Margulis diamond weight bound

Represent each coordinate in $[-m\mathrm{/}2,m\mathrm{/}2)$. Define $z\succ u$ when both absolute coordinates of $z$ are at least those of $u$, with one strict. Put $w(z,u)=5\mathrm{/}4$ if $z\succ u$, $4\mathrm{/}5$ if $u\succ z$, and $1$ otherwise; then $w(z,u)w(u,z)=1$. Squaring $\sqrt{w}a-b\mathrm{/}\sqrt{w}$ gives $2ab\le w{a}^2+w^{-1}{b}^2$. Apply this to each term of $Q$ and reindex the inverse-shear terms; a shear leaves its corresponding cosine coordinate fixed. The coefficient at $g(z{)}^2$ is bounded by $c_1(z)[w(z,T_{2}z)+w(z,T_2^{-1}z)]+c_2(z)[w(z,T_{1}z)+w(z,T_1^{-1}z)]$, where $c_j(z)=\mathrm{\mid }\cos (\pi z_j\mathrm{/}m)\mathrm{\mid }$.

Outside the open diamond $\mathrm{\mid }{z}_1\mathrm{\mid }+\mathrm{\mid }{z}_2\mathrm{\mid }<m\mathrm{/}2$, let $a=\pi \mathrm{\mid }{z}_1\mathrm{\mid }\mathrm{/}m$ and $b=\pi \mathrm{\mid }{z}_2\mathrm{\mid }\mathrm{/}m$. They lie in $[0,\pi \mathrm{/}2]$ with $a+b\ge \pi \mathrm{/}2$, so $\cos a+\cos b\le \cos a+\sin a\le \sqrt{2}$. Each weight is at most $5\mathrm{/}4$, giving coefficient at most $5\sqrt{2}\mathrm{/}2<73\mathrm{/}20$. This includes the diamond boundary and the centered-coordinate endpoints.

Inside the diamond and away from zero, sign changes and coordinate interchange permute the four shear neighbors and preserve their absolute-coordinate order. First suppose the absolute coordinates are $a>b>0$ with $a+b<m\mathrm{/}2$. The change $a\mapsto a-2b$ strictly decreases its absolute value. For $a+2b$, the centered absolute value is $\min (a+2b,m-a-2b)>a$, since $m>2a+2b$. For $b+2a$, both $b+2a$ and $m-b-2a$ exceed $b$ by the same strict inequality. For $b-2a$, both $2a-b$ and $m-2a+b$ exceed $b$, the latter since $2a<m$. Thus exactly three neighbors dominate and one is dominated, including when a coordinate wraps. The coefficient is at most $3(4\mathrm{/}5)+5\mathrm{/}4=73\mathrm{/}20$.

If $a=b>0$, then $a<m\mathrm{/}4$. Two neighbors preserve the pair of absolute coordinates, and the other two replace one coordinate by $\mathrm{\mid }3a<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>m\mathrm{\mid }>a$, because both $3a$ and $m-3a$ exceed $a$. If $a>0,b=0$, two neighbors fix the point and two change the zero coordinate to a nonzero centered residue of $2a$, since $0<2a<m$. In either case there are two weights $1$ and two weights $4\mathrm{/}5$, giving at most $18\mathrm{/}5<73\mathrm{/}20$. The zero point contributes nothing because $g(0)=0$; this also handles $m=1$.

Every point is covered by the preceding cases. Summing the coefficient bounds proves the claim for $Q$. The Fourier reduction gives $\mathrm{\mid }⟨f,Af⟩\mathrm{\mid }\le 2Q(\mathrm{\mid }\widehat{f}\mathrm{\mid })$ and Parseval gives $\sum \mathrm{\mid }\widehat{f}{\mathrm{\mid }}^2=\mathrm{\parallel }f{\mathrm{\parallel }}^2$, proving the stated consequence.