Showing a 2-by-2 matrix is a contraction

Question

Let $S\subseteq\mathbb{T}:=\{z\in\mathbb{C}:\vert z\vert=1\}$ be a compact set such that $\operatorname{conv}S\supseteq\{z\in\mathbb{C}:\vert z\vert\leq\frac{1}{\sqrt{2}}\}$ and $B\in M_2(\mathbb{C})$. Suppose $I_2+\lambda B+\overline{\lambda}B^*\geq 0$ for all $\lambda\in S$. Is $B$ a contraction?

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Comments: I can see this question has an affirmative answer for some particular choice of $S$. So I am hoping this may be true. But I am unable to prove it in general.

I apologize that I forgot to mention $S\subseteq\mathbb{T}$ before.

Any comment is highly appreciated.

Answer 1

Taking $B=\begin{pmatrix}0&\sqrt{2}\\0&0\end{pmatrix}$ and $S=\{z\in\mathbb{C}, |z|\le \frac{1}{\sqrt{2}}\}$ is a counterexample. The bound is $\sigma_1(B)\le \sqrt{2}$.

For any matrix $B,M\in \mathbb{M}_n(\mathbb{C})$ the numerical range of $B$ is the set $W(B)=\{x^*Bx, x\in \mathbb{C}^n, x^*x=1\}$. Also $W(\dfrac{M+M^*}{2})=\mathfrak{R}(W(M))$, $W(\lambda B)=\lambda W(B)$ (rotation and homothety) and $\sigma_1(M)\le 2w(M)$ where $w(M)$ is the numerical radius of $M$, (these are known), with $\lambda B=M$. We see that if $\dfrac{M+M^*}{2}\ge -0.5I$, for any $\lambda\in S$, then $w(M)\le 0.5$. Say $\lambda=e^{i\theta}\frac{1}{\sqrt{2}}$, if there is a point $z\in W(M)$ with $|z|>0.5$ applying a certain rotation to $B$ this point intersect the $y=0$ line contradicting the fact that the real part of $W(M)$ is in $ [-0.5;*[$.