Critical PSD phases: quadratic tests and small-alpha mass bound

Differentiating the displayed finite sum in an angular chart gives ${\mathrm{\partial }}_{y}E={\sum }_x{A}_{xy}\sin ({\theta }_y-{\theta }_x)$ and, for real $h$, $h^{T}Bh=\frac{1}{2}{\sum }_{x,y}{A}_{xy}\cos ({\theta }_x-{\theta }_y)(h_x-h_y{)}^2$. The latter formula gives a real symmetric matrix $B$. For $z=u+iv$ with real $u,v$, symmetry cancels the cross terms, so $z^{\ast }Bz=u^{T}Bu+v^{T}Bv\ge 0$ and $z^{\ast }Bz=\frac{1}{2}{\sum }_{x,y}{A}_{xy}\cos ({\theta }_x-{\theta }_y)\mathrm{\mid }{z}_x-z_y{\mathrm{\mid }}^2$. This justifies complex-vector testing from the assumed real positive semidefiniteness. Criticality makes the imaginary part of $e^{-i{\theta }_y}(A{e}^{i\theta }{)}_y$ zero. Testing $B$ at a real coordinate vector gives ${\sum }_{x\mathrm{\ne }y}{A}_{xy}\cos ({\theta }_x-{\theta }_y)\ge 0$; adding the nonnegative loop entry proves $r_y\ge 0$. The same complex number has real part $r_y$, proving the stated phasor identity.

Write $f_x=e^{i{\theta }_x}$, $g_x=e^{2i{\theta }_x}$, ${\rho }_2=n^{-1}{\sum }_x{g}_x$, and $M=A-(d\mathrm{/}n)J$. The unit-modulus phasors give $\mathrm{\parallel }f{\mathrm{\parallel }}^2=\mathrm{\parallel }g{\mathrm{\parallel }}^2=\mathrm{\parallel }\mathbf{1}{\mathrm{\parallel }}^2=n$, and $v^{\ast }Jv=\mathrm{\mid }{\sum }_x{v}_x{\mathrm{\mid }}^2$. By [F3] and $\mathrm{\parallel }M\mathrm{\parallel }\le ad$, $\mathrm{\mid }{f}^{\ast }Af-dn{\rho }^2\mathrm{\mid }\le adn$, $g^{\ast }Ag\ge dn\mathrm{\mid }{\rho }_2{\mathrm{\mid }}^2-adn$ and ${\mathbf{1}}^{T}A\mathbf{1}\ge dn-adn$. Symmetry of $A$ gives $f^{\ast }Af={\sum }_{x,y}{A}_{xy}\cos ({\theta }_x-{\theta }_y)$ and the analogous identity for $g$. Testing the Hessian with $z=f$ using step 1.1 and $\mathrm{\mid }{f}_x-f_y{\mathrm{\mid }}^2=2(1-\cos ({\theta }_x-{\theta }_y))$ gives ${\sum }_{x,y}{A}_{xy}\cos ({\theta }_x-{\theta }_y)\ge {\sum }_{x,y}{A}_{xy}{\cos }^2({\theta }_x-{\theta }_y)$. The phasor addition formula yields ${\cos }^{2}u=(1+\cos 2u)\mathrm{/}2$, so the right side equals $({\mathbf{1}}^{T}A\mathbf{1}+g^{\ast }Ag)\mathrm{/}2$. Thus $dn{\rho }^2+adn\ge dn(1+\mathrm{\mid }{\rho }_2{\mathrm{\mid }}^2)\mathrm{/}2-adn$. Divide by $dn>0$ and discard $\mathrm{\mid }{\rho }_2{\mathrm{\mid }}^2\mathrm{/}2\ge 0$ to obtain ${\rho }^2\ge 1\mathrm{/}2-2a>7\mathrm{/}16$.

Set $W=Mf$. The phasor identity from step 1.1 gives $e^{-i{\theta }_y}{W}_y=r_y-d\rho e^{-i{\theta }_y}$. Its imaginary part is $d\rho \sin {\theta }_y$. If $\mathrm{\mid }{\theta }_y\mathrm{\mid }\ge \pi \mathrm{/}2$, then $\mathrm{\mid }{\theta }_y\mathrm{\mid }\in [\pi \mathrm{/}2,\pi ]$ by the chosen representatives. The phasor identities give $\cos (-u)=\cos u$ for real $u$, so the decrease of cosine on $[0,\pi ]$ and [F4] imply $\cos {\theta }_y=\cos \mathrm{\mid }{\theta }_y\mathrm{\mid }\le \cos (\pi \mathrm{/}2)=0$. Thus its real part is at least $-d\rho \cos {\theta }_y\ge 0$, since $r_y\ge 0$. Squaring the real and imaginary parts in this region gives $\mathrm{\mid }{W}_y{\mathrm{\mid }}^2\ge (d\rho {)}^2({\cos }^2{\theta }_y+{\sin }^2{\theta }_y)=(d\rho {)}^2$. For $\mathrm{\mid }{\theta }_y\mathrm{\mid }\le \pi \mathrm{/}2$, the imaginary part alone gives $\mathrm{\mid }{W}_y{\mathrm{\mid }}^2\ge (d\rho {)}^2{\sin }^2{\theta }_y$. Consequently $\mathrm{\mid }{W}_y{\mathrm{\mid }}^2\ge (d\rho {)}^{2}H({\theta }_y)$ in both regions. By [F3], $\mathrm{\parallel }W{\mathrm{\parallel }}^2\le \mathrm{\parallel }M{\mathrm{\parallel }}^2\mathrm{\parallel }f{\mathrm{\parallel }}^2\le a^2{d}^{2}n$. Summing the coordinate bounds and using ${\rho }^2>7\mathrm{/}16$ from step 2.1 gives ${\sum }_{x}H({\theta }_x)\le a^{2}n\mathrm{/}{\rho }^2\le (16\mathrm{/}7)a^{2}n$. For $a>0$, $16\mathrm{/}7<5\mathrm{/}2$ gives the final strict bound. No step uses the point being a local minimum.