Do we have that $f \in W^{1, 1}(0, 1)$?

Question

This is a follow up to my previous question here.

How do I see that the function $$f(x) = \begin{cases} x \sin(1/x) & 0 < x \le 1,\\ 0 & x = 0 \end{cases} $$ is continuous on $[0, 1]$?

Do we have that $f \in W^{1, 1}(0, 1)$?

Answer 1

As $f$ is continuous on $[0,1]$, $f$ is integrable on $[0,1]$, therefore $f \in L^1([0,1])$. As $f$ is differentiable everywhere but in $0$, its weak derivative exists and equals its derivative, we have $$ f'(x) = -\frac 1x\cos x^{-1} + \sin x^{-1} $$ which is not integrable, as $$ |f'(x)| \ge \frac 1{2x} $$ for $x \le \frac 14$. Therefore $f' \not\in L^1([0,1])$ and $f \not\in W^{1,1}([0,1])$.