Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 111 / 2

Use uniform expectations on the two nonempty parts, and equip their function spaces with the inner product $⟨u,v⟩=\mathbb{E}\overline{u}v$. Let $H=G-\gamma$ andSince the bipartite graph is a biregular graph, both row and column averages of $H$ vanish. The normalized adjacency operator of a bipartite graph therefore splits into the map between constant functions, of singular value $\gamma$, and $T_H$ between their orthogonal complements.

There is no singular value larger than $\gamma$. Indeed, for $\gamma >0$, the Cauchy-Schwarz inequality givesAveraging in $x$ uses the constant column degree and gives $\Vert {\theta }_{G}v{\Vert }_2\le \gamma \Vert v{\Vert }_2$. For $\gamma =0$ every operator is zero. Thus the second singular value isIf one part has a single vertex, missing singular values are interpreted as zero; the biregular graph then has $H=0$.

The discrepancy in (i) is exactly $\mathbb{E}H{1}_A{1}_B$. Its supremum over $A,B$ is the cut norm $\delta =\Vert H{\Vert }_{\mathrm{cut}}$. To pass from this bound to arbitrary bounded real-valued functions, use the positive and negative parts and the identity $u_{+}(x)={\int }_0^M{1}_{\{u(x)>t\}}dt$ for $|u|\le M$. Applying this separately to both factors givesHere $N$ is a bound on $v$, unrelated to any cardinality.

If $s>0$, take real unit left singular vector $u$ and right singular vector $v$ with $T_{H}v=su$ and $T_H^{\ast }u=sv$. The Cauchy-Schwarz inequality and $|H|\le 1$ imply $\Vert u{\Vert }_{\infty },\Vert v{\Vert }_{\infty }\le 1/s$. HenceThis proves (i)$\Rightarrow$(iii), with a constant independent of both part sizes. The case $s=0$ satisfies the same conclusion immediately.

Let ${\sigma }_1=\gamma ,{\sigma }_2,\dots$ be the singular values of the normalized adjacency operator of a bipartite graph, with the uniform inner products used above. Its ordinary matrix in an orthonormal basis on each side is $G/\sqrt{|X||Y|}$. Expanding the trace of $({\theta }_G{\theta }_G^{\ast }{)}^2$ proves the box norm singular-value identityTherefore (ii) immediately implieswhich is (iii).

Conversely, the square of the Hilbert-Schmidt norm of this operator issince a graph indicator function takes values zero and one. Under (iii), every ${\sigma }_j$ with $j\ge 2$ is at most $c_3$, soThus (iii)$\Rightarrow$(ii) with $\boxed{c_2=c_3^2/4}$. This bound also covers the empty and complete bipartite graphs.

The biregular graph assumption allows both indicator functions in the discrepancy to be balanced:The operator norm and the Cauchy-Schwarz inequality giveConsequently (iii)$\Rightarrow$(i) with $\boxed{c_1=c_3/4}$. Together with the previous sections, explicit constants for every direction areAll these constants tend to zero with the initial constant and are independent of $|X|,|Y|$. This is the required equivalence for a quasirandom bipartite graph.