Hilbert Spaces and Bounded Operators

Abstract

With this chapter we introduce the first mathematical notions relative to Hilbert spaces that we will use to build the mathematical foundations of Quantum Mechanics. A good part is devoted to Hilbert bases (complete orthonormal systems), which we treat in full generality without assuming the Hilbert space be separable. Before that, we discuss a paramount result in the theory of Hilbert spaces: Riesz’s representation theorem, according to which there is a natural anti-isomorphism between a Hilbert space and its dual.

There’s no such thing as a deep theorem, but only theorems we haven’t understood very well.

Nicholas P. Goodman

Exercises

3.1

Let ${\mathsf X}$ be a real vector space and $\langle \cdot , \cdot \rangle : {\mathsf X}\times {\mathsf X}\rightarrow {\mathbb R}$ a bilinear map. Prove that the polarisation identity

$$\langle x, y \rangle = \frac{1}{4}\left( \langle x+y, x+y\rangle - \langle x-y, x-y\rangle \right) $$

holds for all $x, y \in {\mathsf X}$ if and only if $\langle x, y\rangle = \langle y, x \rangle$ for all $x, y \in {\mathsf X}$.

3.2

Let ${\mathsf X}$ be a complex vector space and $\langle \cdot , \cdot \rangle : {\mathsf X}\times {\mathsf X}\rightarrow {\mathbb C}$ a map which is linear in the right entry and antilinear in the left entry. Prove the polarisation identity

$$\langle x, y \rangle = \frac{1}{4}\left( \langle x+y, x+y\rangle - \langle x-y, x-y\rangle - i \langle x+iy, x+iy\rangle + i \langle x-iy, x-iy\rangle \right) \,$$

for every $x, y \in {\mathsf X}$. Next show that $\langle y, x \rangle = \overline{\langle x, y \rangle }$ for all $x, y \in {\mathsf X}$ if and only if $\langle z, z\rangle \in {\mathbb R}$ for every $z\in {\mathsf X}$.

3.3

Definition 3.1 of a (semi-)inner product makes sense on real vector spaces as well, simply by replacing H2 with $S(u,v)=S(v, u)$, and using real linear combinations in H1.

Show that with this definition Proposition 3.3 still holds, provided the polarisation formula is written as in (3.7).

3.4

(Hard.) Consider a real vector space and prove that if a (semi)norm p satisfies the parallelogram rule (3.3):

$$\begin{aligned} p(x+y)^2 + p(x-y)^2 = 2(p(x)^2 +p(y)^2)\,, \end{aligned}$$

(3.94)

then there exists a unique (semi-)inner product S, defined in Exercise 3.3, inducing p via (3.2).

Solution. If S is a (semi-)inner product on the real vector space ${\mathsf X}$, we have the polarisation formula (3.7):

$$S(x, y) = \frac{1}{4}\left( S(x+y,x+y) -S(x-y, x-y)\right) \,.$$

This implies S is unique, for $S(z, z)= p(z)^2$. For the existence from a given norm p set:

$$S(x, y) := \frac{1}{4}\left( p(x+y)^2 - p(x-y)^2\right) \,.$$

We shall prove S is a semi-inner product or an inner product according to whether p is a norm or a seminorm. If this is true and p is a norm, then substituting S to p above, on the right, gives $S(x, x)=0$ and $x=\mathbf{0}$, making S an inner product.

To finish we need to prove, for any $x,y, z \in {\mathsf X}$:

(a) $S(\alpha x,y) = \alpha S(x, y)$ if $\alpha \in {\mathbb R}$,

(b) $S(x+y,z)= S(x,z)+S(y, z)$,

(c) $S(x,y)= S(y, x)$,

(d) $S(x, x) = p(x)^2$.

Properties (c) and (d) are straightforward from the definition of S. Let us prove (a) and (b). By (3.3) and the definition of S:

$$S(x,z) + S(y, z) = 4^{-1}\left( p(x+z)^2 -p(x-z)^2 + p(y+z)^2 -p(y-z)^2 \right) $$

$$= 2^{-1}\left( p\left( \frac{x+y}{2} +z\right) ^2 - p\left( \frac{x+y}{2} -z \right) ^2\right) = 2S\left( \frac{x+y}{2}, z \right) \,.$$

Hence

$$\begin{aligned} S(x,z) + S(y, z) = 2S\left( \frac{x+y}{2}, z \right) \,. \end{aligned}$$

(3.95)

Then (a) clearly implies (b), and we have to prove (a) only. Take $y=\mathbf{0}$ in (3.95) and recall $S(0,z)=0$ by definition of S. Then

$$S(x, z)= 2S(x/2, z)\,.$$

Iterating this formula gives (a) for $\alpha = m/2^n$, $m, n =0,1,2,\ldots$. These numbers are dense in $[0,+\infty )$. At the same time ${\mathbb R}\ni \alpha \mapsto p(\alpha x+ z)$ and ${\mathbb R}\ni \alpha \mapsto p(\alpha x- z)$ are both continuous (in the topology induced by p), so

$$S(x, y) := \frac{1}{4}\left( p(x+y,x+y) - p(x-y, x-y)\right) $$

allows to conclude ${\mathbb R}\ni \alpha \mapsto S(\alpha x, y)$ is continuous in $\alpha$. That is to say, (a) holds for any $\alpha \in [0,+\infty )$. Again by definition of S we have $S(-x,y)= -S(x, y)$, so the previous result is valid for any $\alpha \in {\mathbb R}$, ending the proof.

3.5

(Hard.) Suppose a (semi)norm p satisfies the parallelogram rule (3.3):

$$\begin{aligned} p(x+y)^2 + p(x-y)^2 = 2(p(x)^2 +p(y)^2)\, \end{aligned}$$

(3.96)

on a ${\mathbb C}$-vector space. Show that there is a unique (semi-)inner product S inducing p by means of (3.2).

Solution. If S is a (semi-)inner product on the complex vector space ${\mathsf X}$ we have the polarisation formula (3.4):

$$4S(x,y) = S(x+y,x+y) -S(x-y,x-y) -iS(x+iy,x+iy)+iS(x-iy, x-iy).$$

Since $S(z, z) = p(z)^2$, as in the real case, that implies uniqueness of S for a given norm p on ${\mathsf X}$. Existence: define, for a given (semi)norm p and $x, y\in {\mathsf X}$:

$$S_1(x, y) := 4^{-1}(p(x+y)^2 - p(x-y)^2)\,,\quad S(x, y) := S_1(x, y) - iS_1(x, iy)\,.$$

Notice $S(x, x)=p(x)^2$, and if p is a norm, by construction $S(x, x)=0$ implies $x=\mathbf{0}$. There remains to show that the above S is a Hermitian (semi-)inner product. By Definition 3.1 we have to check:

(a) $S(\alpha x,y) = \alpha S(x, y)$ if $\alpha \in {\mathbb C}$,

(b) $S(x+y,z)= S(x,z)+S(y, z)$,

(c) $S(x,y)= \overline{S(y, x)}$,

(d) $S(x, x) = p(x)^2$.

The last one is true by construction. Proceeding as in the previous exercise, using $S_1$ instead of S, we can prove (b) for $S_1$, (a) for $S_1$ with $\alpha \in {\mathbb R}$, and also $S_1(x, y)=S_1(y, x)$. These, using the definition of S in terms of $S_1$, imply (a), (b) and (c).

3.6

Let ${\mathsf X}$ be a real vector space equipped with a real inner product $\langle \cdot , \cdot \rangle$ and associated norm $||\cdot ||$. Prove that

$$||x_1+ \cdots + x_n|| \le ||x_1|| + \cdots + ||x_n||\,,$$

where

$$||x_1+ \cdots + x_n|| = ||x_1|| + \cdots + ||x_n||$$

if and only if $x_k = \lambda _k x$ for some $x \in {\mathsf X}$ and $\lambda _k \in [0,+\infty )$ for all $k=1,\ldots , n$.

Hint. First show that $||x_1+ \cdots + x_p|| \le ||x_1|| + \cdots + ||x_p||$. Then prove the second part of the thesis for $n=2$, using the fact that $|\langle x, y \rangle |= ||x||\, ||y||$ $\,\Leftrightarrow \;$ x and y are linearly independent, then extend the result using induction. Observe that $||x_1+ \cdots + x_p + x_{p+1}|| = ||x_1|| + \cdots + ||x_p||+ ||x_{p+1}||$ implies $||x_1+ \cdots + x_p|| = ||x_1|| + \cdots + ||x_p||$ since $||x_1+ \cdots + x_p + x_{p+1}|| \le ||x_1+ \cdots + x_p || + ||x_{p+1}||$.

3.7

Prove the claim in Remark 3.4(1) on a (semi-)inner product space $({\mathsf X}, S)$: the (semi-)inner product $S:{\mathsf X}\times {\mathsf X}\rightarrow {\mathbb C}$ is continuous in the product topology of ${\mathsf X}\times {\mathsf X}$, having on ${\mathsf X}$ the topology induced by the (semi-)inner product itself. Consequently S is continuous in both arguments separately.

Hint. Suppose ${\mathsf X}\times {\mathsf X}\ni (x_n,y_n) \rightarrow (x, y) \in {\mathsf X}\times {\mathsf X}$ as $n\rightarrow +\infty$. Use the Cauchy–Schwarz inequality to show that if S is the (semi-)inner product associated to p, then:

$$|S(x,y) - S(x_n, y_n)| \le p(x_n) p(y_n-y) + p(x_n-x) p(y)\,.$$

Recall that $p(x_n) \rightarrow p(x)$ and that the canonical projections are continuous in the product topology.

3.8

Prove Proposition 3.8: a linear operator $L: {\mathsf X}\rightarrow {\mathsf Y}$ between inner product spaces is an isometry, in the sense of Definition 3.6, if and only if

$$||Lx||_{\mathsf Y}=||x||_{\mathsf X}\quad \text{ for } \text{ any } x\in {\mathsf X},$$

where norms are associated to the respective inner products.

Hint. Polarise.

3.9

Consider the Banach space $\ell ^p({\mathbb N})$, $p\ge 1$. Show that for $p\ne 2$ one cannot define any Hermitian inner product inducing the usual norm $||\,\,||_p$. Conclude that $\ell ^p({\mathbb N})$ cannot be rendered a Hilbert space for $p\ne 2$.

Hint. Show there are pairs of vectors f, g violating the parallelogram rule. E.g. $f=(1,1,0, 0,\ldots )$ and $g=(1,-1,0, 0,\ldots )$.

3.10

Prove that the Banach space $(C([0,\pi /2]), ||\,\,||_\infty )$ does not admit a Hermitian inner product inducing $||\,\,||_\infty$, i.e.: $(C([0,\pi /2]), ||\,\,||_\infty )$ cannot be made into a Hilbert space.

Hint. Show there are pairs of vectors f, g violating the parallelogram rule. Consider for example $f(x)= \cos x$ and $g(x)=\sin x$.

3.11

In the Hilbert space ${\mathsf H}$ consider a sequence $\{x_n\}_{n\in {\mathbb N}}\subset {\mathsf H}$ converging to $x\in {\mathsf H}$ weakly: $f(x_n) \rightarrow f(x)$, $n\rightarrow +\infty$, for any $f\in {\mathsf H}'$. Show that, in general, $x_n\not \rightarrow x$ in the topology of ${\mathsf H}$. However, if we additionally assume $||x_n||\rightarrow ||x||$, $n\rightarrow +\infty$, then $x_n \rightarrow x$, $n\rightarrow +\infty$, also in the topology of ${\mathsf H}$.

Hint. Riesz’s theorem implies that $\{x_n\}_{n\in {\mathbb N}}\subset {\mathsf H}$ weakly converges to $x\in {\mathsf H}$ iff $(z|x_n) \rightarrow (z|x)$, $n\rightarrow +\infty$, for any $z\in {\mathsf H}$. Let $\{x_n\}_{n\in {\mathbb N}}$ be a basis of ${\mathsf H}$, thought of as separable. Then $x_n \rightarrow \mathbf{0}$ weakly but not in the topology of ${\mathsf H}$. For the second claim, note $||x-x_n||^2 = ||x||^2 + ||x_n||^2 - 2 Re(x|x_n)$.

3.12

Consider the basis of $L^2([-L/2,L/2], dx)$ formed by the functions (up to zero-measure sets):

$$e_n(x):= \frac{e^{i2\pi n x/L}}{\sqrt{L}} \quad n\in {\mathbb Z}\,.$$

Suppose, for $f\in L^2([-L/2,L/2], dx)$, that the series

$$\sum _{n\in {\mathbb Z}} (e_n|f) e_n(x)$$

converges to some g in norm $||\,\,||_\infty$. Prove $f(x)=g(x)$ a.e.

Hint. Compute the components $(e_n|g)$ using the fact that the integral of an absolutely convergent series on [a, b] is the series of the integrated summands. Check that $(e_n|g)= (e_n|f)$ for any $n\in {\mathbb Z}$.

3.13

Consider the basis of $L^2([-L/2,L/2], dx)$ made by the functions $e_n$ of Exercise 3.12. Suppose $f: [-L/2,L/2] \rightarrow {\mathbb C}$ is continuous, $f(-L/2)= f(L/2)$, and f is piecewise $C^1$ on $[-L/2,L/2]$ (i.e. $[-L/2,L/2]= [a_1, a_2] \cup [a_2, a_3] \cup \cdots \cup [a_{n-2}, a_{n-1}] \cup [a_{n-1}, a_n]$ and $f\upharpoonright _{[a_i, a_{i+1}]} \in C^1([a_i, a_{i+1}])$ for any i, understanding boundary derivatives as left and right derivatives). Show:

$$f(x) = \sum _{n\in {\mathbb Z}} (e_n|f) e_n(x) \quad \text{ for } \text{ any } x\in [-L/2,L/2]$$

where

$$e_n(x):= \frac{e^{i2\pi n x/L}}{\sqrt{L}} \quad n\in {\mathbb Z}\,.$$

Prove the series converges uniformly.

Hint. Compute the components $(e_n|df/dx)$ by integration by parts: $|n (e_n|f)| = 2c |(e_n| df/dx)|$, where $c= L/(4\pi )$. Then

$$|(e_n|f)| = c 2 |(e_n| df/dx)| |1/n| \le c (|(e_n| df/dx)|^2 + 1/n^2),\quad {n\ne 0}\,.$$

Now, df / dx gives an $L^2$ map, the series with generic term $1/n^2$ converges, and $|e_n(x)|=1$ for any x. Therefore the series

$$\sum _{n\in {\mathbb Z}} (e_n|f) e_n(x)$$

converges uniformly, i.e. in norm $||\,\,||_\infty$. Apply Exercise 3.12.

3.14

Rephrase and prove Exercise 3.13, replacing the requirement that f be continuous and piecewise $C^1$ with the demand that f be absolutely continuous on $[-L/2,L/2]$ and either with essentially bounded derivative, or with derivative in ${\mathscr {L}}^2([-L/2,L/2], dx)$.

Hint. Remember Theorem 1.76(a).

3.15

Consider the basis $\{e_n\}$ of $L^2([-L/2,L/2], dx)$ of Exercise 3.12. Let $f: [-L/2,L/2] \rightarrow {\mathbb C}$ be of class $C^{N}$, suppose $d^kf/dx^k|_{-L/2}= d^kf/dx^k|_{L/2}$, $k=0,1,\ldots , N$ and that f is piecewise $C^{N+1}$ on $[-L/2,L/2]$. Prove

$$ \frac{d^kf(x)}{dx^k} = \sum _{n\in {\mathbb Z}} (e_n|f) \frac{d^k}{dx^k}e_n(x) \quad \text{ for } \text{ any } x\in [-L/2,L/2] $$

where

$$e_n(x):= \frac{e^{i2\pi n x/L}}{\sqrt{L}} \quad n\in {\mathbb Z}\,,$$

and the series’ convergence is uniform, $k=0,1,2,\ldots , N$

Hint. Iterate the procedure of Exercise 3.13, bearing in mind that we can swap derivatives and sum in a convergent series of $C^1$ maps whose series of derivatives converges uniformly.

3.16

Prove that the functions $[0,L] \ni x \mapsto s_n(x) := \sqrt{\frac{2}{L}} \sin \left( \frac{\pi n x}{L}\right)$, $n=1,2,3,\ldots$, form an orthonormal system in $L^2([0,L], dx)$.

Sketch. A direct computation tells $||s_n||=1$. Then observe that $\varDelta s_n = \left( \frac{\pi n x}{L}\right) ^2 s_n$ where $\varDelta := -\frac{d^2}{dx^2}$. Therefore if $(\,|\,)$ is the inner product in $L^2([0,L], dx)$:

$$(s_n| s_m) = \frac{1}{n}(\varDelta s_n| s_m) = \frac{1}{n}(s_n| \varDelta s_m) = \frac{m}{n} (s_n| s_m)$$

where, in the middle, we integrated twice by parts in order to shift $\varDelta$ from left to right, and we used $s_k(0)=s_k(L)=0$ to annihilate the boundary terms.

Therefore

$$\left( 1- \frac{m}{n}\right) (s_n| s_m)=0,$$

implying $(s_n| s_m)=0$ if $n\ne m$.

3.17

Prove that the maps $[0,L] \ni x \mapsto c_n(x) := \sqrt{\frac{2}{L}} \cos \left( \frac{\pi n x}{L}\right)$, $n=0,1,2,\ldots$, form an orthonormal system in $L^2([0,L], dx)$.

Hint. Proceed exactly as in Exercise 3.16. If $\varDelta := -\frac{d^2}{dx^2}$ we still have $\varDelta c_n = \left( \frac{\pi n x}{L}\right) ^2 c_n$, but now it is the derivatives of $c_n$ that vanish on the boundary of [0, L].

3.18

Recall that the space ${\mathscr {D}}((0,L))$ of smooth maps with compact support in (0, L) is dense in $L^2([0,L], dx)$ in the latter’s topology. Using Exercise 3.13 prove that the functions $[0,L] \ni x \mapsto s_n(x) := \sqrt{\frac{2}{L}} \sin \left( \frac{\pi n x}{L}\right)$, $n=1,2,3,\ldots$, are a basis of $L^2([0,L], dx)$.

Outline. It suffices to prove that the space $<s_n>_{n=1,2,\ldots }$ of finite linear combinations of the $s_n$ is dense in ${\mathscr {D}}((0,L))$ for $||\,\,||_\infty$, because this would imply, by elementary integral properties, that they are dense in the topology of $L^2([0,L], dx)$. Since ${\mathscr {D}}((0,L))$ is dense in $L^2([0,L], dx)$, we would have $<s_n>_{n=1,2,\ldots }$ dense in $L^2([0,L], dx)$. Because $\{s_n\}_{n=1,2,\ldots }$ is an orthonormal system (Exercise 3.16), this would in turn imply the claim, by Theorem 3.26. To show that $<s_n>_{n=1,2,\ldots }$ is dense in ${\mathscr {D}}((0,L))$ with respect to $||\,\,||_\infty$, fix $f\in {\mathscr {D}}((0,L))$ and extend it to F on $[-L, L]$ by imposing F be an odd map. By construction F is in ${\mathscr {D}}((-L, L))$ and satisfies $F(-L)=F(L)$, because it and its derivatives vanish around $x=0$ and $x=\pm L$. A fortiori F is continuous and piecewise $C^1$ on $[-L, L]$. Applying Exercise 3.13 we conclude:

$$F(x) = \sum _{n\in {\mathbb N}} (F|e_n) \frac{e^{i\pi n x/L}}{\sqrt{2L}} $$

where now

$$e_n(x) := \frac{e^{i\pi n x/L}}{\sqrt{2L}}\,,$$

and the series’ convergence is in norm $||\,\,||_\infty$. Since F is odd:

$$F(x) = -F(-x) = -\sum _{n\in {\mathbb N}} (F|e_n) \frac{e^{-i\pi n x/L}}{\sqrt{2L}} $$

adding which to the previous expression of F(x) gives:

$$2F(x) = \sum _{n\in {\mathbb N}} \frac{2i (F|e_n)}{\sqrt{2L}} \sin \left( \frac{\pi n x}{L}\right) . $$

Restricting to $x\in [0,L]$:

$$f(x) = \sum _{n\in {\mathbb N}} \frac{i(F|e_n)}{\sqrt{2L}} \sin \left( \frac{\pi n x}{L}\right) \,.$$

Since the convergence is in norm $||\,\,||_\infty$, we have the claim.

3.19

Recall that the space ${\mathscr {D}}((0,L))$ of smooth maps with compact support in (0, L) is dense in $L^2([0,L], dx)$ in the latter’s topology. Using Exercise 3.13 prove that the functions $[0,L] \ni x \mapsto c_n(x) := \sqrt{\frac{2}{L}} \cos \left( \frac{\pi n x}{L}\right)$ are a basis of $L^2([0,L], dx)$.

Hint. Proceed as in Exercise 3.18, extending f to an even function on $[-L, L]$.

3.20

Let $C\subset {\mathsf H}$ be a closed subspace in the Hilbert space ${\mathsf H}$. Prove C is weakly closed. Put otherwise, show that if $\{x_n\}_{n \in {\mathbb N}} \subset C$ converges weakly (cf. Exercise 3.11) to $x\in {\mathsf H}$, then $x\in C$.

Hint. If $P_C: {\mathsf H}\rightarrow C$ is the orthogonal projector onto C, show $P_Cx_n \rightarrow P_Cx$ weakly.

3.21

Let ${\mathsf H}$ be a Hilbert space and $T : D(T) \rightarrow {\mathsf H}$ a linear operator, where $D(T)\subset {\mathsf H}$ is a dense subspace in ${\mathsf H}$ ($D(T)={\mathsf H}$, possibly). Prove that if $(u|Tu)=0$ for any $u\in D(T)$ then $T=0$, i.e. T is the null operator (sending everything to $\mathbf{0}$).

Solution. We have

$$0=(u+v|T(u+v)) = (u|Tu) + (v|Tv) + (u|Tv) +(v|Tu) = (u|Tv) + (v|Tu)\,$$

and similarly

$$0=i(u+iv|T(u+iv)) = i(u|Tu) + i(v|Tv) -(u|Tv) + (v|Tu) = - (u|Tv) + (v|Tu)\,. $$

Adding these two gives $(v|Tu)=0$ for any $u, v\in D(T)$. Choose $\{v_n\}_{n\in {\mathbb N}} \subset D(T)$ such that $v_n\rightarrow Tu$, $n\rightarrow +\infty$. Then $||Tu||^2= (Tu|Tu) = \lim _{n\rightarrow +\infty } (v_n|Tu) =0$ for any $u\in D(T)$, i.e. $Tu=\mathbf{0}$ for any $u\in D(T)$, hence $T=0$.

3.22

Consider $L^2([0,1], m)$ where m is the Lebesgue measure, and take $f\in {\mathscr {L}}^2([0,1], m)$. Let $T_f:L^2([0,1], m) \ni g \mapsto f\cdot g$, where $\cdot$ is the standard pointwise product of functions. Prove $T_f$ is well defined, bounded with norm $||T_f|| \le ||f||$ and normal. Moreover, show $T_f$ is self-adjoint iff f is real-valued up to a zero-measure set in [0, 1].

3.23

Let $T\in {\mathfrak B}({\mathsf H})$ be self-adjoint. Given $\lambda \in {\mathbb R}$, consider the series of operators

$$U(\lambda ) := \sum _{n=0}^\infty (i\lambda )^n\frac{T^n}{n!}\,,$$

where $T^0:=I$, $T^1:=T$, $T^2:=TT$ and so on, and the convergence is uniform. Prove the series converges to a unitary operator.

Hint. Proceed as when proving the properties of the exponential map using its definition as a series.

3.24

Referring to the previous exercise, show that $\lambda ,\mu \in {\mathbb R}$ imply $U(\lambda )U(\mu )= U(\lambda +\mu )$.

3.25

Show that the series of Exercise 3.23 converges for any $\lambda \in {\mathbb C}$ to a bounded operator, and that $U(\lambda )$ is always normal.

3.26

Show that the operator $U(\lambda )$ of Exercise 3.23 is positive if $\lambda \in i{\mathbb R}$. Are there values $\lambda \in {\mathbb C}$ for which $U(\lambda )$ is a projector (not necessarily orthogonal)?

3.27

Compute explicitly $U(\lambda )$ in Exercise 3.23 if T is the operator $T_f$ of Exercise 3.22 and $f=\overline{f}$.

3.28

In $\ell ^2({\mathbb N})$ consider the operator $T: \{x_n\} \mapsto \{x_{n+1}/n\}$. Prove T is bounded and compute $T^*$.

3.29

Consider the Volterra operator $T: L^2([0,1], dx) \rightarrow L^2([0,1], dx)$:

$$(Tf)(x) = \int _0^x f(t)dt\,.$$

Prove it is well defined, bounded and its adjoint satisfies:

$$(T^*f)(x) = \int _x^1 f(t) dt \quad \text{ for } \text{ any } f \in L^2([0,1], dx). $$

Hint. Since [0, 1] has finite Lebesgue measure, $L^2([0,1], dx)\subset L^1([0,1], dx)$. Then use Theorem 1.76.

3.30

Let $U: {\mathsf H}\rightarrow {\mathsf H}$ be a bounded operator over a Hilbert space ${\mathsf H}$. Prove that if $(x|y)=0$ implies $(Ux|Uy) =0$ for $x, y \in {\mathsf H}$ and $U\ne 0$, then there is $a>0$ such that $V:= aU$ is isometric.

Solution. The first part of the hypothesis can be rephrased as $y \perp x$ implies $y \perp U^*Ux$. As a consequence $U^*Ux \in \{\{x\}^\perp \}^\perp$ which is the linear span of x. In other words, if $x \in {\mathsf H}$ then $U^*U x = \lambda _x x$ for some $\lambda _x\in {\mathbb C}$. Let us prove that $\lambda _x$ does not depend on x. To this end, consider a couple of vectors $x \perp y$ with $x, y \ne \mathbf{0}$. Using the argument above we have $U^*U x = \lambda _x x\,,\quad U^*U y = \lambda _y y\,, \quad U^*U (x+y) = \lambda _{x+y} (x+y)\,.$ Linearity of $U^*U$ applied to the last identity leads to $U^*Ux + U^*Uy = \lambda _{x+y}x + \lambda _{x+y}y\,,$ namely $U^*Ux- \lambda _{x+y}x = -(U^*Uy- \lambda _{x+y}y)$. Exploiting $U^*U x = \lambda _x x$ and $U^*U y = \lambda _y y$ we get $(\lambda _x- \lambda _{x+y})x = -(\lambda _y- \lambda _{x+y})y$. Since $x \perp y$ and $x, y \ne \{\mathbf{0}\}$, the only possibility is that $\lambda _x = \lambda _{x+y} = \lambda _y$. In summary, a couple of orthogonal non-vanishing vectors x, y satisfies $\lambda _x= \lambda _y$. Next consider a Hilbert basis $\{x_n\}_{n\in J}\subset {\mathsf H}$ so that, if $z\in {\mathsf H}$, $z = \sum _{n \in J} c_n x_n$. for complex numbers $c_n$. Since $U^*U$ is bounded,

$$U^*Uz = U^*U\sum _{n \in J} c_n x_n = \sum _{n \in J} c_n U^*Ux_n = \sum _n c_n \lambda _{x_n}x_n$$

From the previous argument that $\lambda _{x_n} = \lambda _{x_m}$ so that, indicating with c the common value of the $\lambda _{x_n}$, we have

$$U^*Uz = \sum _{n \in J} c_n cx_n = c\sum _{n \in J} c_n x_n = cz\,.$$

Since $z\in {\mathsf H}$ is arbitrary, we have found that

$$U^*U=c I\,.$$

Finally, if $x \in {\mathsf H}$, it must hold $0 \le (Ux | Ux) = (x| U^*U x) = c (x| x )$ so that $c\ge 0$ and $c \ne 0$ because $U\ne 0$. $V:= c^{-1/2}U$ satisfies $V^*V=I$ and therefore is isometric.

3.31

Let ${\mathfrak A}$ be a $C^*$-algebra without unit. Consider the direct sum ${\mathfrak A}\oplus {\mathbb C}$ and define the product:

$$(x, c)\cdot (y,c') := (x\circ y + cy + c'x, cc'),\quad {(x',c'), (x, c)\in {\mathfrak A}\oplus {\mathbb C}}\,,$$

where $\circ$ is the product on ${\mathfrak A}$. Define the norm:

$$||(x, c)|| := \sup \{||cy+ xy||\,|\, y \in {\mathfrak A}, ||y||=1\}$$

and the involution: $(x, c)^* = (x^*, \overline{c})$, where $\overline{c}$ is the complex conjugate of c and the involution on the right is the one of ${\mathfrak A}$. Prove that the vector space ${\mathfrak A}\oplus {\mathbb C}$ with the above structure is a $C^*$-algebra with unit (0, 1).

Hint. The triangle inequality is easy. The proof that $||(x, c)||=0$ implies $c=0$ and $x=0$ goes as follows. If $c=0$, $||(x, 0)||=0$ means $||x||=0$, so $x=0$. If $c\ne 0$, we can simply look at $c=1$. In that case $||y+xy|| \le ||y|| \, ||(x, 1)||$, so $||(x, 1)||=0$ implies $y=xy$ for any $y\in {\mathfrak A}$. Using the involution we have $y= y x^*$ for any $y \in {\mathfrak B}$. In particular $x^*=xx^* =x$, and then $y=xy=yx$ for any $y\in {\mathfrak A}$. Therefore x is the unit of ${\mathfrak A}$, a contradiction. Hence $c=0$ is the only possibility, and we fall back to the previous case. Let us see to the $C^*$ properties of the norm. By definition of norm: $||(c, x)||^2 = \sup \{||cy+ xy||^2\,|\, y \in {\mathfrak A}, ||y||=1\} = \sup \{||y^*(\overline{c}cy + \overline{c}xy + c x^*y + x^* x y)||\,|\, y \in {\mathfrak A}, ||y||=1\}$. Hence $||(c, x)||^2 \le ||(c,x)^*(c,x)|| \le ||(c,x)^*||\, ||(c, x)||$. In particular $||(c,x)|| \le ||(c, x)^*||$, and replacing (c, x) with $(c, x)^*$ gives $||(c,x)^*|| = ||(c, x)||$. The inequality $||(c, x)||^2 \le ||(c,x)^*(c,x)|| \le ||(c,x)^*||\, ||(c, x)||$ implies $||(c, x)||^2 \le ||(c,x)^*(c,x)|| \le ||(c, x)||^2$, and so $||(c, x)||^2 = ||(c,x)^*(c, x)||$.

3.32

Prove Proposition 3.55:

Proposition. Let ${\mathfrak A}$ be a $^*$ -algebra with unit, ${\mathsf H}$ a Hilbert space, and consider a linear map $\phi : {\mathfrak A}\rightarrow {\mathfrak B}({\mathsf H})$ which preserves the products and the involutions. The following facts hold.

(a) ${\mathsf H}_\phi := Ran(\phi ({\mathbb I}))$ and ${\mathsf H}_\phi ^\perp$ are closed subspaces of ${\mathsf H}$ satisfying ${\mathsf H}= {\mathsf H}_\phi \oplus {\mathsf H}_\phi ^\perp$, and each subspace is invariant under $\phi (a)$, for every $a\in {\mathfrak A}$.

(b) $\phi (a)\upharpoonright _{{\mathsf H}_\phi ^\perp } =0$ for every $a\in {\mathfrak A}$.

(c) The restriction to the complement

$$\pi _\phi : {\mathfrak A}\ni a \mapsto \phi (a)\upharpoonright _{{\mathsf H}_\phi } \in {\mathfrak B}({\mathsf H}_\phi )$$

is a representation of ${\mathfrak A}$ over ${\mathsf H}_\phi$ according to Definition 3.52. It also satisfies

(i) $\pi _\phi$ is faithful $\,\Leftrightarrow \;$ $\phi$ is injective;

(ii) $\pi _\phi (a)$ is the zero representation $\,\Leftrightarrow \;$ $\phi ({\mathbb I})=0$ (in this case ${\mathsf H}_\phi = \{\mathbf{0}\}$ );

(iii) $\pi _\phi =\phi$ if $\phi$ is surjective;

(iv) $\pi _\phi =\phi$ if $\phi$ is not the zero map and irreducible (i.e. there are no closed subspaces ${\mathsf M}$ of ${\mathsf H}$ with $\{\mathbf{0}\}\subsetneq {\mathsf M}\subsetneq {\mathsf H}$ and such that $\phi (a) ({\mathsf M}) \subset {\mathsf M}$ for every $a\in {\mathfrak A}$ ).

Solution. (a) First of all we notice that $P:= \phi ({\mathbb I})$ is an orthogonal projector because $PP= \phi ({\mathbb I})\phi ({\mathbb I}) = \phi ({\mathbb I}{\mathbb I}) = \phi ({\mathbb I}) =P$ and $P^* = \phi ({\mathbb I})^* = \phi ({\mathbb I}^*)= \phi ({\mathbb I})=\phi$, where we have used Proposition 3.44(c). Therefore ${\mathsf H}_\phi := P({\mathsf H})$ is a closed subspace by Proposition 2.101(b). Moreover $I-P$ projects orthogonally onto ${\mathsf H}_\phi ^\perp$, and ${\mathsf H}= {\mathsf H}_\phi \oplus {\mathsf H}_\phi ^\perp$ by Proposition 3.64(a, b). Now observe that $P\phi (a) = \phi ({\mathbb I})\phi (a) = \phi ({\mathbb I}a)= \phi (a) = \phi (a{\mathbb I}) = \phi (a) P$, which entails that ${\mathsf H}_\phi$ is invariant under every $\phi (a)$. The same fact holds for ${\mathsf H}^\perp _\phi$ when we replace P by $I-P$ in the previous argument. (b) If $x \in {\mathsf H}_\phi ^\perp$, $\phi (a) x = \phi (a) (I-P)x = \phi (a)x -\phi (a{\mathbb I})x = \phi (a)x -\phi (a)x=\mathbf{0}$. (c) Since P restricts to the identity map I on ${\mathsf H}_\phi$, we have $\phi (a)\upharpoonright _{{\mathsf H}_\phi }({\mathbb I}) = I$. Then the restriction ${\mathfrak A}\ni a \mapsto \pi (a) := \phi (a)\upharpoonright _{{\mathsf H}_\phi }$ is a representation on the Hilbert space ${\mathsf H}_\phi$ in the sense of Definition 3.52, because it is linear and it preserves the product and the involution, as the reader immediately proves using the same properties of $\phi$. (i) It is obvious that $\pi _\phi (a)=0$ $\Leftrightarrow$ $\phi (a)=0$, since $\phi (a)\upharpoonright _{{\mathsf H}_\phi ^\perp }=0$ and this is the same as saying that $\pi _\phi$ is injective (faithful) if and only if $\phi$ is injective. (ii) If $\pi _\phi$ is the zero representation then $\pi _\phi ({\mathbb I})=0$ in particular, and hence $\phi ({\mathbb I}) = 0$ as well, by construction. If $\phi ({\mathbb I})=0$ then $\pi _\phi ({\mathbb I})=0$ and so $\pi _\phi (a)= \pi _\phi ({\mathbb I}a) = \pi _\phi ({\mathbb I})\pi _\phi (a) = 0\pi _\phi (a)=0$ for every $a\in {\mathfrak A}$. (If $\phi ({\mathbb I})=0$, we also have ${\mathsf H}_\phi = P({\mathsf H})= \{\mathbf{0}\}$ since $P=\phi ({\mathbb I})$.) (iii) If $\phi$ is surjective, there exists $a_1 \in {\mathfrak A}$ with $\phi (a_1) = I$ so $I= \phi (a_1) = \phi ({\mathbb I}a_1) = \phi ({\mathbb I}) \phi (a_1) = \phi ({\mathbb I}) I = \phi ({\mathbb I})=P$. Therefore ${\mathsf H}_\phi ={\mathsf H}$, which implies $\phi =\pi _\phi$. (iv) ${\mathsf M}:= P({\mathsf H})$ satisfies $\phi (a)({\mathsf M}) \subset {\mathsf M}$ for every $a\in {\mathfrak A}$, so either ${\mathsf M}={\mathsf H}$, and thus $\phi = \pi _\phi$ as we wanted, or ${\mathsf M}= \{\mathbf{0}\}$. The latter is forbidden by hypothesis, because it would imply $\phi (a)= 0$ for every $a\in {\mathfrak A}$.

3.33

If $\{{\mathfrak R}_\alpha \}_{\alpha \in A}$ is a family of von Neumann algebras on a Hilbert space ${\mathsf H}$, prove that

$$\left( \bigvee _{\alpha \in A}{\mathfrak R}_\alpha \right) ' = \bigwedge _{\alpha \in A}{\mathfrak R}'_\alpha \quad \text{ and }\quad \left( \bigwedge _{\alpha \in A}{\mathfrak R}_\alpha \right) ' = \bigvee _{\alpha \in A}{\mathfrak R}'_\alpha \,.$$

3.34

Prove that a von Neumann algebra ${\mathfrak R}$ on a Hilbert space ${\mathsf H}$ is complete in the strong operator topology. In other words, given $\{A_n\}_{n\in {\mathbb N}} \subset {\mathfrak R}$ such that $\{A_nx\}_{n\in {\mathbb N}}$ is Cauchy in ${\mathsf H}$ for every fixed $x\in {\mathsf H}$, there exists $A\in {\mathfrak R}$ such that $A_n \rightarrow A$ strongly (the converse being trivially true).

Solution. As ${\mathfrak R}$ is strongly closed in ${\mathfrak B}({\mathsf H})$, it is enough to prove that ${\mathfrak B}({\mathsf H})$ is strongly complete. Fix $x\in {\mathsf H}$. As $\{A_nx\}_{n\in {\mathbb N}}$ is Cauchy, $Ax := \lim _{n\rightarrow +\infty }A_nx$ does exist. Linearity of each $A_n$ implies linearity of $A: {\mathsf H}\ni x \mapsto Ax$. There remains to prove that A is bounded. As $A_nx$ admits limit, we have $||A_nx||<+\infty$ for every $x\in {\mathsf H}$ and every $n\in {\mathbb N}$. The Banach–Steinhaus Theorem 2.62 implies that $||A_n||<S <+\infty$ for every $n\in {\mathbb N}$. Moreover, if $x\in {\mathsf H}$, for every $\varepsilon _x >0$ there is $n_{\varepsilon _x}\in {\mathbb N}$ such that $||(A-A_n)x||<\varepsilon _{x}$ if $n> n_{\varepsilon _x}$. Summing up,

$$||Ax|| \le ||Ax-A_nx|| + ||A_nx|| < \varepsilon _{x}+ S||x||\quad \text{ for } n >n_{\varepsilon _x}.$$

Since $\varepsilon _x >0$ is arbitrary,

$$||Ax|| \le S||x||\quad \forall x \in {\mathsf H}\,,$$

which entails $||A||\le S< +\infty$ and therefore $A \in {\mathfrak B}({\mathsf H})$.

3.35

Prove that a von Neumann algebra ${\mathfrak R}$ on a Hilbert space ${\mathsf H}$ is complete in the weak operator topology. In other words, if $\{A_n\}_{n\in {\mathbb N}} \subset {\mathfrak R}$ is such that $\{(y|A_nx)\}_{n\in {\mathbb N}}$ is Cauchy in ${\mathbb C}$ for every fixed $x, y\in {\mathsf H}$, then there exists $A\in {\mathfrak R}$ such that $A_n \rightarrow A$ weakly (the converse being trivially true).

Solution. As ${\mathfrak R}$ is weakly closed in ${\mathfrak B}({\mathsf H})$, it suffices to prove that ${\mathfrak B}({\mathsf H})$ is weakly complete. Fix $x, y\in {\mathsf H}$. As $\{(y|A_nx)\}_{n\in {\mathbb N}}$ is Cauchy, $A_{yx} := \lim _{n\rightarrow +\infty }(y|A_nx)\in {\mathbb C}$ exists. Each $A_n$ is linear and the inner product is Hermitian, so ${\mathsf H}\ni x \mapsto A_{yx}$ is linear and ${\mathsf H}\ni y \mapsto A_{yx}$ antilinear. We claim there exist a linear operator $A: {\mathsf H}\rightarrow {\mathsf H}$ such that $A_{yx} = (y|Ax)$ for every $x, y \in {\mathsf H}$. For $x, y \in {\mathsf H}$ fixed, $|(y|A_nx)|<+\infty$ for every $n \in {\mathbb N}$. Therefore, for fixed x, applying the Banach–Steinhaus Theorem 2.62 to the family of linear functionals ${\mathsf H}\ni y\mapsto \overline{(y|A_nx)}$, we conclude that there is a positive number $S_x <+\infty$ satisfying $||\overline{(\cdot |A_nx)}||< S_x$ uniformly in n. Moreover, if $y\in {\mathsf H}$, for every $\varepsilon _y >0$ there is $n_{\varepsilon _y}\in {\mathbb N}$ such that $|\overline{A_{yx}}-\overline{(y|A_nx)}|<\varepsilon _{y}$ if $n> n_{\varepsilon _y}$. Summing up,

$$|\overline{A_{yx}}|\le |\overline{A_{yx}}-\overline{(y|A_nx)}| + |\overline{(y|A_nx)}| < \varepsilon _{y}+ S_x||y||\quad \text{ for } n >n_{\varepsilon _y}.$$

Since $\varepsilon _y >0$ is arbitrary,

$$|\overline{A_{yx}}| \le S_x||y||\quad \forall y \in {\mathsf H}\,,$$

which entails $||\overline{A_{\cdot x}}||\le S_x< +\infty$. Therefore Riesz’s lemma proves that there exists $A_x \in {\mathsf H}$ with

$$A_{yx} = (y|A_x)\quad \forall y \in {\mathsf H}\,.$$

It easy to prove that ${\mathsf H}\ni x \mapsto A_x=: Ax$ is linear. So now we have a linear operator $A: {\mathsf H}\rightarrow {\mathsf H}$ such that $(y|A_nx) \rightarrow (y|Ax)$ for every $x, y \in {\mathsf H}$. To finish it suffices to prove that A is bounded. Since $(A_n^*y|x)= (y|A_nx)$ we know the sequence $\{(A_n^*y|x)\}_{n\in {\mathbb N}}$ is Cauchy for every choice of $x, y\in {\mathsf H}$. Repeating the procedure above, there exists an operator $B : {\mathsf H}\rightarrow {\mathsf H}$ such that $(A_n^*y|x) \rightarrow (By|x)$ for every $x, y \in {\mathsf H}$, so that $(y|Ax)=(By|x)$ for every $x, y \in {\mathsf H}$. This identity easily implies that B is closed (Definition 2.98). Applying the closed graph Theorem 2.99, we conclude that $B\in {\mathfrak B}({\mathsf H})$. The identity $(y|Ax)=(By|x)$, $x, y\in {\mathsf H}$, rewritten as $(Ax|y)=(x|By)=(B^*x|y)$, implies $A=B^* \in {\mathfrak B}({\mathsf H})$.

3.36

Let ${\mathsf H},{\mathsf K}$ be complex Hilbert spaces, ${\mathfrak A}\subset {\mathfrak B}({\mathsf H})$ and ${\mathfrak B}\subset {\mathfrak B}({\mathsf K})$ unital $C^*$-algebras of operators, and $\phi : {\mathfrak A}\rightarrow {\mathfrak B}$ a continuous linear map relatively to the strong operator topology (or weak operator topology) of both ${\mathfrak A}$ and ${\mathfrak B}$. Prove that $\phi$ extends uniquely to a continuous linear map $\varPhi : {\mathfrak A}'' \rightarrow {\mathfrak B}''$ relatively to the strong operator topology (resp. weak operator topology) both on ${\mathfrak A}''$ and ${\mathfrak B}''$.

Solution. Let us treat the case of strong topologies. Since ${\mathfrak A}$ is strongly dense in ${\mathfrak A}''$, any strongly continuous extension $\varPhi$ must be unique. Let us prove that $\varPhi$ exists by using Exercise 3.34. Since ${\mathfrak A}$ is dense in ${\mathfrak A}''$, if $A\in {\mathfrak A}''$ there is a sequence $\{A_n\}_{n \in {\mathbb N}}\in {\mathfrak A}$ converging to A strongly. Define $\varPhi (A) :=$ s-$\lim _{n \rightarrow +\infty } \phi (A_n)$. First of all we prove that $\{\phi (A_n)\}_{n \in {\mathbb N}}\subset {\mathfrak B}''$ is Cauchy (with respect to the strong operator topology) and hence its limit belongs to ${\mathfrak B}''$, which is complete in that topology as established in Exercise 3.34. Fix a seminorm $p_y(\cdot ) := ||\cdot y||$ in ${\mathfrak B}({\mathsf K})$ associated to the vector $y \in {\mathsf K}$. Since $\phi$ is strongly continuous, for every $\varepsilon >0$ there is an open set $O_{y,\varepsilon }$ containing $0 \in {\mathfrak B}({\mathsf H})$, given by intersecting a finite number of balls defined by seminorms $p_{x^{(y)}_1}, \ldots p_{x^{(y)}_{n_y}}$. The vectors $x^{(y)}_k$ and the number $n_y$ also depend on $\varepsilon$, with $p_y(\phi (A)) < \varepsilon$ if $A \in O_{y,\varepsilon }\cap {\mathfrak A}$. In other words $||\phi (A)y||< \varepsilon$ if $\sum _{k=1}^{n_y}||Ax^{(y)}_k|| < \delta$ for some $\delta >0$ depending on y and $\varepsilon$. If A does not satisfy the last condition, $A' := \frac{\delta A}{\sum _{k=1}^{n_y}||Ax^{(y)}_k||}$ certainly will. Therefore, for every $y \in {\mathsf K}$ there is a constant $C_y\ge 0$ $(= \varepsilon /\delta )$ and there are vectors $x^{(y)}_1, \ldots , x^{(y)}_{n_y} \in {\mathsf H}$, such that

$$||\phi (A)y|| \le C_y \sum _{k=1}^{n_y}||Ax^{(y)}_k|| \quad \forall A \in {\mathfrak A}\,.$$

This inequality implies that the sequence $\{\phi (A_n)\}_{n\in {\mathbb N}}$ is Cauchy in the strong operator topology if $\{A_n\}_{n\in {\mathbb N}}$ is. We conclude that $\varPhi (A) :=$ s-$\lim _{n \rightarrow +\infty } \phi (A_n)$ is well defined. Notice that a different sequence $\{A_n'\}_{n \in {\mathbb N}}$ tending to A (strongly) would produce the same limit, because

$$||\phi (A_n)y-\phi (A_n')y|| \le C_y \sum _{k=1}^{n_y}||(A_n-A'_n)x^{(y)}_k||$$

and $||(A_n-A'_n)x^{(y)}_k|| \rightarrow 0$ (both sequences converge to A strongly). It is therefore clear that $\varPhi$ extends $\phi$, because for any $A \in {\mathfrak A}$ we may define $\varPhi (A) :=$ s-$\lim _{n \rightarrow +\infty }\phi (A_n)$, using a constant sequence $A_n =A$. The limiting process used to define $\varPhi$ immediately proves that $\varPhi$ is linear as well, since $\phi$ is. Regarding the strong continuity of $\varPhi$, observe that, by construction

$$||\varPhi (A)y||= \left| \left| \lim _{n \rightarrow +\infty } \phi (A_n)y\right| \right| = \lim _{n \rightarrow +\infty } ||\phi (A_n)y||\,.$$

Since $A_nx^{(y)}_k \rightarrow Ax^{(y)}_k$ and

$$||\phi (A_n)y|| \le C_y \sum _{k=1}^{n_y}||A_nx^{(y)}_k|| \,, $$

the limit above produces

$$||\varPhi (A)y|| \le C_y \sum _{k=1}^{n_y}||Ax^{(y)}_k||\,,\quad \text{ if } A \in {\mathfrak A}'',$$

valid for every $y \in {\mathsf K}$, for the corresponding $C_y\ge 0$ and for $x^{(y)}_1, \ldots , x^{(y)}_{n_y} \in {\mathsf H}$. This inequality is equivalent to the continuity of $\varPhi$ in the strong topology of the two von Neumann algebras ${\mathfrak A}''$ and ${\mathfrak B}''$.

The case of weak topologies is completely analogous if we replace the seminorms $p_x(\cdot ) = ||\cdot x||$ with the seminorms $p_{x, y}(\cdot ) = |(x|\cdot y)|$. Then $\varPhi$ is the unique map satisfying $(x|\varPhi (A)y) = \lim _{n \rightarrow +\infty }(x|\phi (A_n)y)$ for $x, y \in {\mathsf K}$, $A \in {\mathfrak A}''$ and ${\mathfrak A}\ni A_n \rightarrow A$ weakly.

3.37

Let ${\mathsf H},{\mathsf K}$ be complex Hilbert spaces, ${\mathfrak A}\subset {\mathfrak B}({\mathsf H})$ and ${\mathfrak B}\subset {\mathfrak B}({\mathsf K})$ unital $C^*$-algebras of operators. Prove that a continuous $^*$-homomorphism $\phi : {\mathfrak A}\rightarrow {\mathfrak B}$ (relatively to the weak operator topology of ${\mathfrak A}$ and ${\mathfrak B}$) extends to a unique continuous $^*$-homomorphism of von Neumann algebras $\varPhi : {\mathfrak A}'' \rightarrow {\mathfrak B}''$ (in the weak topology).

Solution. As $\phi$ is linear it defines a unique weakly continuous linear extension $\varPhi : {\mathfrak A}'' \rightarrow {\mathfrak B}''$, as established in Exercise 3.36. We shall exploit the fact that ${\mathfrak A}$ is weakly dense in ${\mathfrak A}''$, by the double commutant theorem. The claim is that $\varPhi (I_{\mathsf H})=I_{\mathsf K}$, $\varPhi (A^{*_{\mathsf H}})= \varPhi (A)^{*_{\mathsf K}}$ and $\varPhi (A\circ _{\mathsf H}B)=\varPhi (A)\circ _{\mathsf K}\varPhi (B)$ for $A, B \in {\mathfrak A}''$. (We henceforth omit the subscripts $_{\mathsf H}, _{\mathsf K}$ for simplicity.) The first requirement automatically holds because it is valid for $\phi$. Let us prove $\varPhi (A^*)= \varPhi (A)^*$. If ${\mathfrak A}\ni A_n \rightarrow A$ weakly, then ${\mathfrak A}\ni A^*_n \rightarrow A^*$ weakly. From now on all limits will be meant in the weak sense. We have

$$\varPhi (A^*)= \lim _{n\rightarrow +\infty } \phi (A_n^*)= \lim _{n\rightarrow +\infty }\phi (A_n)^* = \varPhi (A)^*$$

because $\varPhi$ is weakly continuous and extends the $^*$-homomorphism $\phi$. To conclude, we need to establish $\varPhi (AB) = \varPhi (A) \varPhi (B)$ for $A, B \in {\mathfrak A}''$. Let us start by assuming $A\in {\mathfrak A}''$ but $B \in {\mathfrak A}$. Then ${\mathfrak A}\ni A_n \rightarrow A$ weakly implies ${\mathfrak A}\ni A_nB \rightarrow AB$ weakly. Therefore, by the very definition of $\varPhi$:

$$\varPhi (AB)= \lim _{n \rightarrow +\infty }\varPhi (A_nB)= \lim _{n \rightarrow +\infty }\phi (A_nB)= \lim _{n \rightarrow +\infty } \phi (A_n)\phi (B)$$

$$= \lim _{n \rightarrow +\infty } \phi (A_n)\varPhi (B) = \varPhi (A)\varPhi (B)\,,$$

where we have used the fact that $\phi (A_n)\rightarrow \varPhi (A)$ weakly implies $\phi (A_n)\varPhi (B)\rightarrow \varPhi (A)\varPhi (B)$ weakly. (With a similar proof, $\varPhi (AB) = \varPhi (A) \varPhi (B)$ holds true also if $A\in {\mathfrak A}$ but $B \in {\mathfrak A}''$.) Let us finally focus on the general case $A, B \in {\mathfrak A}''$. Since $\varPhi$ is weakly continuous, and ${\mathfrak A}\ni B_n \rightarrow B\in {\mathfrak A}''$ weakly implies $AB_n \rightarrow AB$, we have:

$$ \varPhi (AB)= \lim _{n \rightarrow +\infty }\varPhi (AB_n)= \lim _{n \rightarrow +\infty }\varPhi (A)\varPhi (B_n) = \varPhi (A)\varPhi (B)\,. $$

3.38

Let ${\mathsf H}$ be a complex Hilbert space and suppose $h : {\mathfrak B}({\mathsf H}) \rightarrow {\mathfrak B}({\mathsf H})$ is an antilinear map preserving the product, the operation $^*$, and fixing the identity: $h(I)=I$. Prove that if h is bijective, then there exists an isometric and surjective antilinear map $U : {\mathsf H}\rightarrow {\mathsf H}$ (called an anti-unitary operator) such that $h(A)= UAU^{-1}$.

Hint. If $N\subset {\mathsf H}$ is a Hilbert basis, define the map $C : {\mathsf H}\rightarrow {\mathsf H}$ by

$$ C : \sum _{x \in N} (x|z)x \mapsto \sum _{x \in N} \overline{(x|z)}x \, $$

for every $z\in {\mathsf H}$. Next, set $\alpha (A) := C h(A)C^{-1}$ for every $A \in {\mathfrak B}({\mathsf H})$, so that the map $\alpha$ becomes a $^*$-automorphism of ${\mathfrak B}({\mathsf H})$. Finally, exploit Theorem 3.96.