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Janzer 2019 improved bounds extremal number subdivisions

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corollary_5: Bounds the extremal number of the one-subdivision of K_{a,b} by C_{a,b} n^{3/2-1/(4a-2)} for integers 2 at most a at most b, an exponent gap depending on the smaller side only.

theorem_3: Gives a gap below the three-halves exponent whose reciprocal grows linearly with the order of the fixed clique being subdivided.

theorem_4: Bounds the extremal number of the one-subdivision of K_{s+t-1} with the edges of a K_s removed by C_{s,t} n^{3/2-1/(4t-6)}, for integers s at least 1 and t at least 3.


Oliver Janzer, Improved bounds for the extremal number of subdivisions, Electronic Journal of Combinatorics 26(3) (2019), Paper P3.3, 6 pp. DOI: 10.37236/8262.

The copy read for this card is the six-page journal version. Its first page records submission on 24 October 2018, acceptance on 10 June 2019 and publication on 5 July 2019; printed and PDF page numbers agree. The five-page arXiv:1809.00468v1 manuscript is a different version and was not compared with it. The journal version prints "© The author. Released under the CC BY-ND license (International 4.0).", the Creative Commons Attribution-NoDerivatives 4.0 license.

Writing HtH_t for the one-subdivision of KtK_t, Theorem 3, on p. 2, proves

ex⁡(n,Ht)≤Ctn1+(t−2)/(2t−3)=Ctn3/2−1/(4t−6),t≥3.\operatorname{ex}(n,H_t) \le C_t n^{1+(t-2)/(2t-3)} =C_t n^{3/2-1/(4t-6)},\qquad t\ge3.

Here tt is fixed and CtC_t is independent of nn. This improves Conlon--Lee's [[extremal_graph_theory/conlon_2021_extremal_number_subdivisions/theorem_5_1|earlier explicit gap 6−t6^{-t}]]. The reciprocal of Janzer's exponent gap is 4t−64t-6, which grows linearly in tt; the gap itself is 1/(4t−6)1/(4t-6). The source's prose immediately before Theorem 3, answering Conlon--Lee's request for a gap δt\delta_t with 1/δt1/\delta_t bounded by a polynomial in tt, speaks of "a linear δt\delta_t"; the displayed theorem makes the dependence exact: it is 1/δt=4t−61/\delta_t=4t-6 that is linear. The source notes tightness when t=3t=3, since H3=C6H_3=C_6 and ex⁡(n,C6)=Θ(n4/3)\operatorname{ex}(n,C_6)=\Theta(n^{4/3}); it does not assert sharpness for every tt.

The same p. 2 gives Theorem 4 for the one-subdivision Ls,t′L'_{s,t} of Ls,t=Ks+t−1∖E(Ks)L_{s,t}=K_{s+t-1}\setminus E(K_s), with fixed s≥1s\ge1, t≥3t\ge3 and the same exponent gap 1/(4t−6)1/(4t-6). Taking s=1s=1 gives Theorem 3. Taking s=bs=b and t=a+1t=a+1 gives Corollary 5: for integers 2≤a≤b2\le a\le b the one-subdivision Ha,bH_{a,b} of Ka,bK_{a,b} has ex⁡(n,Ha,b)≤Ca,bn3/2−1/(4a−2)\operatorname{ex}(n,H_{a,b})\le C_{a,b}n^{3/2-1/(4a-2)}. The paper contrasts this with Conlon--Lee's bound Cn3/2−1/(12b)Cn^{3/2-1/(12b)}, which it calls weak when bb is much larger than aa. These further results are context, not a complete local proof reconstruction.

For Problem 1021, the required graph GkG_k is exactly HkH_k, so one may take ck=1/(4k−6)>0c_k=1/(4k-6)>0. The subdivision definition on pp. 1--2 replaces every edge by a path of length two, with a distinct internal vertex for each edge. This proves the requested existence of a power improvement for each fixed kk.

Reading and proof scope. All six pages were read for identity, definitions, statements, the reduction and the structure of the proof. Section 2, pp. 3--5, proves Theorem 4 using its Theorem 7; Lemmas 6 and 8 cite Conlon--Lee, Lemmas 2.3 and 2.4, and Lemma 10 and Corollary 11 (p. 4) carry the light-edge count. The argument was not reconstructed line by line. The extraction records source statements and a proof pointer, not complete reconstruction, independent proof acceptance or formal verification. The publisher and arXiv records were checked. No code or Lean build was run.

Source: EJC publication record.

Bears on. #1021: Theorem 3 (p. 2) is the problem's bound for Gk=HkG_k=H_k with ck=1/(4k−6)c_k=1/(4k-6), and Theorem 4 (p. 2) contains it as the case s=1s=1. Corollary 5 (p. 2) bears on no problem page of this corpus.

No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.