# Composition of a bounded operator and an unbounded closed operator

Let $$T: \mathcal{D}(T) \to H$$ be a densely defined and closed unbounded linear operator on a Hilbert space $$H$$ and let $$P: H \to H$$ be a non-trivial projection operator, particularly bounded and self-adjoint and closed on $$H$$.

How can I show that the operator $$PT: \mathcal{D}(PT) \to H$$ with $$\mathcal{D}(PT) = \mathcal{D}(T)$$ is a closed operator?

My idea: To show that the operator $$PT: \mathcal{D}(PT) \to H$$ is closed, we need to show that if a sequence $$\{x_n\} \subset \mathcal{D}(T)$$ converges to some $$x \in H$$ and $$PTx_n$$ converges to some $$y \in H$$, then $$x \in \mathcal{D}(T)$$ and $$PTx = y$$.

Suppose $$\{x_n\} \subset \mathcal{D}(T)$$ is a sequence such that $$x_n \to x \quad \text{in } H \quad \text{and} \quad PTx_n \to y \quad \text{in } H.$$

Since $$T$$ is a closed operator and $$\{x_n\}$$ converges to $$x$$, we need to show that $$T x_n$$ converges to some $$z \in H$$. By the closedness of $$T$$, $$x \in \mathcal{D}(T) \quad \text{and} \quad T x_n \to T x \quad \text{in } H.$$

We have $$PT x_n \to y$$. Since $$P$$ is a bounded operator, we can interchange the limit and the bounded operator: $$P T x_n \to P T x \quad \text{in } H.$$ Therefore, we have: $$P T x = y.$$ Since $$x \in \mathcal{D}(T)$$ and $$PT x = y$$, we conclude that $$x \in \mathcal{D}(PT)$$ and $$PT x = y$$.

Is that correct?

No, this is not correct. The error is in the following step:

Since $$T$$ is a closed operator and $$\{x_n\}$$ converges to $$x$$, we need to show that $$Tx_n$$ converges to some $$z \in H$$. By the closedness of $$T$$, $$x \in \mathcal{D}(T) \quad \text{and} \quad Tx_n \to Tx \quad \text{in } H.$$

For this to work, you’re implicitly assuming $$Tx_n$$ converges in the first place, which may not be the case.

Indeed, the result is simply false. Say, $$H = l^2$$, $$T$$ be the diagonal operator $$Te_n = ne_n$$, and $$P$$ be the projection onto $$\text{span}\{e_1\}$$. Then $$PT$$ is simply $$P$$ restricted to $$\mathcal{D}(T) \neq H$$, so it is not closed. (Though, of course, it is closable.)

One can make $$PT$$ not even closable. Say, $$H = l^2$$, $$T$$ be the diagonal operator $$Te_n = 2^ne_n$$, and $$P$$ be the projection onto $$\text{span}\{\sum_n 2^{-n/2}e_n\}$$. Then $$2^{-k/2}e_k \to 0$$, but $$PT(2^{-k/2}e_k) = \sum_n 2^{-n/2}e_n \not\to 0$$.

• I see, thanks! So the statement does not hold even with "$PT$ closable....." Are there any further assumptions I could use to get $PT$ closed? Commented Aug 8 at 2:56
• @kumquat Generally speaking, composing an unbounded operator with a bounded operator to the left is not a good idea. But you can get closability at least in some very specific case, say $T$ is diagonal and $P$ is diagonal (w.r.t. the same ONB) (or, more generally, $P$ can be block diagonal with each block of finite size, w.r.t. that ONB). This won’t necessarily get you a closed operator (with the first example I provided in the answer a counterexample to this), but you do get closability. Another case would be if $T$ is self-adjoint and $P$ is a spectral projection of $T$. Commented Aug 8 at 3:03
• @kumquat If you know about von Neumann algebras, then there is a sort of generalization of both of the above two cases. Namely, if you have a finite vNa s.t. $T$ is affiliated to it and $P$ is in it, then $PT$ is closable (and in fact affiliated to the same finite vNa). Commented Aug 8 at 3:05
• Unfortunately not, but thanks for the cue! On the other hand the construction $TP$ should give me a closed operator at least in the case that $T$ is self-adjoint or skew-adjoint, since the adjoint of $PT$ is well-defined? But than I would stick with the problem that $TP$ is not necessarily densely defined.... Commented Aug 8 at 3:13
• @kumquat $TP$ is always closed, as long as $T$ is closed and $P$ is bounded, no additional assumptions needed. (The proof is just the proof in your question, only this time because $P$ is bounded, you do get $Px_n \to Px$.) But you’re right in saying it may not be densely defined. Adding additional assumptions on $T$ wouldn’t help, since as long as $\mathcal{D}(T) \neq H$, you can choose $z \in H \setminus \mathcal{D}(T)$ and let $P$ be the projection onto the span of $z$. Then $\mathcal{D}(TP) = \{0\}$. Commented Aug 8 at 4:38