Let K be a Hilbert space, let X,Y be commuting positive contractions on K, and let P∈ B(K) be an orthogonal projection such that PXP and PYP commute. Let f:[0,1]^2→R be continuous and strictly convex. Suppose that P f(X,Y) P = P f(PXP,PYP) P. Does it follow that P reduces both X and Y? Equivalently, does it follow that PX=XP,qquad PY=YP?

Planar strictly convex hyperrigidity

The problem asks whether equality in the compressed functional calculus for a continuous strictly convex function forces the projection to reduce both commuting positive contractions. It is attributed to Boris Bilich and identified as Problem 2 in the CUHK-Shenzhen AI Math Problems database.

Known results

For every compact convex set $K\subset\mathbb{R}^2$ , the affine operator system $A(K)$ is hyperrigid in $C(\operatorname{ex}(K))$ .
In strictly convex graph cases, the associated operator system is completely order isomorphic to $\langle 1,t,f\rangle$ , extending earlier one-variable work of Brown.

June 2026 claimed proof

A repository claims that, for commuting positive contractions $X,Y$ , commuting compressions $A=PXP|_{\operatorname{ran}P}$ and $B=PYP|_{\operatorname{ran}P}$ , and continuous strictly convex $f:[0,1]^2\to\mathbb{R}$ , equality $Pf(X,Y)P|_{\operatorname{ran}P}=f(A,B)$ implies $PX=XP$ and $PY=YP$ . This would settle the stated problem affirmatively, but the claim is unverified.

Current status (as of June 2026): An affirmative solution is claimed, while independent verification and publication remain outstanding.

Planar strictly convex hyperrigidity

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