Why is the numerical range of an operator convex? Let $T$ be a Hilbert space operator. Its numerical range is \begin{equation} W(T)=\{\langle Tx,x\rangle:\|x\|=1\}.\end{equation}
It is a well-known fact that $W(T)$ is a convex subset of the complex plane. However, every proof I know is by brute force computation. First for $2\times 2$ matrices, then the general case. 
Even though the computation can be carried out in clever ways, it still fails to provide some explanation why this is true. What is the link between this result and other concepts of the theory?
I wonder whether there is any conceptual explanation for this result. I do not ask the explanation to be rigorous, just some ideas.
 A: Let $\mathcal{H}$ be a complex separable Hilbert space and $T$ be a bounded linear operator on $\mathcal{H}$. Let us assume in advance that, for a $2\times 2$ complex matrix $A$ the $W(A)$ is an elliptic disc (The result is true but I am unable to show here because it will make the proof much bigger).
Now we have to show that for any $\alpha,\beta \in W(T)\implies\lambda \alpha+(1-\lambda)\beta\in W(T)$, where $\lambda\in (0,1)$.
Let $\alpha=\langle Tf,f\rangle$ and $\beta=\langle Tg,g\rangle$ where $\|f\|=\|g\|=1$. Clearly $f,g$ are linearly independent unit vectors otherwise, $\alpha=\beta$. Denote  $V=\mbox{span}\{f,g\}$ and $\dim V=2$. Since $V$ is closed, therefore there exists an orthogonal projection $P_V$ onto $V$.
Now the compression $A=P_V T|_V$ of $T$ is a two dimensional operator. So, $W(A)$ is an elliptic disc from our initial assumption. Again $\langle A f,f\rangle= \langle Tf,f\rangle=\alpha$ and $\langle A g,g\rangle= \langle Tg,g\rangle=\beta$. Therefore, $\lambda \alpha+(1-\lambda)\beta\in W(A)$, where $\lambda\in (0,1)$. Also we know that $W(A)\subseteq W(T)$. Thus, $\lambda \alpha+(1-\lambda)\beta\in W(T)$. Hence we get our required result.
