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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.

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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 $y_n$.

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Thank you. It is indeed clear now. – johnathan Jul 3 '12 at 10:15

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