Numerical range of an operator on Hilbert spaces. 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 


*

*The point and residual spectrum are subsets of $W(T)$.

*The continuous spectrum is a subset of closure of $W(T)$.


Please help me out, Thank you.
 A: $\lambda$ 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 $y_n$. 
