Perturbation of the numerical range by bounded operators Let $H$ be a complex Hilbert space with scalar product $\langle, \rangle$ and $T :D(T)\subset H\to H$ be linear operator. Then numerical range of $T,$ denoted $NR(T),$ is the subset 
$$NR(T) := \{\langle Tx,x \rangle | x\in H, \|x\|=1\}.$$
It is stated here that if $T\in \mathcal{B}(H)$ is normal, then the closure of $NR(T)$ is the smallest convex set containing $\sigma(T)$ the spectrum of $T.$ 
Let now $N\in \mathcal{B}(H)$ be normal such that 
$$\sigma(N) \subset \Gamma^+ := \{z \in \mathbb{C}| {\rm Re}(z) >0\}.$$
Then it is obvious that 
\begin{equation} \tag{1}
\overline{NR(N)} \subset \Gamma^+.
\end{equation}
Is it possible to find  $\delta >0$ such that  (1) is still true  for every $E \in B(N,\delta):= \{A \in \mathcal{B}(H), \|N-A\| <\delta \}.$
My attempt is to argue by continuity of 
$$f_x: \mathcal{B}(H)\to \mathbb{C}: f_x(E) = {\rm Re} \langle Ex,x\rangle,$$
but in this case, we can prove that for every $x\in H,$ there exists $\delta_x$ such that for all $E \in B(N,\delta_x),$ $f_x(E) \geq 0$ 
(not the hole numerical range of $E$.) How can we find $\delta$ independently of $x.$
Thank you for any hint.
 A: $\overline{NR(N)}$ is a compact subset of $\mathbb{C}$, hence there exists $\lambda\in\overline{NR(N)}$ such that $0<\Re(\lambda)=\inf\{\Re(\lambda')\,|\,\lambda'\in\overline{NR(N)}\}$. Put $\delta=\frac{\Re(\lambda)}{2}>0$ and assume that $E\in B(N,\delta)$. Take any unit vector $x\in H$. Then
$|\langle Ex,x\rangle-\langle Nx,x\rangle|\le\delta$,
hence $\Re\langle Ex,x\rangle\ge\Re\langle Nx,x\rangle-\delta\ge\frac{\Re(\lambda)}{2}>0$ and $\overline{NR(E)}\subset\Gamma^+$.
A: Here is my answer:
Since $\overline{NR(N)}$ a compact sube set in $\mathbb{C},$ we have 
$$ \alpha := {\rm dist}(NR(N), i\mathbb{R}) >0. \tag{1}$$
Assume that for all $\delta >0$ there exists $E$ with $\|E-N\| <\delta$ and there exists $x_\delta$ with $\|x_\delta\|=1$ such that ${\rm Re}\langle Ex_\delta,x_\delta\rangle < \frac{\alpha}{2}.$ Take $\delta = \frac 1n.$ There exist $E_n, x_n$ with $\|N-E_n\|<\frac 1n$ and $\|x_n\|=1$ such that 
$${\rm Re}\langle E_nx_n,x_n\rangle < \frac{\alpha}{2}\quad \forall n .$$
Therefore 
$${\rm Re}\langle (E_n -N)x_n,x_n\rangle +{\rm Re}\langle Nx_n,x_n\rangle  < \frac{\alpha}{2}$$
and $$\liminf_{n\to \infty}{\rm Re}\langle Nx_n,x_n\rangle  \leq \frac{\alpha}{2}$$ 
which contradicts (1).
