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For spanning slab normals, apply the maximum of finitely many Rademacher linear forms bound to $u_i=(V{\Lambda }^{1/2}{)}^T{a}_i$, whose squared norms are $a_i^{T}X{a}_i\le 1$. Some sign vector has scaling denominator squared at most $2\log (2m)$. Rademacher rounding for a semidefinite relaxation then produces a feasible vector of squared norm at least the displayed fraction of the SDP optimum. This is an existence guarantee from the probabilistic method.

Set $u_i=(V{\Lambda }^{1/2}{)}^T{a}_i$. SDP feasibility impliesChoose $\alpha =\sqrt{2\log (2m)}$. For the uniform sign vector, whose coordinates are independent Rademacher random variables, the supplied bound for the maximum of finitely many Rademacher linear forms givesA positive-probability event in this finite space contains at least one sign vector. ThusThe strict inequality in the supplied probability estimate matters: at the chosen threshold its right-hand side is zero.

For that sign vector, part (b) gives a feasible original point satisfying $\Vert x(\xi ){\Vert }_2^2\ge p_{\mathrm{SDP}}^{\ast }/[2\log (2m)]$. Taking the best original objective proves the unheaded concluding request as well:This is the logarithmic approximation bound for slab-constrained quadratic maximization, obtained by the probabilistic method. It is a finite-value rounding guarantee under the spanning/attainment conditions explained above.

A $k$-colouring assigns one of $k$ colours to every vertex so that adjacent vertices have different colours. The chromatic number $\chi (G)$ is the least such $k$. An independent set contains no adjacent pair, and the independence number $\alpha (G)$ is the maximum size of such a set.

For a concrete first example, start with the five-cycle $C_5$ and apply the Mycielski construction $r-2$ times. Each application raises the chromatic number by one and preserves triangle-freeness. The resulting graph has chromatic number $r+1>r$ and contains no triangle, hence no complete graph $K_r$ for $r\ge 3$.

We next prove the stronger high-girth assertion by the probabilistic method. For a large integer $n$, takeIf $X$ counts cycles of lengths $3,\dots ,g$, thenBy the first moment method, with probability tending to one $X<n/2$.

Let $m=\lfloor n/(2k)\rfloor$. The expected number of independent sets of size $m$ is at mostbecause $p{m}^2=\Theta (n^{1+1/(2g)})$ dominates $m\log (en/m)=O(n)$. Hence with positive probability $G$ has fewer than $n/2$ short cycles and $\alpha (G)<m$.

Delete one vertex from each cycle of length at most $g$. The remaining graph $H$ has no such cycle, has at least $n/2$ vertices, and still has $\alpha (H)<n/(2k)$. Since each colour class is independent,This proves that there are graphs of arbitrarily high girth and chromatic number.

For the final claim, begin instead withThe expected number $T$ of triangles isso with probability tending to one $T<n$. Put $m=\lceil n^{0.7}\rceil$. The expected number of independent $m$-sets is bounded bybecause the positive term is $O(n^{0.7}\log n)$ whereas the negative term has order $n^{0.71}$.

Thus some such $G$ has fewer than $n$ triangles and no independent set of size $m$. Delete one vertex from every triangle; more than $n$ vertices remain and the graph is triangle-free. Take any induced subgraph on exactly $n$ vertices. Vertex deletion cannot increase the independence number, so the resulting graph satisfiesThis is the triangle-free graph with sub-power independence number construction.