Neighborhood whose vertices have above average degree












2












$begingroup$


Let $G$ be a simple, connected, undirected graph with more than one vertex. Proof that a that $G$ contains a vertex $v$ such that



$frac{1}{|delta(v)|}
displaystylesum_{win{}N(v)} |delta(w)| geq frac{2|E(G)|}{|V(G)|}$



,where $delta(v)$ is the degree of a vertex, $N(v)$ is the neighborhood of $v$, $E(G)$ is the number of edges in $G$ and $V(G)$ the number of vertices in $G$.



So far if have shown this to bee true for trees where there are always $V(G)-1$ edges,
cycle graphs where there are $V(G)$ edges and I have looked at quite a few examples.



The intuition is that the right side is the average degree for the vertices in $G$ while the left is the average degree of vertices in the neighborhood of $v$.
So proof that there is a neighborhood with above average degree.



I have also tried proving it by removing edges from a fully connected graph or adding edges to a tree but it didn't really work out.



It would be really nice if someone could give me a hint.






graph-theory







share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Let $G$ be a simple, connected, undirected graph with more than one vertex. Proof that a that $G$ contains a vertex $v$ such that



    $frac{1}{|delta(v)|}
    displaystylesum_{win{}N(v)} |delta(w)| geq frac{2|E(G)|}{|V(G)|}$



    ,where $delta(v)$ is the degree of a vertex, $N(v)$ is the neighborhood of $v$, $E(G)$ is the number of edges in $G$ and $V(G)$ the number of vertices in $G$.



    So far if have shown this to bee true for trees where there are always $V(G)-1$ edges,
    cycle graphs where there are $V(G)$ edges and I have looked at quite a few examples.



    The intuition is that the right side is the average degree for the vertices in $G$ while the left is the average degree of vertices in the neighborhood of $v$.
    So proof that there is a neighborhood with above average degree.



    I have also tried proving it by removing edges from a fully connected graph or adding edges to a tree but it didn't really work out.



    It would be really nice if someone could give me a hint.






    graph-theory







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      Let $G$ be a simple, connected, undirected graph with more than one vertex. Proof that a that $G$ contains a vertex $v$ such that



      $frac{1}{|delta(v)|}
      displaystylesum_{win{}N(v)} |delta(w)| geq frac{2|E(G)|}{|V(G)|}$



      ,where $delta(v)$ is the degree of a vertex, $N(v)$ is the neighborhood of $v$, $E(G)$ is the number of edges in $G$ and $V(G)$ the number of vertices in $G$.



      So far if have shown this to bee true for trees where there are always $V(G)-1$ edges,
      cycle graphs where there are $V(G)$ edges and I have looked at quite a few examples.



      The intuition is that the right side is the average degree for the vertices in $G$ while the left is the average degree of vertices in the neighborhood of $v$.
      So proof that there is a neighborhood with above average degree.



      I have also tried proving it by removing edges from a fully connected graph or adding edges to a tree but it didn't really work out.



      It would be really nice if someone could give me a hint.






      graph-theory







      share|cite|improve this question









      $endgroup$




      Let $G$ be a simple, connected, undirected graph with more than one vertex. Proof that a that $G$ contains a vertex $v$ such that



      $frac{1}{|delta(v)|}
      displaystylesum_{win{}N(v)} |delta(w)| geq frac{2|E(G)|}{|V(G)|}$



      ,where $delta(v)$ is the degree of a vertex, $N(v)$ is the neighborhood of $v$, $E(G)$ is the number of edges in $G$ and $V(G)$ the number of vertices in $G$.



      So far if have shown this to bee true for trees where there are always $V(G)-1$ edges,
      cycle graphs where there are $V(G)$ edges and I have looked at quite a few examples.



      The intuition is that the right side is the average degree for the vertices in $G$ while the left is the average degree of vertices in the neighborhood of $v$.
      So proof that there is a neighborhood with above average degree.



      I have also tried proving it by removing edges from a fully connected graph or adding edges to a tree but it didn't really work out.



      It would be really nice if someone could give me a hint.






      graph-theory




      graph-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 1 '18 at 23:06









      Ymi_YugyYmi_Yugy

      204




      204






















          1 Answer
          1






          active

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          2












          $begingroup$

          If you sum the quantity
          $$
          frac{1}{|delta(v)|}
          displaystylesum_{win{}N(v)} |delta(w)|
          $$
          over all vertices $v$, you will get
          $$
          sum_{vw in E(G)} left(frac{|delta(w)|}{|delta(v)|} + frac{|delta(v)|}{|delta(w)|}right).
          $$
          You can show that this sum exceeds $2|E(G)|$. Therefore its average over all $v$ is at least $frac{2|E(G)|}{|V(G)|}$, and there must be some vertex $v$ which achieves at least this average.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            can you elaborate how the first step works?
            $endgroup$
            – Ymi_Yugy
            Dec 3 '18 at 1:34










          • $begingroup$
            If you have a sum over all vertices $v$ and over all $w in N(v)$, then your sum considers every edge $vw$ once from both points of view (once as $vw$ and once as $wv$).
            $endgroup$
            – Misha Lavrov
            Dec 3 '18 at 1:37










          • $begingroup$
            Ok. That makes sense. Regarding the second step. Is it non obvious or just me not seeing it?
            $endgroup$
            – Ymi_Yugy
            Dec 3 '18 at 2:05












          • $begingroup$
            ok. I got it. It was obvious.
            $endgroup$
            – Ymi_Yugy
            Dec 3 '18 at 2:29











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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          If you sum the quantity
          $$
          frac{1}{|delta(v)|}
          displaystylesum_{win{}N(v)} |delta(w)|
          $$
          over all vertices $v$, you will get
          $$
          sum_{vw in E(G)} left(frac{|delta(w)|}{|delta(v)|} + frac{|delta(v)|}{|delta(w)|}right).
          $$
          You can show that this sum exceeds $2|E(G)|$. Therefore its average over all $v$ is at least $frac{2|E(G)|}{|V(G)|}$, and there must be some vertex $v$ which achieves at least this average.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            can you elaborate how the first step works?
            $endgroup$
            – Ymi_Yugy
            Dec 3 '18 at 1:34










          • $begingroup$
            If you have a sum over all vertices $v$ and over all $w in N(v)$, then your sum considers every edge $vw$ once from both points of view (once as $vw$ and once as $wv$).
            $endgroup$
            – Misha Lavrov
            Dec 3 '18 at 1:37










          • $begingroup$
            Ok. That makes sense. Regarding the second step. Is it non obvious or just me not seeing it?
            $endgroup$
            – Ymi_Yugy
            Dec 3 '18 at 2:05












          • $begingroup$
            ok. I got it. It was obvious.
            $endgroup$
            – Ymi_Yugy
            Dec 3 '18 at 2:29
















          2












          $begingroup$

          If you sum the quantity
          $$
          frac{1}{|delta(v)|}
          displaystylesum_{win{}N(v)} |delta(w)|
          $$
          over all vertices $v$, you will get
          $$
          sum_{vw in E(G)} left(frac{|delta(w)|}{|delta(v)|} + frac{|delta(v)|}{|delta(w)|}right).
          $$
          You can show that this sum exceeds $2|E(G)|$. Therefore its average over all $v$ is at least $frac{2|E(G)|}{|V(G)|}$, and there must be some vertex $v$ which achieves at least this average.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            can you elaborate how the first step works?
            $endgroup$
            – Ymi_Yugy
            Dec 3 '18 at 1:34










          • $begingroup$
            If you have a sum over all vertices $v$ and over all $w in N(v)$, then your sum considers every edge $vw$ once from both points of view (once as $vw$ and once as $wv$).
            $endgroup$
            – Misha Lavrov
            Dec 3 '18 at 1:37










          • $begingroup$
            Ok. That makes sense. Regarding the second step. Is it non obvious or just me not seeing it?
            $endgroup$
            – Ymi_Yugy
            Dec 3 '18 at 2:05












          • $begingroup$
            ok. I got it. It was obvious.
            $endgroup$
            – Ymi_Yugy
            Dec 3 '18 at 2:29














          2












          2








          2





          $begingroup$

          If you sum the quantity
          $$
          frac{1}{|delta(v)|}
          displaystylesum_{win{}N(v)} |delta(w)|
          $$
          over all vertices $v$, you will get
          $$
          sum_{vw in E(G)} left(frac{|delta(w)|}{|delta(v)|} + frac{|delta(v)|}{|delta(w)|}right).
          $$
          You can show that this sum exceeds $2|E(G)|$. Therefore its average over all $v$ is at least $frac{2|E(G)|}{|V(G)|}$, and there must be some vertex $v$ which achieves at least this average.






          share|cite|improve this answer











          $endgroup$



          If you sum the quantity
          $$
          frac{1}{|delta(v)|}
          displaystylesum_{win{}N(v)} |delta(w)|
          $$
          over all vertices $v$, you will get
          $$
          sum_{vw in E(G)} left(frac{|delta(w)|}{|delta(v)|} + frac{|delta(v)|}{|delta(w)|}right).
          $$
          You can show that this sum exceeds $2|E(G)|$. Therefore its average over all $v$ is at least $frac{2|E(G)|}{|V(G)|}$, and there must be some vertex $v$ which achieves at least this average.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 2 '18 at 3:54









          Mike

          3,532411




          3,532411










          answered Dec 1 '18 at 23:42









          Misha LavrovMisha Lavrov

          45.1k656107




          45.1k656107












          • $begingroup$
            can you elaborate how the first step works?
            $endgroup$
            – Ymi_Yugy
            Dec 3 '18 at 1:34










          • $begingroup$
            If you have a sum over all vertices $v$ and over all $w in N(v)$, then your sum considers every edge $vw$ once from both points of view (once as $vw$ and once as $wv$).
            $endgroup$
            – Misha Lavrov
            Dec 3 '18 at 1:37










          • $begingroup$
            Ok. That makes sense. Regarding the second step. Is it non obvious or just me not seeing it?
            $endgroup$
            – Ymi_Yugy
            Dec 3 '18 at 2:05












          • $begingroup$
            ok. I got it. It was obvious.
            $endgroup$
            – Ymi_Yugy
            Dec 3 '18 at 2:29


















          • $begingroup$
            can you elaborate how the first step works?
            $endgroup$
            – Ymi_Yugy
            Dec 3 '18 at 1:34










          • $begingroup$
            If you have a sum over all vertices $v$ and over all $w in N(v)$, then your sum considers every edge $vw$ once from both points of view (once as $vw$ and once as $wv$).
            $endgroup$
            – Misha Lavrov
            Dec 3 '18 at 1:37










          • $begingroup$
            Ok. That makes sense. Regarding the second step. Is it non obvious or just me not seeing it?
            $endgroup$
            – Ymi_Yugy
            Dec 3 '18 at 2:05












          • $begingroup$
            ok. I got it. It was obvious.
            $endgroup$
            – Ymi_Yugy
            Dec 3 '18 at 2:29
















          $begingroup$
          can you elaborate how the first step works?
          $endgroup$
          – Ymi_Yugy
          Dec 3 '18 at 1:34




          $begingroup$
          can you elaborate how the first step works?
          $endgroup$
          – Ymi_Yugy
          Dec 3 '18 at 1:34












          $begingroup$
          If you have a sum over all vertices $v$ and over all $w in N(v)$, then your sum considers every edge $vw$ once from both points of view (once as $vw$ and once as $wv$).
          $endgroup$
          – Misha Lavrov
          Dec 3 '18 at 1:37




          $begingroup$
          If you have a sum over all vertices $v$ and over all $w in N(v)$, then your sum considers every edge $vw$ once from both points of view (once as $vw$ and once as $wv$).
          $endgroup$
          – Misha Lavrov
          Dec 3 '18 at 1:37












          $begingroup$
          Ok. That makes sense. Regarding the second step. Is it non obvious or just me not seeing it?
          $endgroup$
          – Ymi_Yugy
          Dec 3 '18 at 2:05






          $begingroup$
          Ok. That makes sense. Regarding the second step. Is it non obvious or just me not seeing it?
          $endgroup$
          – Ymi_Yugy
          Dec 3 '18 at 2:05














          $begingroup$
          ok. I got it. It was obvious.
          $endgroup$
          – Ymi_Yugy
          Dec 3 '18 at 2:29




          $begingroup$
          ok. I got it. It was obvious.
          $endgroup$
          – Ymi_Yugy
          Dec 3 '18 at 2:29


















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