Proof explanation: convexity of the numerical range of an operator (Toeplitz-Hausdorff Theorem) Let $\mathcal{H}$ be a complex Hilbert space and $\mathcal{B}(\mathcal{H})$  be the algebra of all bounded linear operators on $\mathcal{H}$.
Theorem:
Let $T\in \mathcal{B}(\mathcal{H})$, then $W(T)$ is a convex subset of $\mathbb{C}$, where
$$W(T):=\{\langle T x\mid x\rangle;\;x \in \mathcal{H}\;\;\text{with}\;\|x\|=1\}.$$
Proof:
Given $\lambda,\mu\in W(T)$ such that $\lambda\neq\mu$ and we will prove that $\eta=t\lambda+(1-t)\mu \in W(T)$ whenever $t\in [0,1]$. Clearly we have
$$\eta\in W(T)\Leftrightarrow t\in W(\alpha I+\beta T),$$
with $\alpha=-\frac{\mu}{\lambda-\mu}$ and $\beta=\frac{1}{\lambda-\mu}$. Let $S=\alpha I+\beta T$. We observe that $\eta\in W(T)$ if and only if $t\in W(S)$ for all $t\in [0,1]$. Since $\lambda,\mu\in W(T)$, then there exist unit vectors $x,y\in \mathcal{H}$ such that
$$\lambda=\langle Tx\mid x\rangle\;\text{and}\;\mu=\langle Ty\mid y\rangle.$$
A simple calculation shows that
$$\langle Sx\mid x\rangle=1\;\text{and}\;\langle Sy\mid y\rangle=0.$$
Define $g:\mathbb{R}\to \mathbb{C}$ by
$$g(\theta)=\langle Sy\mid x\rangle e^{-i\theta}+\langle Sx\mid y\rangle e^{i\theta},\;\theta\in \mathbb{R}.$$
Obviously $g(\theta+\pi) = -g(\theta)$ for all $\theta\in \mathbb{R}$. Further, $\Im m g(0)=-\Im m g(\pi)$. Since $g$ is a continuous function, then there exists $\theta_0\in [0,\pi]$ such that $\Im m g(\theta_0)=0$.
Now observe that the vectors $y$ and $\hat{x}=e^{i\theta_0}x$ are linearly independent. Otherwise, we write $y=\alpha \hat{x}$ for some $\alpha\in \mathbb{C}$, then $|\alpha|=1$ and $0=\langle Sy\mid y\rangle=|\alpha|^2\langle S\hat{x}\mid \hat{x}\rangle=\langle Sx\mid x\rangle=1$ which is a contradiction. Hence, $\|(1-t)y+t\hat{x}\|\neq 0$ for all $t\in [0,1]$. Put
$$z_t=\frac{(1-t)y+t\hat{x}}{\|(1-t)y+t\hat{x}\|},\;t\in [0,1].$$
Clearly $\|z_t\|=1$ for all $t\in [0,1]$. Moreover, one can see that $\langle Sz_0\mid z_0\rangle=0$ and $\langle Sz_1\mid z_1\rangle=1$. This implies that $0,1\in W(S)$.
To finish the proof, define a continuous function $f$ on $[0,1]$ by
$$f(t)=\langle Sz_t\mid z_t\rangle,\;t\in [0,1].$$
A straightforward calculation shows that $f$ is a real-valued function with $f(0)=0$ and $f(1) = 1$. Hence $[0,1]\subset W(S)$.

In this proof I don't understant the following facts:

*

*$\eta\in W(T)\Leftrightarrow t\in W(\alpha I+\beta T).$


*Why we use $\Im m g$ in order to find $\theta_0$.


*Why we use $f$ in order to conclude that $[0,1]\subset W(S)$?

 A: *

*Note first that, for $x$ with $\|x\|=1$,  $$ \langle (\alpha I+\beta T)x,x\rangle =\alpha+\beta\langle Tx,x\rangle.$$ Since $\eta=t\lambda+(1-t)\mu$, 
$$
t=\frac{\eta-\mu}{\lambda-\mu}.
$$
So, if $\mu\in W(T)$, then $\mu=\langle Ty,y\rangle$ for some $y$ with $\|y\|=1$, and then 
$$
t=\frac{\eta-\mu}{\lambda-\mu}=\alpha+\beta\mu=\alpha+\beta\langle Ty,y\rangle=\langle (\alpha I+\beta T)y,y\rangle \in W(\alpha I+\beta T).
$$
Conversely, if $t\in W(\alpha I+\beta T)$, then $t=\alpha+\beta\langle Tz,z\rangle$ for some $z$ with $\|z\|=1$, and 
\begin{align}
\eta=t\lambda+(1-t)\mu&=\lambda\alpha+\lambda\beta\langle Tz,z\rangle+\mu-\mu\alpha-\mu\beta\langle Tz,z\rangle\\ \ \
&=(\lambda-\mu)(\alpha+\beta\langle Tz,z\rangle)+\mu\\ \ \\
&=-\mu+\langle Tz,z\rangle+\mu\\ \ \\
&=\langle Tz,z\rangle\in W(T). 
\end{align}

*The fact that $g(\theta_0)$ is real is used in the "straightforward calculation" that shows that $f$ is real. 

*The trick in the proof is to change the question of whether $\eta\in W(T)$ into $t\in W(S)$.  At the end of the proof the function $f$ is used to show that all of $[0,1]\subset W(S)$, so in particular the $t$ from the beginning satisfies $t\in W(S)$, which in turns implies $\eta\in W(T)$. 
