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All graphs here are finite and simple. For the strong graph product $G⊠H$, two distinct graph vertex pairs are adjacent if in each coordinate the entries are equal or adjacent. Use $G^{⊠m}$ for its strong graph power; the exam's $G^m$ means this product power, not the distance-based graph power. The Shannon capacity of a graph iswhere $\alpha$ is the independence number. Products of independent sets are independent, so $a_{r+s}\ge a_r{a}_s$ for $a_m=\alpha (G^{⊠m})$. To see why the limit exists, fix $r$ and write $m=qr+s$, $0\le s<r$. Then $a_m\ge a_r^q$, using $a_s\ge 1$ and $a_0=1$, so ${\mathrm{liminf}}_m(\log a_m)/m\ge (\log a_r)/r$. Take the supremum over $r$; the reverse upper bound holds termwise. This proves the required form of Fekete lemma. For the empty graph on no graph vertices one sets graph capacity zero separately.

Associativity of the strong product givesAn orthonormal representation of a graph assigns a real unit vector $u_v$ to each graph vertex, with $u_v\perp u_w$ whenever distinct graph vertices $v,w$ are nonadjacent. Adjacent graph vertices need not receive orthogonal vectors. A handle is a unit vector $h$. The handle formulation of the Lovász theta function iswith a zero denominator giving infinity. Representations may use any finite-dimensional Euclidean space. This is the usual Lovász number in its orthonormal-representation formulation.

For an independent set $I$, the vectors $u_v$, $v\in I$, are mutually orthonormal. Directly,If $T={\max }_v|h\cdot u_v{|}^{-2}$ is finite, every summand is at least $1/T$, giving $|I|\le T$. For a strong power, use the vectors and handleTheir norms are one. Nonadjacency of two distinct words means a coordinate has distinct nonadjacent entries, so the corresponding tensor inner product has a zero factor. The handle inner products satisfyApplying the preceding calculation gives $\alpha (G^{⊠m})\le T^m$. Taking roots, then the infimum over representations and handles, provesThis is the tensor-product capacity bound from an orthonormal representation, and the displayed projection calculation supplies Bessel inequality explicitly.

For $C_5$, label graph vertices by $\mathbb{Z}/5\mathbb{Z}$, with edges at differences $\pm 1$. The five wordsare independent in its strong square: a difference $\pm 1$ in the first coordinate produces a nonedge difference $\pm 2$ in the second, while a first-coordinate difference $\pm 2$ is already a nonedge. Hence $c(C_5)\ge \sqrt{5}$.

For the matching upper bound set $\rho =1/\sqrt{5}$, $h=(0,0,1)$, andEach vector is unit. For a nonedge, the angle difference is $\pm 4\pi /5$, andbecause $\cos (4\pi /5)=-(1+\sqrt{5})/4$ and $\rho =1/\sqrt{5}$. Thus these vectors form an orthonormal representation. The handle has $|h\cdot u_j{|}^2=\rho$, giving $\vartheta (C_5)\le 1/\rho =\sqrt{5}$. Together the two bounds prove the pentagon capacity code result