# Differential operator is not bounded.

$$E=\{ f\in L^{2}(\mathbb{R}) : f' \in L^{2}(\mathbb{R})\}$$ . Then differential operator is not bounded on $$E$$.

I can't find out sequence of function in whose derivative norm in $$L^{2}$$ tend to infinity. What's the way to construct such sequence?Help.

Consider the following sequence of functions: $$f_n(x)= \begin{cases} 0&\text{ if }|x|\ge 1+\frac{1}{n}\\ 1&\text{ if }|x|\le 1\\ 1+n\left(x+1\right) & \text{ if } x\in \left[-1-\frac{1}{n}, -1\right]\\ 1-n(x-1) & \text{ if } x\in \left[1,1+\frac{1}{n}\right]\\ \end {cases}$$ Each member of the sequence is compactly supported and piecewise (thus weakly) differentiable, and lies in $$L^2(\Bbb R)$$ jointly with its derivative. However, while $$\lim_{n\to\infty}f_n(x)=\chi_{[-1,1]}\in L^2(\Bbb R)$$ where $$\chi_{[-1,1]}$$ is the indicator function of the closed interval $$[-1,+1]$$, $$\lim_{n\to\infty}\frac{\mathrm{d}f_n(x)}{\mathrm{d} x}\notin L^2(\Bbb R)$$ since its squared absolute values grow unboundedly in a non integrable manner. The functions in this example are not smooth: however, you can slightly modify their definition near the points $$-1-1/n, -1, 1$$ and $$1+1/n$$ in order to get a sequence of smooth functions.
$$\sigma(A)=\sigma_c(A)=\{\lambda\in\mathbb{C}:\mathscr{R}\lambda=0\},$$ where $$\mathscr{R}\lambda$$ stands for the real part of $$\lambda$$.
And combining the lemma $$\sigma(A)\subset\overline{B\left(\theta,\Vert A\Vert_{\mathscr{L}(L^2(\mathbb{R}))}\right)},$$ where $$\theta$$ is the zero in $$\mathbb{C}.$$