Capped DIIS coefficients have a unique minimum-norm least-squares selection

The vector $e_1=(1,0,\dots ,0)$ belongs to ${\mathcal{C}}_{s,C}$ because its entries sum to $1$ and its ${\mathrm{\ell }}^1$ norm is $1\le C$. Thus the coefficient set is nonempty. For $c,d\in {\mathbb{R}}^s$, each $\mathrm{\mid }{c}_i-d_i\mathrm{\mid }\le \mathrm{\parallel }c-d{\mathrm{\parallel }}_2$, and [F3] gives $\mid \mathrm{\mid }{c}_i\mathrm{\mid }-\mathrm{\mid }{d}_i\mathrm{\mid }\mid \le \mathrm{\mid }{c}_i-d_i\mathrm{\mid }$; hence both $c\mapsto {\sum }_i{c}_i$ and $c\mapsto {\sum }_i\mathrm{\mid }{c}_i\mathrm{\mid }$ are continuous. Their closed level sets show that ${\mathcal{C}}_{s,C}$ is closed. It is bounded because $\mathrm{\parallel }c{\mathrm{\parallel }}_1^2={\sum }_i{c}_i^2+2{\sum }_{i<j}\mathrm{\mid }{c}_i\mathrm{\mid }\mathrm{\mid }{c}_j\mathrm{\mid }\ge \mathrm{\parallel }c{\mathrm{\parallel }}_2^2$ and $\mathrm{\parallel }c{\mathrm{\parallel }}_1\le C$. Therefore [F2] makes ${\mathcal{C}}_{s,C}$ compact.

Put $T(c)={\sum }_i{c}_i{R}_i$. By the triangle inequality and homogeneity of the Frobenius norm, $\mathrm{\parallel }T(c)-T(d){\mathrm{\parallel }}_F\le {\sum }_i\mathrm{\mid }{c}_i-d_i\mathrm{\mid }\mathrm{\parallel }{R}_i{\mathrm{\parallel }}_F\le \mathrm{\parallel }c-d{\mathrm{\parallel }}_2{\sum }_i\mathrm{\parallel }{R}_i{\mathrm{\parallel }}_F$, so $T$ is continuous; the reverse triangle inequality then makes $c\mapsto J(c)=\mathrm{\parallel }T(c){\mathrm{\parallel }}_F^2$ continuous. By [F2], $J$ attains a minimum $m$ on the nonempty compact set ${\mathcal{C}}_{s,C}$. The set $M={\mathcal{C}}_{s,C}\cap J^{-1}(\{m\})$ is closed and bounded in ${\mathbb{R}}^s$, hence compact by [F2], and it is nonempty by attainment.

For $c,d\in {\mathcal{C}}_{s,C}$ and $0\le t\le 1$, the coefficients $e=(1-t)c+td$ satisfy ${\sum }_i{e}_i=(1-t){\sum }_i{c}_i+t{\sum }_i{d}_i=1$. By [F3], ${\sum }_i\mathrm{\mid }{e}_i\mathrm{\mid }\le (1-t){\sum }_i\mathrm{\mid }{c}_i\mathrm{\mid }+t{\sum }_i\mathrm{\mid }{d}_i\mathrm{\mid }\le C$, so ${\mathcal{C}}_{s,C}$ is convex. Expanding each complex matrix entry gives $\mathrm{\parallel }(1-t)A+tB{\mathrm{\parallel }}_F^2=(1-t)\mathrm{\parallel }A{\mathrm{\parallel }}_F^2+t\mathrm{\parallel }B{\mathrm{\parallel }}_F^2-t(1-t)\mathrm{\parallel }A-B{\mathrm{\parallel }}_F^2$ for real $t$. Applying this identity to $A=T(c)$ and $B=T(d)$ shows $J((1-t)c+td)\le (1-t)J(c)+tJ(d)=m$ when $c,d\in M$; since the combination is feasible and $m$ is the minimum, it also lies in $M$. Thus $M$ is convex. The function $c\mapsto \mathrm{\parallel }c{\mathrm{\parallel }}_2^2={\sum }_i{c}_i^2$ is continuous, so compactness of $M$ gives a minimizer $c_{\ast }\in M$.

If distinct $c,d\in M$ both had minimum Euclidean norm, their midpoint would lie in $M$ and satisfy ${\parallel \frac{c+d}{2}\parallel }_2^2=\frac{\mathrm{\parallel }c{\mathrm{\parallel }}_2^2+\mathrm{\parallel }d{\mathrm{\parallel }}_2^2}{2}-\frac{\mathrm{\parallel }c-d{\mathrm{\parallel }}_2^2}{4}<\frac{\mathrm{\parallel }c{\mathrm{\parallel }}_2^2+\mathrm{\parallel }d{\mathrm{\parallel }}_2^2}{2},$ contradicting minimality. Thus $c_{\ast }$ is the unique minimum-norm least-squares minimizer. No nonsingularity of the Gram matrix is needed for existence or tie-breaking.