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)?

  • $\begingroup$ 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. $\endgroup$
    – Berci
    Commented Nov 6, 2019 at 12:20
  • $\begingroup$ @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. $\endgroup$
    – 0xbadf00d
    Commented Nov 6, 2019 at 16:01
  • $\begingroup$ @Berci Please take note of my answer. $\endgroup$
    – 0xbadf00d
    Commented Nov 6, 2019 at 18:46
  • $\begingroup$ Instead of "We easily see that T is unique" I guess you mean "We easily see that T*(y) is unique". $\endgroup$ Commented Apr 3, 2021 at 21:32

1 Answer 1


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 $$\langle z,x\rangle_{H_1}=\langle y,Tx\rangle_{H_2}\;\;\;\text{for all }x\in\mathcal D(T)\tag{10}.$$

  • $\begingroup$ In (7) and (8) it is not D(A) but D(T)? $\endgroup$ Commented Apr 3, 2021 at 22:50
  • $\begingroup$ 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? $\endgroup$ Commented Apr 3, 2021 at 23:02

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