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$E=\{ f\in L^{2}(\mathbb{R}) : f' \in L^{2}(\mathbb{R})\}$ . 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.

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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.

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It happened that I was studying the spectrum theory of operators recently.Hence this question is a good application of what I have learned.

From this question, Spectrum of the derivative operator, we can get

$$\sigma(A)=\sigma_c(A)=\{\lambda\in\mathbb{C}:\mathscr{R}\lambda=0\},$$ 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 $\mathbb{C}.$

Therefore, we have that A is a unbounded operator.

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