Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $H$ is a Hilbert space and $T$ is in $\mathcal{L}(H)$, the numerical range of $T$ is defined by $$W(T) := \left\{(Tx; x) \mid x \in H,\ \|x\| = 1 \right\}.$$ We have to prove that

  1. The point and residual spectrum are subsets of $W(T)$.
  2. The continuous spectrum is a subset of closure of $W(T)$.

Please help me out, Thank you.

share|cite|improve this question
We can prove it in any way we want. However, there was a hint to the second part: If s does not belong to closure of W(T). Prove that norm(sId-T) is greater or equal to dist(s;W(T))norm(x). Conclude that (sId - T) : H to R(sId - T) is continuously invertible. – johnathan Jun 30 '12 at 22:46
up vote 3 down vote accepted

$\lambda$ is in the point spectrum.

By definition it exists a $x\in H$ such that $\Vert x \Vert = 1$ and $Tx = \lambda x$. We have $$ (Tx,x) = (\lambda x, x) = \lambda $$ so $\lambda$ belongs to $W(T)$.

$\lambda$ is in the residual spectrum.

Then it exists $x\in H$ such that $\Vert x \Vert = 1$ and $x$ is orthogonal to the range of $T - \lambda$.
For each $y\in H$ we have $((T - \lambda)y, x) = 0$, in particular $$ 0 = ((T - \lambda)x, x) = (Tx, x) - \lambda $$ Also in this case $\lambda$ belongs to $W(T)$.

$\lambda$ is in the continuous spectrum.

$(T- \lambda)^{-1}$ is not bounded, therefore it exists a sequence $\{x_n\}_{n\in \mathbb N}$ of elements of $H$, with $\Vert x_n\Vert = 1$ and $\Vert (T - \lambda)^{-1} x_n\Vert \to \infty$.
Let's consider the sequence $$ y_n := \frac {(T - \lambda)^{-1} x_n} {\Vert (T - \lambda)^{-1} x_n \Vert} $$ we have $\Vert y_n \Vert = 1$ and $$ (T - \lambda)y_n = \frac {x_n} {\Vert (T - \lambda)^{-1} x_n \Vert} \to 0 $$ If $\lambda \notin \overline{W(T)}$ then it exists a $M > 0$ such that $\vert \lambda - (Tx, x)\vert > M$ for each $x\in H$, $\Vert x \Vert = 1$. As conseguence $$ \Vert (T-\lambda)x \Vert = \Vert (T-\lambda)x \Vert \Vert x \Vert \geq \vert ((T - \lambda)x, x)\vert = \vert \lambda - (Tx, x)\vert > M $$ for each $x\in H$, $\Vert x \Vert = 1$. But that contradicts the existence of the sequence $y_n$.

share|cite|improve this answer
Thank you. It is indeed clear now. – johnathan Jul 3 '12 at 10:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.