Composition of a bounded operator and an unbounded closed operator

Question

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?

Answer 1

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.)

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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 $PT(2^{-k/2}e_k) = \sum_n 2^{-n/2}e_n \not\to 0$.