Example of a function in $L^2(\mathbb{R})$ with derivative not in $L^2(\mathbb{R})$.

We know examples of a function which doesn't lie in $L^2(\mathbb{R})$ with derivatives in $L^2(\mathbb{R})$: $$f_1(x) = \mathrm{arctg}(x) \notin L^2(\mathbb{R}), \qquad f_1'(x) = \frac{1}{x^2+1}\in L^2(\mathbb{R}): \qquad \int\limits_{\mathbb{R}}\left(\frac{1}{x^2+1}\right)^2\,dx = \frac{\pi}{2};$$ $$f_2(x) = \mathrm{Si}(x)\notin L^2(\mathbb{R}), \qquad f_2'(x) = \frac{\sin(x)}{x}\in L^2(\mathbb{R}): \qquad \int\limits_{\mathbb{R}}\left(\frac{\sin(x)}{x}\right)^2\,dx = \pi;$$ $$f_3(x) = \mathrm{erf}(x)\notin L^2(\mathbb{R}), \qquad f_3'(x) = e^{-x^2}\in L^2(\mathbb{R}): \qquad \int\limits_{\mathbb{R}}e^{-2x^2}\,dx = \sqrt{\frac{\pi}{2}}.$$ Are there any classical examples of such function $f\in L^2(\mathbb{R})$ that its strong derivative $f'$ doesn't lie in $L^2(\mathbb{R})$?

Solution 1:

First, note that some of $\,L^2\,$ functions are not differentiable everywhere in $\,\mathbb R$.

Second, for the more general case of weak derivatives one can easily come up with such a function.

Now, as an example let us take $\ f(x) = \sin\left(1/x^2\right) \in L^2\left(\mathbb R\right)\,$. The derivative $\,f'$ looks like $\ f'(x) = -2\cos\left(1/x^2\right)\big/x^3$. Clearly $\ f'(x)\not\in L^2\left(\mathbb R\right)$.