Kovari-Sos-Turan Theorem - proof












0












$begingroup$


recently I'm facing problems while trying to prove the following theorem:



Theorem: Let Ka,b denote the complete bipartite graph with a and b vertices in its color-classes. Then:
$$ex(n,K_{a,b}) leq frac{1}{2} sqrt[a]{b-1} cdot n^{2-(1/a)} + frac{a-1}{2}n.$$



I have only the sketch of the proof, which is the following one:



Sketch of the proof: The number of a-stars $K_{a,1}$ in a graph
$G_{n}$ is $sum binom{d_{i}}{a}$, where d1, . . . , dn are the degrees in $G_{n}$. If $G_{n}$ contains no $K_{a,b}$ then at most b −1 of these a-stars can share the same set of endpoints. We obtain
$$sum binom{d_{i}}{a} = text{the number of a−stars} ≤ (b − 1)binom{n}{a}$$



Extending $binom{n}{a}$ to all $x>0$ by



$binom{x}{a} = x(x-1)...(x-a+1)$ for $x>= a-1$ and $0$ for $x < a-1$



we have a convex function. Then Jensen’s Inequality implies that, the left
hand side in exponated inequality above is at least $binom{e(G)/n}{a}$, and the result follows by an easy
calculation.
End of sketch.



I understand the beginning, the proof of convexity is also obvious for me. But I can't follow how the Jensen's Inequality was used here, and how further calculations gives us final result.



Maybe somebody will be interested in the topic? Thank you in advance for the discussion.






graph-theory proof-explanation jensen-inequality







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is $ex(cdot,cdot)$?
    $endgroup$
    – Mike
    Jan 9 at 21:31






  • 1




    $begingroup$
    Well, ex(n, H) is the Turan number for H, i.e. the maximum number of edges of an H-free graph on n vertices.
    $endgroup$
    – Olga
    Jan 10 at 9:09
















0












$begingroup$


recently I'm facing problems while trying to prove the following theorem:



Theorem: Let Ka,b denote the complete bipartite graph with a and b vertices in its color-classes. Then:
$$ex(n,K_{a,b}) leq frac{1}{2} sqrt[a]{b-1} cdot n^{2-(1/a)} + frac{a-1}{2}n.$$



I have only the sketch of the proof, which is the following one:



Sketch of the proof: The number of a-stars $K_{a,1}$ in a graph
$G_{n}$ is $sum binom{d_{i}}{a}$, where d1, . . . , dn are the degrees in $G_{n}$. If $G_{n}$ contains no $K_{a,b}$ then at most b −1 of these a-stars can share the same set of endpoints. We obtain
$$sum binom{d_{i}}{a} = text{the number of a−stars} ≤ (b − 1)binom{n}{a}$$



Extending $binom{n}{a}$ to all $x>0$ by



$binom{x}{a} = x(x-1)...(x-a+1)$ for $x>= a-1$ and $0$ for $x < a-1$



we have a convex function. Then Jensen’s Inequality implies that, the left
hand side in exponated inequality above is at least $binom{e(G)/n}{a}$, and the result follows by an easy
calculation.
End of sketch.



I understand the beginning, the proof of convexity is also obvious for me. But I can't follow how the Jensen's Inequality was used here, and how further calculations gives us final result.



Maybe somebody will be interested in the topic? Thank you in advance for the discussion.






graph-theory proof-explanation jensen-inequality







share|cite|improve this question









$endgroup$












  • $begingroup$
    What is $ex(cdot,cdot)$?
    $endgroup$
    – Mike
    Jan 9 at 21:31






  • 1




    $begingroup$
    Well, ex(n, H) is the Turan number for H, i.e. the maximum number of edges of an H-free graph on n vertices.
    $endgroup$
    – Olga
    Jan 10 at 9:09














0












0








0





$begingroup$


recently I'm facing problems while trying to prove the following theorem:



Theorem: Let Ka,b denote the complete bipartite graph with a and b vertices in its color-classes. Then:
$$ex(n,K_{a,b}) leq frac{1}{2} sqrt[a]{b-1} cdot n^{2-(1/a)} + frac{a-1}{2}n.$$



I have only the sketch of the proof, which is the following one:



Sketch of the proof: The number of a-stars $K_{a,1}$ in a graph
$G_{n}$ is $sum binom{d_{i}}{a}$, where d1, . . . , dn are the degrees in $G_{n}$. If $G_{n}$ contains no $K_{a,b}$ then at most b −1 of these a-stars can share the same set of endpoints. We obtain
$$sum binom{d_{i}}{a} = text{the number of a−stars} ≤ (b − 1)binom{n}{a}$$



Extending $binom{n}{a}$ to all $x>0$ by



$binom{x}{a} = x(x-1)...(x-a+1)$ for $x>= a-1$ and $0$ for $x < a-1$



we have a convex function. Then Jensen’s Inequality implies that, the left
hand side in exponated inequality above is at least $binom{e(G)/n}{a}$, and the result follows by an easy
calculation.
End of sketch.



I understand the beginning, the proof of convexity is also obvious for me. But I can't follow how the Jensen's Inequality was used here, and how further calculations gives us final result.



Maybe somebody will be interested in the topic? Thank you in advance for the discussion.






graph-theory proof-explanation jensen-inequality







share|cite|improve this question









$endgroup$




recently I'm facing problems while trying to prove the following theorem:



Theorem: Let Ka,b denote the complete bipartite graph with a and b vertices in its color-classes. Then:
$$ex(n,K_{a,b}) leq frac{1}{2} sqrt[a]{b-1} cdot n^{2-(1/a)} + frac{a-1}{2}n.$$



I have only the sketch of the proof, which is the following one:



Sketch of the proof: The number of a-stars $K_{a,1}$ in a graph
$G_{n}$ is $sum binom{d_{i}}{a}$, where d1, . . . , dn are the degrees in $G_{n}$. If $G_{n}$ contains no $K_{a,b}$ then at most b −1 of these a-stars can share the same set of endpoints. We obtain
$$sum binom{d_{i}}{a} = text{the number of a−stars} ≤ (b − 1)binom{n}{a}$$



Extending $binom{n}{a}$ to all $x>0$ by



$binom{x}{a} = x(x-1)...(x-a+1)$ for $x>= a-1$ and $0$ for $x < a-1$



we have a convex function. Then Jensen’s Inequality implies that, the left
hand side in exponated inequality above is at least $binom{e(G)/n}{a}$, and the result follows by an easy
calculation.
End of sketch.



I understand the beginning, the proof of convexity is also obvious for me. But I can't follow how the Jensen's Inequality was used here, and how further calculations gives us final result.



Maybe somebody will be interested in the topic? Thank you in advance for the discussion.






graph-theory proof-explanation jensen-inequality




graph-theory proof-explanation jensen-inequality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 9 at 19:23









OlgaOlga

176




176












  • $begingroup$
    What is $ex(cdot,cdot)$?
    $endgroup$
    – Mike
    Jan 9 at 21:31






  • 1




    $begingroup$
    Well, ex(n, H) is the Turan number for H, i.e. the maximum number of edges of an H-free graph on n vertices.
    $endgroup$
    – Olga
    Jan 10 at 9:09


















  • $begingroup$
    What is $ex(cdot,cdot)$?
    $endgroup$
    – Mike
    Jan 9 at 21:31






  • 1




    $begingroup$
    Well, ex(n, H) is the Turan number for H, i.e. the maximum number of edges of an H-free graph on n vertices.
    $endgroup$
    – Olga
    Jan 10 at 9:09
















$begingroup$
What is $ex(cdot,cdot)$?
$endgroup$
– Mike
Jan 9 at 21:31




$begingroup$
What is $ex(cdot,cdot)$?
$endgroup$
– Mike
Jan 9 at 21:31




1




1




$begingroup$
Well, ex(n, H) is the Turan number for H, i.e. the maximum number of edges of an H-free graph on n vertices.
$endgroup$
– Olga
Jan 10 at 9:09




$begingroup$
Well, ex(n, H) is the Turan number for H, i.e. the maximum number of edges of an H-free graph on n vertices.
$endgroup$
– Olga
Jan 10 at 9:09










1 Answer
1






active

oldest

votes


















0












$begingroup$

There are a couple errors in your proof sketch which may be preventing you from successfully filling out the details. Here are some points to consider.




  • We define $binom{x}{a}$ for real $x$ and integer $a$ to be $x(x-1)cdots (x-a+1)/a!$ (and only for $x geq a-1$ as you've stated).


  • Jensen's inequality states the following: let $f$ be a convex function and let $x_1,dots, x_n$ be real. For any arithmetic (weighted) average $bar{x} = sum lambda_i x_i$, we have $f(bar{x}) leq sum lambda_i f(x_i)$. For the unweighted arithmetic average, this means $nf(bar{x}) leq f(x_1) + cdots f(x_n)$.


  • Since $binom{x}{a}$ is convex in $x$, we may lower bound $sum binom{d_i}{a}$ by $nbinom{bar{d}}{a}$, where $bar{d} = e(G)/n$ is the average degree on one side of the bipartite graph $G$.


  • To follow the further calculation, we need to simplify the expressions involving the binomial coefficients. We first multiply both sides by $a!$. We bound $binom{x}{a}$ below by $(x-a+1)^a$ and above by $x^a$.


  • The final error is that I believe the result of the theorem should be multiplied by a factor of $2$.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    You're right! Now I can fully understand, everything can be calculated to the final result :) I've been missing the part with bounding both sides. Thank you very much, what a relief!
    $endgroup$
    – Olga
    Jan 10 at 12:43












  • $begingroup$
    You’re welcome! This simple technique of bounding binomial coefficients is very common, and rather accurate if $x$ is large and $a$ is fixed.
    $endgroup$
    – Bob Krueger
    Jan 10 at 15:23












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

There are a couple errors in your proof sketch which may be preventing you from successfully filling out the details. Here are some points to consider.




  • We define $binom{x}{a}$ for real $x$ and integer $a$ to be $x(x-1)cdots (x-a+1)/a!$ (and only for $x geq a-1$ as you've stated).


  • Jensen's inequality states the following: let $f$ be a convex function and let $x_1,dots, x_n$ be real. For any arithmetic (weighted) average $bar{x} = sum lambda_i x_i$, we have $f(bar{x}) leq sum lambda_i f(x_i)$. For the unweighted arithmetic average, this means $nf(bar{x}) leq f(x_1) + cdots f(x_n)$.


  • Since $binom{x}{a}$ is convex in $x$, we may lower bound $sum binom{d_i}{a}$ by $nbinom{bar{d}}{a}$, where $bar{d} = e(G)/n$ is the average degree on one side of the bipartite graph $G$.


  • To follow the further calculation, we need to simplify the expressions involving the binomial coefficients. We first multiply both sides by $a!$. We bound $binom{x}{a}$ below by $(x-a+1)^a$ and above by $x^a$.


  • The final error is that I believe the result of the theorem should be multiplied by a factor of $2$.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    You're right! Now I can fully understand, everything can be calculated to the final result :) I've been missing the part with bounding both sides. Thank you very much, what a relief!
    $endgroup$
    – Olga
    Jan 10 at 12:43












  • $begingroup$
    You’re welcome! This simple technique of bounding binomial coefficients is very common, and rather accurate if $x$ is large and $a$ is fixed.
    $endgroup$
    – Bob Krueger
    Jan 10 at 15:23
















0












$begingroup$

There are a couple errors in your proof sketch which may be preventing you from successfully filling out the details. Here are some points to consider.




  • We define $binom{x}{a}$ for real $x$ and integer $a$ to be $x(x-1)cdots (x-a+1)/a!$ (and only for $x geq a-1$ as you've stated).


  • Jensen's inequality states the following: let $f$ be a convex function and let $x_1,dots, x_n$ be real. For any arithmetic (weighted) average $bar{x} = sum lambda_i x_i$, we have $f(bar{x}) leq sum lambda_i f(x_i)$. For the unweighted arithmetic average, this means $nf(bar{x}) leq f(x_1) + cdots f(x_n)$.


  • Since $binom{x}{a}$ is convex in $x$, we may lower bound $sum binom{d_i}{a}$ by $nbinom{bar{d}}{a}$, where $bar{d} = e(G)/n$ is the average degree on one side of the bipartite graph $G$.


  • To follow the further calculation, we need to simplify the expressions involving the binomial coefficients. We first multiply both sides by $a!$. We bound $binom{x}{a}$ below by $(x-a+1)^a$ and above by $x^a$.


  • The final error is that I believe the result of the theorem should be multiplied by a factor of $2$.







share|cite|improve this answer









$endgroup$













  • $begingroup$
    You're right! Now I can fully understand, everything can be calculated to the final result :) I've been missing the part with bounding both sides. Thank you very much, what a relief!
    $endgroup$
    – Olga
    Jan 10 at 12:43












  • $begingroup$
    You’re welcome! This simple technique of bounding binomial coefficients is very common, and rather accurate if $x$ is large and $a$ is fixed.
    $endgroup$
    – Bob Krueger
    Jan 10 at 15:23














0












0








0





$begingroup$

There are a couple errors in your proof sketch which may be preventing you from successfully filling out the details. Here are some points to consider.




  • We define $binom{x}{a}$ for real $x$ and integer $a$ to be $x(x-1)cdots (x-a+1)/a!$ (and only for $x geq a-1$ as you've stated).


  • Jensen's inequality states the following: let $f$ be a convex function and let $x_1,dots, x_n$ be real. For any arithmetic (weighted) average $bar{x} = sum lambda_i x_i$, we have $f(bar{x}) leq sum lambda_i f(x_i)$. For the unweighted arithmetic average, this means $nf(bar{x}) leq f(x_1) + cdots f(x_n)$.


  • Since $binom{x}{a}$ is convex in $x$, we may lower bound $sum binom{d_i}{a}$ by $nbinom{bar{d}}{a}$, where $bar{d} = e(G)/n$ is the average degree on one side of the bipartite graph $G$.


  • To follow the further calculation, we need to simplify the expressions involving the binomial coefficients. We first multiply both sides by $a!$. We bound $binom{x}{a}$ below by $(x-a+1)^a$ and above by $x^a$.


  • The final error is that I believe the result of the theorem should be multiplied by a factor of $2$.







share|cite|improve this answer









$endgroup$



There are a couple errors in your proof sketch which may be preventing you from successfully filling out the details. Here are some points to consider.




  • We define $binom{x}{a}$ for real $x$ and integer $a$ to be $x(x-1)cdots (x-a+1)/a!$ (and only for $x geq a-1$ as you've stated).


  • Jensen's inequality states the following: let $f$ be a convex function and let $x_1,dots, x_n$ be real. For any arithmetic (weighted) average $bar{x} = sum lambda_i x_i$, we have $f(bar{x}) leq sum lambda_i f(x_i)$. For the unweighted arithmetic average, this means $nf(bar{x}) leq f(x_1) + cdots f(x_n)$.


  • Since $binom{x}{a}$ is convex in $x$, we may lower bound $sum binom{d_i}{a}$ by $nbinom{bar{d}}{a}$, where $bar{d} = e(G)/n$ is the average degree on one side of the bipartite graph $G$.


  • To follow the further calculation, we need to simplify the expressions involving the binomial coefficients. We first multiply both sides by $a!$. We bound $binom{x}{a}$ below by $(x-a+1)^a$ and above by $x^a$.


  • The final error is that I believe the result of the theorem should be multiplied by a factor of $2$.








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 at 9:14









Bob KruegerBob Krueger

4,1902822




4,1902822












  • $begingroup$
    You're right! Now I can fully understand, everything can be calculated to the final result :) I've been missing the part with bounding both sides. Thank you very much, what a relief!
    $endgroup$
    – Olga
    Jan 10 at 12:43












  • $begingroup$
    You’re welcome! This simple technique of bounding binomial coefficients is very common, and rather accurate if $x$ is large and $a$ is fixed.
    $endgroup$
    – Bob Krueger
    Jan 10 at 15:23


















  • $begingroup$
    You're right! Now I can fully understand, everything can be calculated to the final result :) I've been missing the part with bounding both sides. Thank you very much, what a relief!
    $endgroup$
    – Olga
    Jan 10 at 12:43












  • $begingroup$
    You’re welcome! This simple technique of bounding binomial coefficients is very common, and rather accurate if $x$ is large and $a$ is fixed.
    $endgroup$
    – Bob Krueger
    Jan 10 at 15:23
















$begingroup$
You're right! Now I can fully understand, everything can be calculated to the final result :) I've been missing the part with bounding both sides. Thank you very much, what a relief!
$endgroup$
– Olga
Jan 10 at 12:43






$begingroup$
You're right! Now I can fully understand, everything can be calculated to the final result :) I've been missing the part with bounding both sides. Thank you very much, what a relief!
$endgroup$
– Olga
Jan 10 at 12:43














$begingroup$
You’re welcome! This simple technique of bounding binomial coefficients is very common, and rather accurate if $x$ is large and $a$ is fixed.
$endgroup$
– Bob Krueger
Jan 10 at 15:23




$begingroup$
You’re welcome! This simple technique of bounding binomial coefficients is very common, and rather accurate if $x$ is large and $a$ is fixed.
$endgroup$
– Bob Krueger
Jan 10 at 15:23


















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