# Existence and uniqueness of the adjoint of a linear operator between Hilbert spaces

Let $$H_i$$ be a $$\mathbb C$$-Hilbert space, $$T$$ be a linear operator from $$H_1\to H_2$$ and $$T^\ast$$ be a linear operator from $$H_2$$ to $$H_1$$ with $$\langle T^\ast y,x\rangle_{H_1}=\langle y,Tx\rangle_{H_2}\;\;\;\text{for all }x\in\mathcal D(T)\text{ and }y\in\mathcal D(T^\ast)\tag1.$$

We easily see that $$T$$ is unique if and only $$\mathcal D(T)^\perp=\{0\}$$:

• Let $$y\in H_2$$ and $$z,\tilde z\in H_1$$ with $$\langle\tilde z,x\rangle_{H_1}=\langle z,x\rangle_{H_1}=\langle y,Tx\rangle_{H_2}\;\;\;\text{for all }x\in\mathcal D(T)\tag2.$$
• Then, $$\langle\tilde z-z,x\rangle_{H_1}\;\;\;\text{for all }x\in\mathcal D(T)\tag3$$ and hence $$\tilde z-z\in\mathcal D(T)^\perp\tag4.$$

Now I've read that $$T^\ast$$ exists if and only if $$T$$ is continuous, but is the necessity really true?

Clearly, if $$T$$ is continuous, then $$T^\ast$$ exists:

• Let $$y\in H_2$$.
• If $$T$$ is continuous, then $$\langle y,\;\cdot\;\rangle_{H_2}\circ T\in H_1'\tag5$$ and ence there is a unique $$z\in H_1$$ with $$\langle z,\;\cdot\;\rangle_{H_1}=\langle y,\;\cdot\;\rangle_{H_2}\circ T\tag6$$ by Riesz' representation theorem.
• It's easy to see that the dependence of $$z$$ on $$y$$ is linear.

Question 1: Does the existence of $$T^\ast$$ really imply the continuity of $$T$$? If not, is there an other equivalent criterion to the existence of $$T^\ast$$?

Question 2: Am I missing something or does the continuity of $$T$$ even imply the uniqueness of $$T^\ast$$ (since the $$z$$ in $$(6)$$ is uniquely determined by Riesz' representation theorem)?

• Q1: No, and yes, there is. Q2: if $T$ is continuous, then it is defined everywhere, so your condition applies. See also en.m.wikipedia.org/wiki/Unbounded_operator. Commented Nov 6, 2019 at 12:20
• @Berci Regarding Q1: And what is this equivalent criterion? Q2: I know that a continuous operator can be extended to the whole space, but that wasn't the question. Commented Nov 6, 2019 at 16:01
• @Berci Please take note of my answer. Commented Nov 6, 2019 at 18:46
• Instead of "We easily see that T is unique" I guess you mean "We easily see that T*(y) is unique". Commented Apr 3, 2021 at 21:32

We need to state this differently: If $$T^\ast y$$ exists for some $$y\in H_2$$, then $$\left|\langle y,Tx\rangle_{H_2}\right|=\left|\langle T^\ast y,x\rangle_{H_1}\right|\le\left\|T^\ast y\right\|_{H_1}\left\|x\right\|_{H_1}\;\;\;\text{for all }x\in\mathcal D(A).\tag7$$ Thus, $$\mathcal D(A)\ni x\mapsto\langle y,Tx\rangle_{H_2}\tag8$$ is continuous.
On the other hand, if $$y\in H_2$$ and $$(8)$$ is continuous, then (by the Hahn-Banach theorem) there is a (non-unique) extension $$\varphi\in H_1'$$ of $$(8)$$. By Riesz' representation theorem, there is a $$z\in H_1$$ with $$\langle z,\;\cdot\;\rangle_{H_1}=\varphi\tag9$$ and hence $$\langle z,x\rangle_{H_1}=\langle y,Tx\rangle_{H_2}\;\;\;\text{for all }x\in\mathcal D(T)\tag{10}.$$
• So you get a local (or, by linearity, directional) criterion. If $x\mapsto\langle y,Tx\rangle$ is continuous for all y, can we deduce that T is continuous? Commented Apr 3, 2021 at 23:02