The composition of two unbounded linear operator is also unbounded or not?

Question

Let $H$ be a Hilbert space, and consider some linear operators $ A, B: H\rightarrow H $. In functional analysis, I knew that if $A$ and $B$ are both bounded, then the composition $AB$ is bounded.

I'm curious about the situation when $A$ and $B$ are unbounded and linear. Now, is $AB$ also unbounded or not? I tried to prove it or find a counterexample to show it's wrong, but I failed. Could someone give me a hand?

I've found an example, but not in Hilbert space:

$H=l^\infty,\|x\|=\max_n{x_n}$

Let $a=(1,1,1/2,2,1/3,3,\cdots), Ax=(a_nx_n)_{n=1}^\infty$

and $b=(1,1,2,1/2,3,1/3,\cdots), Bx=(b_nx_n)_{n=1}^\infty$

It seems that $AB=I$; hence, $AB$ is bounded, but both $A, B$ are unbounded.

Am I right? Are there any examples in Hilbert space?

Answer 1

The composition $AB$ can be bounded even if $A$ and $B$ are unbounded. Let $H$ be an infinite-dimensional Hilbert space with normalized Hamel basis $(e_i)_{i\in I}$ and let $\phi\colon \mathbb N\to I$ be an injection. Define \begin{align*} A e_i&=\begin{cases}ne_i&\text{if }i=\phi(n),\,n\text{ even},\\0&\text{otherwise},\end{cases}\\ B e_i&=\begin{cases}n e_i&\text{if }i=\phi(n),\,n\text{ odd},\\0&\text{otherwise}.\end{cases} \end{align*} As $\lim_{n\to \infty}\lVert A e_{\phi(2n)}\rVert=\lim_{n\to\infty} \lVert B e_{\phi(2n+1)}\rVert=\infty$, both $A$ and $B$ are unbounded. However, $AB=0$.

Answer 2

MaoWao's answer provides an example where you require the domain to actually be the entirety of $H$, which matches your question description. However, unbounded operators in practice usually mean something else, namely unbounded closed operators that are defined on a dense subspace (where closed means the graph of the operator is closed in $H \times H$). Such an operator can never be defined on the entirety of $H$ (or any Banach space, for that matter) due to closed graph theorem. If that is what you want instead, the following is an example: $A$ is multiplication by $x$ on $L^2(\mathbb{R})$ and $B$ is multiplication by $1/x$ on $L^2(\mathbb{R})$. Then $AB$ is the identity operator. (More precisely, the closure of $AB$ is the identity operator. $AB$ itself is only densely defined.)