Answer to Question #306271 in Real Analysis for Fresh

Question

Consider the function

F(x)= xsin(π/x) 0<x≤1

0 x=0

Show that f(x) is continuous but not of bounded variation

Answer 1

First, let us prove that $f$ is continuous. It is continuous at any $0<x\leq 1$, as for any such point we can find a neighbourhood not containing zero, and thus $x\mapsto \sin(\pi/x)$ is well-defined and continuous. For $x=0$ let us consider the limit $\lim_{x\to 0} x\sin(\pi/x) = 0$, as $|\sin(\pi/x)|\leq 1$ and $x\to 0$. Therefore, $f$ is continuous.

Now, let us prove that it is not of bounded variation. For this we will consider the following subdivisions:

$x_0 = 0, x_1= \frac{1}{n+1/2}, x_2 = \frac{1}{n-1/2}, ... , x_{n+1}=1$ (for example, for $n=3$ the subdivision is $\{0, \frac{2}{7}, \frac{2}{5}, \frac{2}{3}, 1\}$ )

For this subdivision the variation is given by :

$V_n(f) = \sum_{0\leq i \leq n} |f(x_{i+1})-f(x_i)|$

For $i=0$ we have $|f(x_1)-f(x_0)|=\frac{1}{n+1/2}$ and for $0<i<n$ we have

$|\frac{1}{n-(i-1)+1/2}\cdot (-1)^{n-i+1}-\frac{1}{n-i+1/2} \cdot (-1)^{n-i}|$

We remark that these expressions have opposite signs so we have

$|\frac{1}{n-(i-1)+1/2}\cdot (-1)^{n-i+1}-\frac{1}{n-i+1/2} \cdot (-1)^{n-i}| = \frac{1}{n-(i-1)+1/2}+\frac{1}{n-i+1/2}$

Which we can rewrite as $\frac{1}{n-i+1/2}=\frac{2}{2n-2i+1}$

And for $i=n$ we have $|f(x_{n+1})-f(x_n)|=\frac{2}{3}$.

From all this we deduce that the total variation is of a form (as we counted each term in the sum two times : once for $i$ and once for $i+1$) :

$V_n(f)=2\cdot \sum_{1\leq i \leq n}\frac{2}{2n+3-2i}=4\cdot \sum_{0\leq i\leq n}\frac{1}{2i+1}$

We see that for $n\to \infty$ the variation behaves like the harmonic series, so it diverges, $V_n(f)\to \infty$ when $n\to \infty$. Therefore, it is not of bounded variation.