Does $U^* U=UU^*=I$ imply that $U$ is bounded (and thus unitary)?

Question

Wikipedia defines a unitary operator as "a bounded linear operator $U:H\to H$ on a Hilbert space $H$ such that $U^* U =U U^* =I$, where $U^*$ is the adjoint of $U$, and $I:H\to H$ is the identity operator".

I would naively think that the property $UU^*=U^*U=I$ automatically implies that $U$ is bounded, but unitary operators seem to be always defined as a subclass of bounded operators, which makes me doubt such assumption.

Why are unitary operators defined only as a subclass of bounded operators? Does this defining property imply boundedness, or can unbounded operators also satisfy $U^* U =U U^* =I$ (but are not as interesting to study for some reason)?

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Some related posts I found on the site are:

• Definition of Unitary Operators: Why do we need surjectivity or boundedness?
• definition of unitary operator
• Unitary Operator bounded?

Answer 1

Here is a summary of my comments:

This depends on what exactly “$U^\ast U = UU^\ast = I$” means. The first possible interpretation is that $I$ must be $I_H$, i.e., the identity operator defined on the entire space. This implicitly implies that both $U$ and $U^\ast$ are everywhere defined, in which case $U$ must be bounded (see, for example, this or this answer of mine) and thus a unitary.

The second possible interpretation is that $U^\ast U = I_{D(U)}$ and $UU^\ast = I_{D(U^\ast)}$. This is also enough to imply $U$ is bounded - in fact, the first equation alone is enough to imply $U$ is an isometry. Indeed, for any $h \in D(U)$, one then have $\|Uh\|^2 = \langle Uh, Uh \rangle = \langle h, U^\ast Uh \rangle = \langle h, h \rangle = \|h\|^2$, where in the second equality we have used the fact that $h \in D(U) = D(I_{D(U)}) = D(U^\ast U)$. Thus, $U$ extends to an isometry defined on the entirety of $H$, so the closure of $U$, $\overline{U}$, is an isometry. $U^\ast = \overline{U}^\ast$ is everywhere defined, so $UU^\ast = I_{D(U^\ast)} = I_H$ implies $U^\ast$ is both an isometry and a co-isomery, i.e., a unitary. In fact, this also implies $U$ is everywhere defined, as $U^\ast$ is a unitary, so if $U$ were not everywhere defined we could not have had $UU^\ast = I_H$. Thus, $U$ is a unitary in the first place, without even requiring extensions.

The third possible interpretation is that $U^\ast U = I_{D(U^\ast U)}$ and $UU^\ast = I_{D(UU^\ast)}$. This is somewhat more complicated. In case $U$ is closed, the first equation is still enough to imply $U$ is bounded. This is because, when $U$ is closed and densely defined, $U^\ast U$ is densely defined as well (see, for example, Proposition X.4.2 in Conway's "A Course in Functional Analysis"). Then for $h \in D(U^\ast U)$, using the same argument as in the second case, we can still get $\|Uh\| = \|h\|$. So $U$ is isometric on a dense subspace of $H$, whence by its closedness it must be an everywhere defined isometry. Then $UU^\ast = I_{D(UU^\ast)}$ implies $U$ is a unitary.

I don't know what happened when $U$ is just assumed closable - my assumption is that $D(U^\ast U)$ may not be dense and $U$ may not be bounded, but I can't construct an example at the moment. When $U$ is not necessarily closable, however, there are counterexamples. In an extreme case, we can construct an injective $U$ that is everywhere defined but $D(U^\ast) = \{0\}$, in which case we trivially have $D(U^\ast U) = D(UU^\ast) = \{0\}$ and $U^\ast U = UU^\ast = I_{\{0\}}$. To construct such an $U$, we first prove that,

Lemma: Let $H = \ell^2$. Then there is a linearly independent subset $\{f_\kappa\}_{\kappa < \mathfrak{c}} \subset H$ which is dense in $H$. Here, $\mathfrak{c}$ is the cardinality of continuum.

Proof: Fix a bijection $\pi: \mathfrak{c} \to H \times \mathbb{N}$ and write $\pi(\kappa) = (h_\kappa, n_\kappa)$ for each $\kappa < \mathfrak{c}$. Proceed by induction. For each $\kappa < \mathfrak{c}$, suppose $f_\lambda$ for all $\lambda < \kappa$ has been chosen. Since $\ell^2$ has Hamel dimension $\mathfrak{c}$ (see, for example, this paper), $\text{span}\{f_\lambda\}_{\lambda < \kappa}$ is a proper subspace of $H$. Hence, it has no interior. Thus, the open ball $O_\kappa$ of radius $\frac{1}{n_\kappa}$ around $h_\kappa$ contains a point outside $\text{span}\{f_\lambda\}_{\lambda < \kappa}$. Let $f_\kappa$ be this point. It is then easy to see that $\{f_\kappa\}_{\kappa < \mathfrak{c}}$ is linearly independent and its closure is the entirety of $H$. $\square$

Now, let $H = \ell^2$, $\{e_\kappa\}_{\kappa < \mathfrak{c}}$ be a normalized Hamel basis of $H$, and $\{f_\kappa\}_{\kappa < \mathfrak{c}}$ be a linearly independent dense subset of $H$ given by the lemma above. Define $U(e_\kappa) = f_\kappa$ and extend by linearity. Since $\{f_\kappa\}_{\kappa < \mathfrak{c}}$ is linearly independent, $U$ is injective. To show that $D(U^\ast) = \{0\}$, recall that $v \in D(U^\ast)^\perp$ iff $(0, v)$ is in the closure of the graph of $U$ (a proof can be found on Wikipedia). For any $v \in H$ and $n \geq 1$, by density of $\{f_\kappa\}_{\kappa < \mathfrak{c}}$ there exists a $\kappa_{v, n} < \mathfrak{c}$ s.t. $\|f_{\kappa_{v, n}} - nv\| < 1$. Then,

$$\|U(\frac{1}{n}e_{\kappa_{v, n}}) - v\| = \frac{1}{n}\|f_{\kappa_{v, n}} - nv\| < \frac{1}{n}$$

So, $(\frac{1}{n}e_{\kappa_{v, n}}, U(\frac{1}{n}e_{\kappa_{v, n}})) \to (0, v)$ as $n \to \infty$. This shows $(0, v)$ is in the closure of the graph of $U$ for all $v \in H$, whence $D(U^\ast)^\perp = H$ and so $D(U^\ast) = \{0\}$.

Answer 2

The only problem here is merely technical. Modulo these technical issues, it is true that $U^\star U=I$ implies that $U$ is bounded and $UU^\star=I$ implies that $U^\star$ is bounded.

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Warning there is a flaw here.

Let me explain. Suppose $U\colon H\to H$ is a linear map. (Let's avoid the word "operator" as much as we can, since that word has a precise meaning in unbounded functional analysis, which we don't want to wander into). The adjoint map $U^\star$ is thus defined by means of the following observation; given $f\in H$, $U^\star f$ is the unique vector that satisfies $$\tag{1} \langle U^\star f | g \rangle = \langle f|Ug\rangle, \qquad \forall g\in H.$$

(The flaw is in (1). This only defines $U^\star f$ whenever $\langle f|Ug\rangle$ is uniformly bounded. Which can fail to be the case, unless $U$ is bounded, of course. But that is exactly what we want to prove).

Ok, so $U^\star\colon H\to H$ is a linear map as well. Now if $U^\star U=I$ we have, for all $f\in H$, $$ \lVert f\rVert^2= \langle U^\star U f | f\rangle= \lVert Uf\rVert^2, $$ so, in particular, $\lVert U\rVert=1$ and $U$ is bounded. Similarly, $UU^\star=I$ implies that $$ \lVert f\rVert^2=\lVert U^\star f\rVert, $$ and so $U^\star$ is bounded.

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IMPORTANT REMARK. The key property of $U$ here is that it is everywhere defined. Indeed, we started by assuming $U\colon H\to H$. This is much less innocent that it looks, since in many situations of practical importance this can only happen when $U$ is bounded. There are many ways in which this remark can be made precise; one is outlined in this comment. Another is the following: in functional analysis, an "operator" is not just a linear map, but a pair $(T, D(T))$ where $D(T)$ is a linear subspace of $H$ and $T\colon D(T)\to H$ is a linear map. In the case $D(T)=H$, if additionally $T$ is closed, whatever that means, $T$ is bounded by the closed graph theorem.