### Introduction to von Neumann algebras, addendum to Lecture 1 (solution of Exercise B: the norm of a selfadjoint operator)

One of the challenges I had in preparing this course, was to find a quick route to the modern theory that is different from the standard modern route, in order to save time and be able to reach significant results and examples in the limited time of a one semester course. A main issue was to avoid the (beautiful, beautiful, beautiful) Gelfand theory of commutative Banach and C*-algebras, and base everything on the spectral theorem for a single selfadjoint operator (which is significantly simpler than the one for normal operators). In the previous lecture, I stated Exercise B, which gave some important properties of the spectrum of a selfadjoint operator. Since my whole treatment is based on this, I felt that for completeness I should give the details.

Spoiler alert: If you are a student in the course and you plan to submit the solution of this exercise, then you shouldn’t read the rest of this post.

For every $T \in B(H)$, we define the numerical range of $T$ to be the set

$W(T) = \{\langle T h, h \rangle : \|h\|=1\}$,

and we define the numerical radius  to be

$r(T) = \sup \{|\langle T h, h \rangle| : \|h\|=1\}$.

Recall that the operator norm is given by $\|T\| = \sup_{\|g\|=\|h\|=1}|\langle Tg, h \rangle |$.

From here onwards, let us fix a selfadjoint operator $T \in B(H)_{sa}$. If $h \in H$, then we have

$\langle h, T h \rangle = \overline{\langle T h, h \rangle}$

while, on the other hand, using $T = T^*$, we get

$\langle h, T h \rangle = \langle h, T^* h \rangle = \langle T h, h \rangle$,

so $\langle T h, h \rangle$ is a real number. Thus, the numerical range of a selfadjoint operator is real : $W(T) \subseteq \mathbb{R}$.

Now, set

$m_T = \inf_{\|h\|=1}\langle Th, h \rangle$

and

$M_T = \sup_{\|h\|=1}\langle Th, h \rangle$.

Lemma 1: $\sigma(T) \subseteq [m_T, M_T] \subseteq \mathbb{R}$

Proof: Let $\lambda \notin [m_T,M_T]$. Then, on the one hand,

$|\langle (T-\lambda)h, h \rangle| \leq \|(T - \lambda)h\| \|h\|$,

while on the other hand,

$\langle (T-\lambda)h, h \rangle = \langle T h, h \rangle - \lambda \langle h,h \rangle = (\mu - \lambda)\|h\|^2$

for some $\mu \in W(T) \subseteq [m_T,M_T]$.

So, $\|(T- \lambda) h \| \geq c \|h\|$ where $c:= \inf\{|\mu - \lambda| : \mu \in [m_T,M_T]\}$. In other words, $T- \lambda$ is bounded below. In particular, $Ker (T-\lambda) = \{0\}$.

On the other hand, if $\lambda \notin [m_T, M_T]$, then $\overline{\lambda} \notin [m_T,M_T]$ too, so by the same argument, $T - \overline{\lambda} = (T- \lambda)^*$ is also bounded below. Thus $\overline{Im(T-\lambda)} = Ker (T - \overline{\lambda})^\perp = H$. Since $T- \lambda$ is bounded below, it follows that $Im(T-\lambda) = H$, so$T- \lambda$ is invertible, and $\lambda \notin \sigma(T)$.

Lemma 2: $\|T\| = r(T)$ (this was phrased in Exercise B as $\|T\| = \max \{|m_T|, M_T\}$).

Proof: By the Cauchy-Schwarz inequality,

$|\langle Th,h \rangle| \leq \|Th\|\|h\| \leq \|T\| \|h\|^2$,

so $r(T) \leq \|T\|$, and this holds for any operator. Now, since our $T$ is also assumed selfadjoint, we will be able to get equality. For this, we first require the following fact:

Claim: For every operator $T \in B(H)$

$\|T\| = \sup\{\|PTP\| : P$ is a projection onto a finite dimensional subspace $\}$.

Proof of claim: Suppose that $T \neq 0$, and let $\epsilon >0$, and for peace of mind assume also that $\epsilon < \|T\|$. Let $h \in H$, $\|h\| = 1$, such that $\|Th\| > \|T\|-\epsilon$. Now let $g$ be a normalized vector in the direction of $Th$, thus $|\langle Th, g \rangle| = \|Th\| > \|T\|- \epsilon$.

Now, if $P$ is the orthogonal projection onto the space spanned by $g$ and $h$, then $\langle PTP h, g \rangle = \langle Th, g \rangle$, and so

$\|PTP\| = \sup\{|\langle PTPx,y\rangle| : \|x\|=\|y\|=1\} \geq |\langle PTP h, g \rangle| > \|T\|-\epsilon$.

This proves the claim. To complete the proof of the lemma, we take an orthogonal projection $P$ onto a finite dimensional subspace $M \subseteq H$. Then $PTP$ can be considered as a bounded operator on $M$, and $(PTP)^* = PTP$, so it is selfadjoint. By the spectral theorem for selfadjoint operators on a finite dimensional vector space (over $\mathbb{C}$, in this case), we have that $PTP$ is diagonalizable, and in this case it is easy to see that $\|PTP\|$ is equal to the modulus of the eigenvalue $\lambda_1$ of $PTP$ which has maximal modulus. Moreover, $\|PTP\| = |\langle PTP v_1, v_1\rangle|$, where $v_1 \in M$ is a unit eigenvector corresponding to $\lambda_1$. (Oh do I have to chew it? Well, if $v_i$ corresponds to eigenvalue $\lambda_i$, then $\|T \sum_i c_i v_i\|^2 = \|\sum_i \lambda_i c_i v_i\|^2 = \sum_i |\lambda_i|^2 |c_i|^2$, and the maximal value of this expression for all $\sum_i |c_i|^2 =1$ occurs when $c_1 = 1$ and the rest are zero). Therefore

$\|PTP\| = |\langle PTPv_1,v_1\rangle| = |\langle T v_1 , v_1 \rangle |$.

Applying the claim, this proves the lemma.

Lemma 3: $m_T, M_T \in \sigma(T)$.

Proof: Assume, without loss of generality, that $|m_T| \geq |M_T|$, so that $\|T\| = |m_T|$. Let $h_n$ be a sequence of unit vectors such that $\langle Th_n, h_n \rangle \to m_T$. Then

$\|(T-m_T)h_n\|^2 = \|Th_n\|^2 - 2\langle T h_n, m_T h_n \rangle + |m_T|^2 \|h_n\|^2 \leq 2|m_T|^2 - 2m_T \langle T h_n , h_n \rangle \to 0$.

This shows that $T- m_T$ is not bounded below, so $m_T \in \sigma(T)$. To show that $M_T$ is also in $\sigma(T)$, we consider the operator $T' := T - m_T I$. Then $m_{T'} = 0$ and $M_{T'} = M_T - m_T > 0$. So $\|T'\| = M_{T'}$. Now, by a very similar argument to that given above, we find that $M_{T'} \in \sigma(T') = \sigma(T) - m_T$. Therefore $M_T \in \sigma(T)$.

Conclusion: Putting Lemmas 1, 2 and 3 together, we obtain the following important characterization of the operator of a selfadjoint operator in terms of its spectrum:

(*)  $\|T\| = \sup\{|\lambda| : \lambda \in \sigma(T)\}$.