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With $r$ balanced cyclically ordered vertex classes and $2\le \ell \le r+1$, include an $\ell$-set if some start has at least $k+1$ vertices in its first $k$ classes for every $1\le k\le \ell -1$. The cycle lemma implies every $(r+1)$-set contains an edge. Its asymptotic density is $((\ell -1)/r{)}^{\ell -1}$; taking the hypergraph complement gives a lower bound for the Turán density of a complete uniform hypergraph.

Extremal hypergraph theory studies the largest or smallest possible size of a hypergraph under prescribed forbidden-subhypergraph or covering conditions. It generalizes extremal graph theory. The Turán density describes the asymptotic maximum edge proportion for a fixed uniform forbidden hypergraph.

Normalized hypergraph extremal numbers decrease under vertex sampling and converge to Turán density. At a fixed sample size where the extremal density is close to that limit, excess density forces a positive fraction of samples to contain $H$. Counting incidences of copies and samples proves the displayed lower bound. Shrinking $\delta$ also handles the finitely many small orders when the required count is rounded down.

For a fixed uniform hypergraph $H$ with $h$ vertices, uniformity $\ell$, and at least one edge, define its Turán density bywhere $\mathrm{ex}(n,H)$ is the maximum edge count of an $H$-free hypergraph. The limit exists: averaging the edge count over all $m$-vertex subsets of an extremal $n$-vertex hypergraph, with $n\ge m\ge \ell$, shows that the displayed normalized extremal numbers are nonincreasing. Thus their limit is their infimum.

For hypergraph supersaturation by sampling, choose $m\ge h$ so that $\mathrm{ex}(m,H)/\left(\genfrac{}{}{0ex}{}{m}{\ell }\right)\le \pi (H)+ϵ/2$. The expected edge count in a uniform $m$-vertex subset of $G$ is at least $(\pi (H)+ϵ)\left(\genfrac{}{}{0ex}{}{m}{\ell }\right)$. A subset containing no copy of $H$ has at most $(\pi (H)+ϵ/2)\left(\genfrac{}{}{0ex}{}{m}{\ell }\right)$ edges, while any subset has at most $\left(\genfrac{}{}{0ex}{}{m}{\ell }\right)$. Hence at least an $ϵ/2$ fraction of the subsets contain $H$. Every particular copy of $H$ lies in exactly $\left(\genfrac{}{}{0ex}{}{n-h}{m-h}\right)$ such subsets, so their number is at leastfor $n\ge N=\max (m,2h)$. TakeFor $n<N$, the requested integer lower bound is zero; for $n\ge N$, the calculation proves it. This positive $\delta$ depends only on $H$ and $ϵ$. If the density hypothesis is impossible there is nothing to prove. For an edgeless $H$, every $h$-set is already a copy and the conclusion follows directly.

For the edge-triangle objective, let $N(v)$ be the open neighbourhood of $v$ and define its local scoreIt counts the contribution to $k_2-c{k}_3$ of edges and triangles in a graph containing $v$. If $u,v$ are nonadjacent, making $u$ a false twin of $v$ changes the objective by $s(v)-s(u)$, since no edge or triangle in a graph uses both. More generally, partition vertices into false-twin classes, meaning equal open neighbourhoods. If distinct classes $U,V$ have no edges between them, making every vertex of $U$ a twin of a representative of $V$ changes the objective by $|U|(s(v)-s(u))$. Choose the direction with nonnegative change.

Among graphs maximizing the objective, choose one maximizing the sum of squared false-twin class sizes. The operation cannot split any old false-twin class and merges $U,V$, so it would strictly increase that secondary quantity. Therefore no two distinct classes can be nonadjacent. An objective maximizer is a complete multipartite graph. This edge-triangle symmetrization works for every real $c$, with no assumption on its sign.

For positive part sizes $a_1,\dots ,a_q$, the correct multipartite formula isThe first sum in the printed intermediate formula has inconsistent indices; the edge count is the pairwise-product sum above. Fix two part sizes with sum $A$ and let $B$ be the sum of all other sizes. All terms varying with the pair have the formChoose a maximizing partition with the fewest nonempty parts. If $1-cB\le 0$, merging those two parts cannot decrease $F$ and reduces the number of parts, a contradiction. Therefore this coefficient is positive for every pair. If two integer sizes differ by at least two, moving one vertex from the larger to the smaller increases their product and hence $F$, again a contradiction. All sizes differ by at most one, so a maximizer is a Turan graph. This is balancing a multipartite edge-triangle objective.

Next maximize the same polynomial on the compact simplex of nonnegative real sizes summing to $n$, starting with at least two possible parts. Choose a maximizer with the smallest positive support. The merging argument still applies, and now varying two positive unequal sizes towards equality improves the product. Thus every positive size is $n/q$ for some integer $q$. A two-part choice has value $n^2/4>0$ when $n>0$, so $q\ge 2$. The optimum is attained at rational sizes, and thereforeOne may take the initial simplex to have $\max (n,2)$ coordinates, so it includes every multipartite graph obtained above; zero coordinates are discarded.

Set $c=9/(4n)$. Dividing the right side by $n^2$ givesbecause $q$ is an integer at least two. Rearranging provesEquality in the continuous calculation occurs at two or three equal parts, explaining the triangle support line between bipartite and tripartite Turan graphs.

There is a genuine error in the printed threshold. With $\ell =3$, its only nontrivial requirement is that some two consecutive classes contain at least one vertex of the triple. Every triple meets some class, so every triple qualifies. For example, when $r=4$, the printed construction is the complete 3-uniform hypergraph, of edge density $1$, whereas the claimed density is $(2/4{)}^2=1/4$. Thus the stated edge-count assertion is false even in a nondegenerate parameter range.

The intended cyclic Turán covering construction, which gives the claimed density and the stated lower bound, uses threshold $k+1$, in the natural range $2\le \ell \le r+1$. We solve this repaired version explicitly. Denote it by $G^{+}$ to keep it separate from the literal printed construction.

A cyclic prefix argument gives the covering property. We first prove the relevant cycle lemma. If integers $b_1,\dots ,b_m$ satisfy ${\sum }_i{b}_i=1$, exactly one cyclic starting position has every nonempty partial sum strictly positive. Existence follows by starting after the last minimum among the proper partial sums $0,b_1,\dots ,b_1+\cdots +b_{m-1}$. Subsequent nonwrapped sums are strictly above that minimum, and wrapped sums add the total $1$. For uniqueness, if two starts split the cycle into arcs with sums $x$ and $1-x$, both positive, their integrality would require $x\ge 1$ and $1-x\ge 1$, a contradiction.

Take any $r+1$ vertices and let $a_i$ count them in class $V_i$. The increments $b_i=a_i-1$ sum to one. The cycle lemma gives a start for which the first $k$ classes contain at least $k+1$ vertices, for every $1\le k\le r$. Take the first $\ell$ vertices encountered from that start in class order. The first $\ell -1$ classes already contain at least $\ell$ vertices; the chosen vertices therefore satisfy all the repaired prefix inequalities. HenceEvery repaired edge also satisfies the weaker printed inequalities. Thus this argument proves the covering assertion for the printed construction too, although it cannot repair its false edge count.

Count the repaired edges by labelled class assignments. Put $m=\ell -1$. For a fixed cyclic start, an edge satisfying the repaired inequalities has all $\ell$ vertices in its first $m$ classes. Assign $\ell$ labelled vertices to these $m$ positions. There are $m^{\ell }$ assignments in total. Their occupancy counts $a_1,\dots ,a_m$ satisfy ${\sum }_i(a_i-1)=1$; the cycle lemma says exactly one of their $m$ cyclic starts has the required prefixes. Rotational symmetry therefore gives exactly $m^{\ell -1}$ good assignments for a specified start.

We must check that summing over the $r$ starts does not count an edge twice. Suppose two starts cut the circle into arcs of lengths $h$ and $r-h$. If an arc has length at most $m$, goodness of its starting point forces that arc to contain at least its length plus one vertices. If its length is greater than $m$, it contains all $\ell$ vertices, while the other good start demands vertices in the disjoint complementary arc, impossible. If both lengths are at most $m$, their combined demand is at least $r+2>\ell$ vertices, again impossible. Thus a repaired edge has a unique good start.

There are consequently $r{m}^{\ell -1}$ good assignments of class labels to $\ell$ labelled vertices. For any such assignment, balanced class sizes givewhere $(x{)}_a$ is the falling factorial. Divide the total count of ordered distinct vertices by $\ell !$. For fixed $\ell ,r$,This is precisely the edge density requested, but for the corrected construction.

Pass from covering to a forbidden-hypergraph clique construction. Use $r-1$ balanced classes in $G^{+}$ and take its hypergraph complement. Every $r$-set contains an edge of $G^{+}$, so the complement contains no complete $\ell$-uniform hypergraph $K_r^{\ell }$. The preceding count givesHere Turán density means $\pi (F)={\lim }_{n\to \infty }\mathrm{ex}(n,F)/\left(\genfrac{}{}{0ex}{}{n}{\ell }\right)$ for a fixed $\ell$-uniform forbidden hypergraph; the limit exists because these normalized extremal densities are nonincreasing under averaging over smaller vertex subsets.

A corresponding classical upper bound is the de Caen bound for hypergraph Turán density:Equivalently, an $\ell$-uniform covering hypergraph in which every $r$-set contains an edge has asymptotic edge density at least $1/\left(\genfrac{}{}{0ex}{}{r-1}{\ell -1}\right)$. Its proof recursively counts complete subhypergraphs by their one-vertex extensions. Double counting the extensions and applying the Cauchy-Schwarz inequality to common extension neighbourhoods gives a lower bound for the ratio of successive hypergraph clique counts. If the missing-edge density were smaller than $1/\left(\genfrac{}{}{0ex}{}{r-1}{\ell -1}\right)$, those bounds would force a positive number of $r$-vertex hypergraph cliques, a contradiction. Applying the resulting covering bound to the complement of a $K_r^{\ell }$-free hypergraph gives the displayed upper bound. This incidence argument supplies the extra factor beyond the simpler direct count of one covering edge in each $r$-set, which alone yields only $1-1/\left(\genfrac{}{}{0ex}{}{r}{\ell }\right)$.

The lower bound is therefore valid via $G^{+}$, while the printed enumeration cannot be proved with threshold $k-1$. Both the counterexample and the necessary repair are part of the solution, rather than a silent change of the definition.

The general upper bound is recorded, for example, in Turán densities of some hypergraphs related to complete uniform hypergraphs.