Let the operator $L: dom(L) \subset L^2(\mathbb{R} ) \to L^2 (\mathbb{R})$ defined by $Lf(x)=x^2f(x)$ where $dom(L)=\{ f \in L^2(\mathbb{R} ) : x^2 f(x) \in L^2 (\mathbb{R}) \}$. Prove that $L$ is unbounded.
My approach is to find a sequence $f_n \in dom(L)$ such that $\| f_n \| =1$ and $\| Lf_n \| $ is unbounded.
I was thinking of two possible candidates for $f_n$.
The first is the constant sequence $f_n(x)=\begin{cases} \frac{1}{x^2} & x \geq 1 \\ 0 & x < 1 \end{cases}$, so we have $$\| Lf_n(x) \|_2^2 =\| x^2 f_n(x) \|_2^2 =\int_1^{\infty } \left| \frac{x^2}{x^2} \right|^2 dx =\int_1^{\infty } dx =x|_1^{\infty }$$ so $\| Lf_n (x) \|$ cannot be bounded.
Another potential sequence $f_n$ that I thought is the sequence $f_n(x)=\begin{cases} 1 & x \in [n,n+1] \\ 0 & \text{otherwise} \end{cases}$, then $$\| f_n(x) \|_2 =1$$ and $$\| Lf_n(x) \|_2 =\int_{n}^{n+1} |x^2|^2 dx =\left. \frac{x^5}{5} \right|_n^{n+1} = \frac{(n+1)^5}{5} -\frac{n^5}{5}$$ then again $\|Lf_n \| $ is unbounded.
Are my two choices for $f_n$ correct? I think there must be a mistake somewhere.