +\infty)$$ and differentiable in $$(0, +\infty)$$.