Show that $sum_{k=0}^{N} |P(k)| leq C(N) int_{0} ^{1} |P(t)| dt $. [on hold]












2












$begingroup$


I’m attending a functional analysis course and I am given to solve this problem as an exercise but I’m a little bit disoriented and I don’t know what tools I can use to get it.




Show that, for each $N in mathbb{N}$ , there exists a constant $C(N) in mathbb{R}^+$ such that if $P$ in a polynomial of degree $N$ with complex coefficients, we have
$$sum_{k=0}^{N} |P(k)| leq C(N) int_{0} ^{1} |P(t)| dt,.$$




Any idea or hint? Thank you all in advance










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put on hold as off-topic by user21820, RRL, Did, Xander Henderson, José Carlos Santos 16 hours ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, RRL, Did, Xander Henderson, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.





















    2












    $begingroup$


    I’m attending a functional analysis course and I am given to solve this problem as an exercise but I’m a little bit disoriented and I don’t know what tools I can use to get it.




    Show that, for each $N in mathbb{N}$ , there exists a constant $C(N) in mathbb{R}^+$ such that if $P$ in a polynomial of degree $N$ with complex coefficients, we have
    $$sum_{k=0}^{N} |P(k)| leq C(N) int_{0} ^{1} |P(t)| dt,.$$




    Any idea or hint? Thank you all in advance










    share|cite|improve this question











    $endgroup$



    put on hold as off-topic by user21820, RRL, Did, Xander Henderson, José Carlos Santos 16 hours ago


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, RRL, Did, Xander Henderson, José Carlos Santos

    If this question can be reworded to fit the rules in the help center, please edit the question.



















      2












      2








      2





      $begingroup$


      I’m attending a functional analysis course and I am given to solve this problem as an exercise but I’m a little bit disoriented and I don’t know what tools I can use to get it.




      Show that, for each $N in mathbb{N}$ , there exists a constant $C(N) in mathbb{R}^+$ such that if $P$ in a polynomial of degree $N$ with complex coefficients, we have
      $$sum_{k=0}^{N} |P(k)| leq C(N) int_{0} ^{1} |P(t)| dt,.$$




      Any idea or hint? Thank you all in advance










      share|cite|improve this question











      $endgroup$




      I’m attending a functional analysis course and I am given to solve this problem as an exercise but I’m a little bit disoriented and I don’t know what tools I can use to get it.




      Show that, for each $N in mathbb{N}$ , there exists a constant $C(N) in mathbb{R}^+$ such that if $P$ in a polynomial of degree $N$ with complex coefficients, we have
      $$sum_{k=0}^{N} |P(k)| leq C(N) int_{0} ^{1} |P(t)| dt,.$$




      Any idea or hint? Thank you all in advance







      functional-analysis polynomials norm normed-spaces upper-lower-bounds






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      share|cite|improve this question













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      edited Dec 9 '18 at 8:14









      Batominovski

      33.1k33293




      33.1k33293










      asked Dec 8 '18 at 23:33









      Maggie94Maggie94

      1246




      1246




      put on hold as off-topic by user21820, RRL, Did, Xander Henderson, José Carlos Santos 16 hours ago


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, RRL, Did, Xander Henderson, José Carlos Santos

      If this question can be reworded to fit the rules in the help center, please edit the question.







      put on hold as off-topic by user21820, RRL, Did, Xander Henderson, José Carlos Santos 16 hours ago


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, RRL, Did, Xander Henderson, José Carlos Santos

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Let $V_N$ be the complex vector space of polynomials in $mathbb{C}[t]$ of degree at most $Ninmathbb{Z}_{geq 0}$. Show that $|_|_1$ and $|_|_2$ defined by
          $$|P|_1:=sum_{k=0}^N,big|P(k)big|$$
          and
          $$|P|_2:=int_0^1,big|P(t)big|,text{d}t$$
          for all $Pin V_N$ are norms on $V_N$.




          To show that $|_|_1$ is a norm, we note that $|a, P|_1=|a|,|P|_1$ trivially holds for all $ainmathbb{C}$ and $Pin V_N$. If $|P|_1=0$ for some $Pin V_N$, then $P(0)=P(1)=P(2)=ldots=P(N)=0$, so $$P(t)=Q(t),prod_{k=0}^N,(t-k)$$ for some $Q(t)inmathbb{C}[t]$. As $P$ has degree at most $N$, $Q$ must be identically $0$, so $Pequiv 0$. If $P_1,P_2in V_N$, then $P_1+P_2in V_N$, and $$(P_1+P_2)(k)=P_1(k)+P_2(k)text{ for every }kinmathbb{C},,$$ whence $$big|(P_1+P_2)(k)big|=big|P_1(k)+P_2(k)big|leq big|P_1(k)big|+big|P_2(k)big|text{ for every }kinmathbb{C},.$$ Taking the sum over $k=0,1,2,ldots,N$, we get $|P_1+P_2|_1leq |P_1|_1+|P_2|_1$, justifying the triangle inequality condition. Similarly, $|_|_2$ is a norm on $V_N$.




          Note that all norms on a finite-dimensional complex vector space are equivalent. That means, there exist $c(N),C(N)inmathbb{R}_{>0}$ such that
          $$c(N),|P|_2leq |P|_1leq C(N),|P|_2$$
          for all $Pin V_N$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you! I can observe that the polynomial space I am considering has $N+1$ as dimension because it is generated, for example, by ${x^{N},x^{N-1},...,x,1}$ that has dimension $N+1$, which is finite. Right?
            $endgroup$
            – Maggie94
            Dec 10 '18 at 13:03












          • $begingroup$
            Yes, you are right.
            $endgroup$
            – Batominovski
            Dec 10 '18 at 20:03


















          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2












          $begingroup$

          Let $V_N$ be the complex vector space of polynomials in $mathbb{C}[t]$ of degree at most $Ninmathbb{Z}_{geq 0}$. Show that $|_|_1$ and $|_|_2$ defined by
          $$|P|_1:=sum_{k=0}^N,big|P(k)big|$$
          and
          $$|P|_2:=int_0^1,big|P(t)big|,text{d}t$$
          for all $Pin V_N$ are norms on $V_N$.




          To show that $|_|_1$ is a norm, we note that $|a, P|_1=|a|,|P|_1$ trivially holds for all $ainmathbb{C}$ and $Pin V_N$. If $|P|_1=0$ for some $Pin V_N$, then $P(0)=P(1)=P(2)=ldots=P(N)=0$, so $$P(t)=Q(t),prod_{k=0}^N,(t-k)$$ for some $Q(t)inmathbb{C}[t]$. As $P$ has degree at most $N$, $Q$ must be identically $0$, so $Pequiv 0$. If $P_1,P_2in V_N$, then $P_1+P_2in V_N$, and $$(P_1+P_2)(k)=P_1(k)+P_2(k)text{ for every }kinmathbb{C},,$$ whence $$big|(P_1+P_2)(k)big|=big|P_1(k)+P_2(k)big|leq big|P_1(k)big|+big|P_2(k)big|text{ for every }kinmathbb{C},.$$ Taking the sum over $k=0,1,2,ldots,N$, we get $|P_1+P_2|_1leq |P_1|_1+|P_2|_1$, justifying the triangle inequality condition. Similarly, $|_|_2$ is a norm on $V_N$.




          Note that all norms on a finite-dimensional complex vector space are equivalent. That means, there exist $c(N),C(N)inmathbb{R}_{>0}$ such that
          $$c(N),|P|_2leq |P|_1leq C(N),|P|_2$$
          for all $Pin V_N$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you! I can observe that the polynomial space I am considering has $N+1$ as dimension because it is generated, for example, by ${x^{N},x^{N-1},...,x,1}$ that has dimension $N+1$, which is finite. Right?
            $endgroup$
            – Maggie94
            Dec 10 '18 at 13:03












          • $begingroup$
            Yes, you are right.
            $endgroup$
            – Batominovski
            Dec 10 '18 at 20:03
















          2












          $begingroup$

          Let $V_N$ be the complex vector space of polynomials in $mathbb{C}[t]$ of degree at most $Ninmathbb{Z}_{geq 0}$. Show that $|_|_1$ and $|_|_2$ defined by
          $$|P|_1:=sum_{k=0}^N,big|P(k)big|$$
          and
          $$|P|_2:=int_0^1,big|P(t)big|,text{d}t$$
          for all $Pin V_N$ are norms on $V_N$.




          To show that $|_|_1$ is a norm, we note that $|a, P|_1=|a|,|P|_1$ trivially holds for all $ainmathbb{C}$ and $Pin V_N$. If $|P|_1=0$ for some $Pin V_N$, then $P(0)=P(1)=P(2)=ldots=P(N)=0$, so $$P(t)=Q(t),prod_{k=0}^N,(t-k)$$ for some $Q(t)inmathbb{C}[t]$. As $P$ has degree at most $N$, $Q$ must be identically $0$, so $Pequiv 0$. If $P_1,P_2in V_N$, then $P_1+P_2in V_N$, and $$(P_1+P_2)(k)=P_1(k)+P_2(k)text{ for every }kinmathbb{C},,$$ whence $$big|(P_1+P_2)(k)big|=big|P_1(k)+P_2(k)big|leq big|P_1(k)big|+big|P_2(k)big|text{ for every }kinmathbb{C},.$$ Taking the sum over $k=0,1,2,ldots,N$, we get $|P_1+P_2|_1leq |P_1|_1+|P_2|_1$, justifying the triangle inequality condition. Similarly, $|_|_2$ is a norm on $V_N$.




          Note that all norms on a finite-dimensional complex vector space are equivalent. That means, there exist $c(N),C(N)inmathbb{R}_{>0}$ such that
          $$c(N),|P|_2leq |P|_1leq C(N),|P|_2$$
          for all $Pin V_N$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you! I can observe that the polynomial space I am considering has $N+1$ as dimension because it is generated, for example, by ${x^{N},x^{N-1},...,x,1}$ that has dimension $N+1$, which is finite. Right?
            $endgroup$
            – Maggie94
            Dec 10 '18 at 13:03












          • $begingroup$
            Yes, you are right.
            $endgroup$
            – Batominovski
            Dec 10 '18 at 20:03














          2












          2








          2





          $begingroup$

          Let $V_N$ be the complex vector space of polynomials in $mathbb{C}[t]$ of degree at most $Ninmathbb{Z}_{geq 0}$. Show that $|_|_1$ and $|_|_2$ defined by
          $$|P|_1:=sum_{k=0}^N,big|P(k)big|$$
          and
          $$|P|_2:=int_0^1,big|P(t)big|,text{d}t$$
          for all $Pin V_N$ are norms on $V_N$.




          To show that $|_|_1$ is a norm, we note that $|a, P|_1=|a|,|P|_1$ trivially holds for all $ainmathbb{C}$ and $Pin V_N$. If $|P|_1=0$ for some $Pin V_N$, then $P(0)=P(1)=P(2)=ldots=P(N)=0$, so $$P(t)=Q(t),prod_{k=0}^N,(t-k)$$ for some $Q(t)inmathbb{C}[t]$. As $P$ has degree at most $N$, $Q$ must be identically $0$, so $Pequiv 0$. If $P_1,P_2in V_N$, then $P_1+P_2in V_N$, and $$(P_1+P_2)(k)=P_1(k)+P_2(k)text{ for every }kinmathbb{C},,$$ whence $$big|(P_1+P_2)(k)big|=big|P_1(k)+P_2(k)big|leq big|P_1(k)big|+big|P_2(k)big|text{ for every }kinmathbb{C},.$$ Taking the sum over $k=0,1,2,ldots,N$, we get $|P_1+P_2|_1leq |P_1|_1+|P_2|_1$, justifying the triangle inequality condition. Similarly, $|_|_2$ is a norm on $V_N$.




          Note that all norms on a finite-dimensional complex vector space are equivalent. That means, there exist $c(N),C(N)inmathbb{R}_{>0}$ such that
          $$c(N),|P|_2leq |P|_1leq C(N),|P|_2$$
          for all $Pin V_N$.






          share|cite|improve this answer











          $endgroup$



          Let $V_N$ be the complex vector space of polynomials in $mathbb{C}[t]$ of degree at most $Ninmathbb{Z}_{geq 0}$. Show that $|_|_1$ and $|_|_2$ defined by
          $$|P|_1:=sum_{k=0}^N,big|P(k)big|$$
          and
          $$|P|_2:=int_0^1,big|P(t)big|,text{d}t$$
          for all $Pin V_N$ are norms on $V_N$.




          To show that $|_|_1$ is a norm, we note that $|a, P|_1=|a|,|P|_1$ trivially holds for all $ainmathbb{C}$ and $Pin V_N$. If $|P|_1=0$ for some $Pin V_N$, then $P(0)=P(1)=P(2)=ldots=P(N)=0$, so $$P(t)=Q(t),prod_{k=0}^N,(t-k)$$ for some $Q(t)inmathbb{C}[t]$. As $P$ has degree at most $N$, $Q$ must be identically $0$, so $Pequiv 0$. If $P_1,P_2in V_N$, then $P_1+P_2in V_N$, and $$(P_1+P_2)(k)=P_1(k)+P_2(k)text{ for every }kinmathbb{C},,$$ whence $$big|(P_1+P_2)(k)big|=big|P_1(k)+P_2(k)big|leq big|P_1(k)big|+big|P_2(k)big|text{ for every }kinmathbb{C},.$$ Taking the sum over $k=0,1,2,ldots,N$, we get $|P_1+P_2|_1leq |P_1|_1+|P_2|_1$, justifying the triangle inequality condition. Similarly, $|_|_2$ is a norm on $V_N$.




          Note that all norms on a finite-dimensional complex vector space are equivalent. That means, there exist $c(N),C(N)inmathbb{R}_{>0}$ such that
          $$c(N),|P|_2leq |P|_1leq C(N),|P|_2$$
          for all $Pin V_N$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 9 '18 at 8:08

























          answered Dec 8 '18 at 23:54









          BatominovskiBatominovski

          33.1k33293




          33.1k33293












          • $begingroup$
            Thank you! I can observe that the polynomial space I am considering has $N+1$ as dimension because it is generated, for example, by ${x^{N},x^{N-1},...,x,1}$ that has dimension $N+1$, which is finite. Right?
            $endgroup$
            – Maggie94
            Dec 10 '18 at 13:03












          • $begingroup$
            Yes, you are right.
            $endgroup$
            – Batominovski
            Dec 10 '18 at 20:03


















          • $begingroup$
            Thank you! I can observe that the polynomial space I am considering has $N+1$ as dimension because it is generated, for example, by ${x^{N},x^{N-1},...,x,1}$ that has dimension $N+1$, which is finite. Right?
            $endgroup$
            – Maggie94
            Dec 10 '18 at 13:03












          • $begingroup$
            Yes, you are right.
            $endgroup$
            – Batominovski
            Dec 10 '18 at 20:03
















          $begingroup$
          Thank you! I can observe that the polynomial space I am considering has $N+1$ as dimension because it is generated, for example, by ${x^{N},x^{N-1},...,x,1}$ that has dimension $N+1$, which is finite. Right?
          $endgroup$
          – Maggie94
          Dec 10 '18 at 13:03






          $begingroup$
          Thank you! I can observe that the polynomial space I am considering has $N+1$ as dimension because it is generated, for example, by ${x^{N},x^{N-1},...,x,1}$ that has dimension $N+1$, which is finite. Right?
          $endgroup$
          – Maggie94
          Dec 10 '18 at 13:03














          $begingroup$
          Yes, you are right.
          $endgroup$
          – Batominovski
          Dec 10 '18 at 20:03




          $begingroup$
          Yes, you are right.
          $endgroup$
          – Batominovski
          Dec 10 '18 at 20:03



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