Question about numerical range of a linear operator I am trying to solve the following question, but I did not reach to any answer, I would be so glad if anyone could help me on that.
Let $E$ be a Hilbert space over $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, with inner product $\langle\cdot\;| \;\cdot\rangle$ and the norm $\|\cdot\|$. Let $T\in \mathcal{L}(E)$ be  an operator. It is well known that the numerical range of $T$
$$W(T)=\{\langle Tx\;|\;x\rangle:\;x \in E,\;\;\|x\|=1\},$$
is convex. If $\lambda=\langle T x_1\; |\;x_1\rangle\in W(T)$, and $\mu=\langle T y_1\; |\;y_1\rangle\in W(T)$, where $\|x_1\|=\|y_1\|=1$. Why for any point $\eta$ on the line segment joining $\lambda$ and $\mu$, there exist complex numbers $\alpha$ and $\beta$ such that $\|\alpha x_1+\beta y_1\|=1$ and $\langle T (\alpha x_1+\beta y_1)\; |\;\alpha x_1+\beta y_1\rangle=\eta.$ ??
I try as follows:
Let $M$ be a subspace spanned by $x_1$ and $y_1$ and $P_{M}$ be a projection of $E$ onto ${M}$. Consider $S=P_{M}TP_{M}$. We deduce that $\langle S x_1\; |\;x_n\rangle=\langle T x_1\; |\;x_1\rangle$ and $\langle S y_1\; |\;y_1\rangle=\langle T y_1\; |\;y_1\rangle$. Hence, $\langle T x_1\; |\;x_1\rangle,\langle T y_1\; |\;y_1\rangle\in W(S)=\{\langle Sx\;|\;x\rangle=\langle Tx\;|\;x\rangle:\;x \in M,\;\;\|x\|=1\},$ which is convex. As a consequence, any point $\eta$ on the line segment joining $\lambda$ and $\mu$ belongs to $W(S)$. So, there exist $z\in M$ such that $\|z\|=1$ and $\langle Tz\; |\;z\rangle=\eta$. This implies that there exist complex numbers $\alpha$ and $\beta$ such that $\|\alpha x_1+\beta y_1\|=1$ and $\langle T (\alpha x_1+\beta y_1)\; |\;\alpha x_1+\beta y_1\rangle=\eta.$
But why
$$W(S)=\{\langle Sx\;|\;x\rangle=\langle Tx\;|\;x\rangle:\;x \in M,\;\;\|x\|=1\}?$$
Thank you everyone !!
 A: The statement is not evident but not difficult to prove. It is (also) known as the Toeplitz-Hausdorff Theorem for linear operators. One of the shortest proofs I have seen is one page by K Gustafson
As you are sort of hinting at it suffices to look at the (possibly complex) plane spanned by unit vectors $x$ and $y$ in order to find for any $0<\lambda<1$ a unit element $z=\alpha x + \beta y$ so that $$(Tz|z) = \lambda (Tx|x) + (1-\lambda) (Ty|y)$$
The construction of $\alpha,\beta$ is, however, not completely trivial but you better look at the above reference (which is open access), rather than I reproduce it here.
Regarding $W(S)$ (if I understand your question correctly): Provided $P=P_M$ is the orthogonal projection onto $M$, indeed $(Sx|x)=(Tx|x)$ for all $x\in M$. This follows from $P^2=P^*=P$ and $Px=x$ for $x\in M$ so that for such $x$:
  $$ (Sx|x)=(PTPx|x)=(PTx|x)=(Tx|P^*x)=(Tx|Px)=(Tx|x).$$
On the other hand it does not simplify the problem since showing that $W(S)$ is convex is as hard (or easy depending on taste) as showing that $W(T)$ is convex, since the latter in any case is a 2-dimensional problem (as you will realize by looking in the above reference).
