The foci conjecture for elliptical numerical ranges of block matrices

Counterexample on a separable Hilbert space; every focal distance $r>1$ occurs.

Gau–Hong–Li–Wang, arXiv:2606.08576, Conjecture 3.5, asks whether

$$T=\begin{pmatrix}A&B\\0&0\end{pmatrix}\in B(H\oplus H), \qquad A=A^*,\quad A^2=I,$$

and the assumption that $\overline{W(T)}$ is an elliptical disk force both foci to lie in $\{-1,0,1\}$. The source already proves the finite-dimensional case. The following infinite-dimensional separable construction disproves the general conjecture.

Let

$$I=[-1/8,1/8],\qquad D=I\cap\mathbb Q,\qquad H=\ell^2(D;\mathbb C^2),$$

and define

$$G(x)=240\frac{1-x^2}{1-4x^2}, \qquad d(x)=G'(x)=\frac{1440x}{(1-4x^2)^2}, \qquad s(x)=G(x)-xG'(x).$$

Explicitly,

$$s(x)=240\frac{1-11x^2+4x^4}{(1-4x^2)^2}.$$

Throughout $I$, one has $s(x)>|d(x)|$: for $t=|x|\le1/8$, this reduces to

$$1-6t-11t^2+4t^4\ge\frac{81}{1024}>0.$$

Hence the continuous bounded functions

$$u(x)=\sqrt{\frac{s(x)+d(x)}2}, \qquad v(x)=\sqrt{\frac{s(x)-d(x)}2}$$

are strictly positive. Define bounded operators fiberwise by

$$A_x=\begin{pmatrix}1&0\\0&-1\end{pmatrix}, \qquad B_x=\begin{pmatrix}u(x)&0\\v(x)&0\end{pmatrix}, \qquad x\in D.$$

In particular, $A=A^*$, $A^2=I$, and $\|B\|^2=240$.

Write $c=\cos\theta$. After a unitary removal of the off-diagonal phase, each fiber of $\operatorname{Re}(e^{-i\theta}T)$ equals

$$K_x(c)\oplus[0], \qquad K_x(c)= \begin{pmatrix} c&0&u(x)/2\\ 0&-c&v(x)/2\\ u(x)/2&v(x)/2&0 \end{pmatrix}.$$

Set

$$h(c)=\sqrt{60+4c^2}, \qquad y=\frac{c}{h(c)}\in I.$$

Since $h(c)>|c|$, the Schur-complement criterion gives

$$h(c)I-K_x(c)\succeq0 \iff s(x)+d(x)y\le4\bigl(h(c)^2-c^2\bigr) \iff G(x)+G'(x)(y-x)\le G(y),$$

where

$$4\bigl(h(c)^2-c^2\bigr)=240+12c^2=G(y).$$

But

$$G''(x)=\frac{1440(1+12x^2)}{(1-4x^2)^3}>0.$$

Therefore the required inequality is precisely the supporting-tangent inequality for the strictly convex function $G$, with equality exactly when $x=y$. Since $D$ is dense, continuity gives

$$\sup\sigma\!\left(\operatorname{Re}(e^{-i\theta}T)\right) =\sup_{x\in D}\lambda_{\max}(K_x(\cos\theta)) =\sqrt{60+4\cos^2\theta}.$$

This is the support function of

$$\boxed{\overline{W(T)} =\left\{a+ib:\frac{a^2}{64}+\frac{b^2}{60}\le1\right\}.}$$

Its semiaxes are $8$ and $\sqrt{60}$, so its foci are

$$\boxed{-\sqrt{64-60}=-2,\qquad \sqrt{64-60}=2,}$$

neither of which belongs to $\{-1,0,1\}$.

More generally, let any $r>1$ be prescribed and put

$$R=r^2,\quad \alpha=4R,\quad \beta^2=\alpha^2-R,\quad I_R=[-1/\alpha,1/\alpha],$$ $$G_R(x)=4\beta^2\frac{1-x^2}{1-Rx^2},\qquad d_R=G_R',\qquad s_R=G_R-xG_R'.$$

One has

$$G_R''(x)= \frac{8\beta^2(R-1)(1+3Rx^2)}{(1-Rx^2)^3}>0.$$

For $t=|x|\le1/(4R)$, positivity $s_R>|d_R|$ follows from

$$1-2(R-1)t-(3R-1)t^2+Rt^4 >1-\frac12-\frac3{16R}>\frac5{16}.$$

Repeating the countable dense-fiber construction gives support function

$$h_R(\cos\theta)=\sqrt{\beta^2+R\cos^2\theta}.$$

The resulting elliptical disk has semiaxes $\alpha,\beta$, and therefore foci

$$\boxed{\{-\sqrt{\alpha^2-\beta^2},\sqrt{\alpha^2-\beta^2}\} =\{-r,r\}.}$$

Thus every prescribed focal distance $r>1$ furnishes a separable counterexample to the exact closure formulation of Conjecture 3.5.