Conjugate phases norm a three-atom function

Direct multiplication gives ${\omega }^2=(-1-i\sqrt{3})\mathrm{/}2$, ${\omega }^3=1$, $1+\omega +{\omega }^2=0$ and $\mathrm{\mid }\omega \mathrm{\mid }=1$. Thus all three values of $g$ have modulus one, so F3 gives $\mathrm{\parallel }g{\mathrm{\parallel }}_1=(1+1+1)\mathrm{/}3=1$. The measure is a probability measure: disjoint sets just partition three atoms, and their weighted cardinalities add to one on $X$.

For a real test $t=(t_0,t_1,t_2)\in [-1,1{]}^3$, set $w_{\sigma }={\prod }_{j=0}^2(1+{\sigma }_j{t}_j)\mathrm{/}2$ for $\sigma \in \{-1,1{\}}^3$. Each weight is nonnegative, summing the product over all signs gives ${\sum }_{\sigma }{w}_{\sigma }=1$, and summing ${\sigma }_j{w}_{\sigma }$ gives $t_j$. Thus $t={\sum }_{\sigma }{w}_{\sigma }\sigma$. For the linear expression $L(t)=(t_0+\omega t_1+{\omega }^2{t}_2)\mathrm{/}3=\int gt$, F2 implies $\mathrm{\mid }L(t)\mathrm{\mid }\le {\sum }_{\sigma }{w}_{\sigma }\mathrm{\mid }L(\sigma )\mathrm{\mid }\le {\max }_{\sigma }\mathrm{\mid }L(\sigma )\mathrm{\mid }$.

The conjugated test has modulus one on each atom, so its essential infinity norm is one. Coordinatewise $gs=(1,1,1)$, and F3–F4 give $\int gs=1$. Every complex test of norm at most one has integral modulus at most $\mathrm{\parallel }g{\mathrm{\parallel }}_1=1$ by F1; all tests here have finite-measure support. Hence this test attains the full complex supremum.

At the two equal-sign vertices, $L(\sigma )=0$ because $1+\omega +{\omega }^2=0$. Every other sign vertex has one exceptional sign, at some coordinate $j$, so $L(\sigma )=\pm 2{\omega }^j\mathrm{/}3$ and has modulus $2\mathrm{/}3$. For instance $t=(1,-1,-1)$ gives $L(t)=2\mathrm{/}3$. This proves both the real upper bound and its attainment. Finally F4 gives $\int g^2=(1+{\omega }^2+{\omega }^4)\mathrm{/}3=(1+{\omega }^2+\omega )\mathrm{/}3=0$, proving the claimed failure of the unconjugated phase.