Showing that the derivative operator is not bounded on $L^2 (\mathbb{R})$

If we define $D$ as the set of functions $f \in L^2 (\mathbb{R})$ such that $f' \in L^2 (\mathbb{R})$ too, then it can be shown (I think) that the linear operator $\frac{d}{dt}: D \rightarrow L^2 (\mathbb{R})$ is not bounded with respect to the norm on $L^2 (\mathbb{R})$, which is $||f|| = \sqrt{\int_{-\infty}^{\infty}|f(t)|^2 dt}$.

But I'm having trouble proving this. What I'd like to do is find a sequence $(f_n)$ in $D$ with $||f_n||=1$ $\forall n \in \mathbb{N}$ but with $||f_n '||$ increasing and becoming arbitrarily large with $n$.

I understand that if we were working with the space of continuous functions on $\mathbb{R}$ instead of $L^2 (\mathbb{R})$, with the nice supremum norm, we could just use $f_n (t) = \sin(nt)$, but that won't work here. I tried the sequence $f_n(t) = \sqrt{\frac{\sin(nt)}{t}}$ because that's got a handy $L^2$-norm of $\sqrt{\pi}$ for all $n\in\mathbb{N}$, but its derivative isn't in $L^2 (\mathbb{R})$ so I feel stuck.

I think I'm approaching the problem correctly, but the norm in question here isn't easy to work with -- is there a simple sequence I could use?

Solution 1:

You may also consider $f_n(x) = \sqrt{n}\, e^{-n^2 x^2}$. In explicit terms, $\int_{-\infty}^{+\infty} f_n(x)^2\,dx = \sqrt{\frac{\pi}{2}}$ while $$ \int_{-\infty}^{+\infty}f_n'(x)^2\,dx = \color{red}{n^2} \sqrt{\frac{\pi}{2}}.$$

Solution 2:

You're right that $f_n(t)=\sin(nt)$ isn't in $L^2(\mathbb R)$, but multiplying this by a smooth bump function $g$ with $g(t)=1$ on $[0,2\pi]$ and $\text{supp}(g)=[-\varepsilon,2\pi+\varepsilon]$ for some $\varepsilon>0$ does the trick.

Solution 3:

Choose any smooth $f$ on $\mathbb R$ with both $\int_{\mathbb R}f^2, \int_{\mathbb R}(f')^2$ finite and nonzero. Set $f_n(x)=f(nx).$ Then

$$\int_{\mathbb R}f_n^{\,2} = \frac{1}{n}\int_{\mathbb R}f^2 \to 0,\,\,\text{while }\int_{\mathbb R}(f_n\,')^2 = n\int_{\mathbb R}(f\,')^2 \to \infty.$$