Given the fact that the Laplacian operator \begin{align*} -\Delta\Psi = \mathcal{F}^{-1}(|k|^{2} \hat{\Psi}), \end{align*}
where $\mathcal{F}^{-1}$ denotes the Fourier Inverse Transform, is defined as self-adjoint in the next domain
\begin{align*} Dom(\Delta) = \{\Psi\in L^{2}(\mathbb{R},\mathbb{C}) \:|\: \: |k|^{2}\hat{\Psi}(k) \in L^{2}(\mathbb{R},\mathbb{C})\}, \end{align*} it is natural to ask about if there are functions in $L^{2}(\mathbb{R},\mathbb{C})$ whose second derivate those not belongs to this space. Currently, this post covers the case for the first derivate. Next, I would provide an answer to this question and generalize the case to $L^{m}(\mathbb{R},\mathbb{C})$, where $m\geq 2$.
By considering a not continuous function defined by intervals $[n, n+1]$ where $n\in\mathbb{N}\:\cup\:\{0\}$. We're seeking that the integral in every interval is greater than one. Thus, we propose
\begin{align*} \mathcal{M}(x) = \left\{ \begin{array}{ll} \frac{1}{n+2}(x-n)^{n+2},& \:\:\: x\in[n,n+1), \\ 0,& \:\:\: x < 0, \end{array} \right. \end{align*}
whose second derivate is \begin{align*} \mathcal{M}''(x)= (n+1)(x-n)^{n} \end{align*} for $x\in[n,n+1)$. Now, by evaluating the integral \begin{align*} \int_{n}^{n+1}(n+1)^2(x-n)^{2n}dx = \dfrac{(n+1)^2}{2n+1} \geq 1\:\: \text{for}\: n\geq 1 \end{align*} thus $\mathcal{M}''(x)$ does not belong to $L^2(\mathbb{R},\mathbb{C})$.
Finally, these results generalizes with the following function \begin{align*} \dfrac{1}{(n+m)}(x-n)^{n+m}, \end{align*} where $m\geq 2$ denotes the $L^m(\mathbb{R},\mathbb{C})$ space to whom it doesn't belong it's $m$-derivate.