Is a space of polynomials over Real numbers complete?












-1












$begingroup$


Let $P$ be the space of all polynomials over $mathbb{R}$ normed by, $|P|= max {|a_0|,|a_1|,|a_2|,...,|a_n|}$ where $p(x)=sum_{k=0}^{n}a_kx^k$. Is this space complete?



Actually this problem is regarding the open mapping theorem. The map from $p$ to $p$ is not open. But this map isn't contradict the open mapping theorem. I thought its because, $P$ is not a Banach space.






functional-analysis polynomials complete-spaces







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  • $begingroup$
    Hi @Miraj Chamara, welcome to MSE. Did you already try to prove this? It's no problem if you make any mistakes, but we can help you more if you try first.
    $endgroup$
    – Math Girl
    Dec 4 '18 at 8:04










  • $begingroup$
    yes. I have tried this. Actually this problem is regarding the open mapping theorem. The map from p to p is not open. But this map isn't contradict the open mapping theorem. I thought its because, P is not a Banach space.
    $endgroup$
    – Miraj Chamara
    Dec 4 '18 at 8:09










  • $begingroup$
    It is not complete in any norm, by Baire's theorem.
    $endgroup$
    – Math_QED
    Dec 4 '18 at 16:25
















-1












$begingroup$


Let $P$ be the space of all polynomials over $mathbb{R}$ normed by, $|P|= max {|a_0|,|a_1|,|a_2|,...,|a_n|}$ where $p(x)=sum_{k=0}^{n}a_kx^k$. Is this space complete?



Actually this problem is regarding the open mapping theorem. The map from $p$ to $p$ is not open. But this map isn't contradict the open mapping theorem. I thought its because, $P$ is not a Banach space.






functional-analysis polynomials complete-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Hi @Miraj Chamara, welcome to MSE. Did you already try to prove this? It's no problem if you make any mistakes, but we can help you more if you try first.
    $endgroup$
    – Math Girl
    Dec 4 '18 at 8:04










  • $begingroup$
    yes. I have tried this. Actually this problem is regarding the open mapping theorem. The map from p to p is not open. But this map isn't contradict the open mapping theorem. I thought its because, P is not a Banach space.
    $endgroup$
    – Miraj Chamara
    Dec 4 '18 at 8:09










  • $begingroup$
    It is not complete in any norm, by Baire's theorem.
    $endgroup$
    – Math_QED
    Dec 4 '18 at 16:25














-1












-1








-1





$begingroup$


Let $P$ be the space of all polynomials over $mathbb{R}$ normed by, $|P|= max {|a_0|,|a_1|,|a_2|,...,|a_n|}$ where $p(x)=sum_{k=0}^{n}a_kx^k$. Is this space complete?



Actually this problem is regarding the open mapping theorem. The map from $p$ to $p$ is not open. But this map isn't contradict the open mapping theorem. I thought its because, $P$ is not a Banach space.






functional-analysis polynomials complete-spaces







share|cite|improve this question











$endgroup$




Let $P$ be the space of all polynomials over $mathbb{R}$ normed by, $|P|= max {|a_0|,|a_1|,|a_2|,...,|a_n|}$ where $p(x)=sum_{k=0}^{n}a_kx^k$. Is this space complete?



Actually this problem is regarding the open mapping theorem. The map from $p$ to $p$ is not open. But this map isn't contradict the open mapping theorem. I thought its because, $P$ is not a Banach space.






functional-analysis polynomials complete-spaces




functional-analysis polynomials complete-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 4 '18 at 12:53









Davide Giraudo

126k16150261




126k16150261










asked Dec 4 '18 at 7:56









Miraj ChamaraMiraj Chamara

1




1












  • $begingroup$
    Hi @Miraj Chamara, welcome to MSE. Did you already try to prove this? It's no problem if you make any mistakes, but we can help you more if you try first.
    $endgroup$
    – Math Girl
    Dec 4 '18 at 8:04










  • $begingroup$
    yes. I have tried this. Actually this problem is regarding the open mapping theorem. The map from p to p is not open. But this map isn't contradict the open mapping theorem. I thought its because, P is not a Banach space.
    $endgroup$
    – Miraj Chamara
    Dec 4 '18 at 8:09










  • $begingroup$
    It is not complete in any norm, by Baire's theorem.
    $endgroup$
    – Math_QED
    Dec 4 '18 at 16:25


















  • $begingroup$
    Hi @Miraj Chamara, welcome to MSE. Did you already try to prove this? It's no problem if you make any mistakes, but we can help you more if you try first.
    $endgroup$
    – Math Girl
    Dec 4 '18 at 8:04










  • $begingroup$
    yes. I have tried this. Actually this problem is regarding the open mapping theorem. The map from p to p is not open. But this map isn't contradict the open mapping theorem. I thought its because, P is not a Banach space.
    $endgroup$
    – Miraj Chamara
    Dec 4 '18 at 8:09










  • $begingroup$
    It is not complete in any norm, by Baire's theorem.
    $endgroup$
    – Math_QED
    Dec 4 '18 at 16:25
















$begingroup$
Hi @Miraj Chamara, welcome to MSE. Did you already try to prove this? It's no problem if you make any mistakes, but we can help you more if you try first.
$endgroup$
– Math Girl
Dec 4 '18 at 8:04




$begingroup$
Hi @Miraj Chamara, welcome to MSE. Did you already try to prove this? It's no problem if you make any mistakes, but we can help you more if you try first.
$endgroup$
– Math Girl
Dec 4 '18 at 8:04












$begingroup$
yes. I have tried this. Actually this problem is regarding the open mapping theorem. The map from p to p is not open. But this map isn't contradict the open mapping theorem. I thought its because, P is not a Banach space.
$endgroup$
– Miraj Chamara
Dec 4 '18 at 8:09




$begingroup$
yes. I have tried this. Actually this problem is regarding the open mapping theorem. The map from p to p is not open. But this map isn't contradict the open mapping theorem. I thought its because, P is not a Banach space.
$endgroup$
– Miraj Chamara
Dec 4 '18 at 8:09












$begingroup$
It is not complete in any norm, by Baire's theorem.
$endgroup$
– Math_QED
Dec 4 '18 at 16:25




$begingroup$
It is not complete in any norm, by Baire's theorem.
$endgroup$
– Math_QED
Dec 4 '18 at 16:25










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$sum _{k=1}^{n} frac {x^{k}} {k!}$ is a Cauchy sequence which does not converge.






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    1 Answer
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    $begingroup$

    $sum _{k=1}^{n} frac {x^{k}} {k!}$ is a Cauchy sequence which does not converge.






    share|cite|improve this answer









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      3












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      $sum _{k=1}^{n} frac {x^{k}} {k!}$ is a Cauchy sequence which does not converge.






      share|cite|improve this answer









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        3





        $begingroup$

        $sum _{k=1}^{n} frac {x^{k}} {k!}$ is a Cauchy sequence which does not converge.






        share|cite|improve this answer









        $endgroup$



        $sum _{k=1}^{n} frac {x^{k}} {k!}$ is a Cauchy sequence which does not converge.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 8:01









        Kavi Rama MurthyKavi Rama Murthy

        54.8k32055




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