Supnorm Defined on set of all uniformly continuous functions is Banach space












0












$begingroup$


If we define Supnorm on $C(bar{U})$ is



$$||f||_{C(bar{U})}:=sup_{xin U} |f(x)|.$$ on $C(bar{U})$ is Banach space



what I know is :



I'm proving this normed space



since $||f||_{C(bar{U})}=sup_{xin U} |f(x)| =0 iff f=0$



and $||af||_{C(bar{U})}=sup_{xin U} |af(x)| =|a| ||f||$



finally $||f+g||_{C(bar{U})}=sup_{xin U} |f(x)+g(x)|le sup _{xin U} |f(x)|+sup _{xin U} |g(x)|$



how to prove completeness ..thank you






functional-analysis sobolev-spaces







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    If we define Supnorm on $C(bar{U})$ is



    $$||f||_{C(bar{U})}:=sup_{xin U} |f(x)|.$$ on $C(bar{U})$ is Banach space



    what I know is :



    I'm proving this normed space



    since $||f||_{C(bar{U})}=sup_{xin U} |f(x)| =0 iff f=0$



    and $||af||_{C(bar{U})}=sup_{xin U} |af(x)| =|a| ||f||$



    finally $||f+g||_{C(bar{U})}=sup_{xin U} |f(x)+g(x)|le sup _{xin U} |f(x)|+sup _{xin U} |g(x)|$



    how to prove completeness ..thank you






    functional-analysis sobolev-spaces







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      If we define Supnorm on $C(bar{U})$ is



      $$||f||_{C(bar{U})}:=sup_{xin U} |f(x)|.$$ on $C(bar{U})$ is Banach space



      what I know is :



      I'm proving this normed space



      since $||f||_{C(bar{U})}=sup_{xin U} |f(x)| =0 iff f=0$



      and $||af||_{C(bar{U})}=sup_{xin U} |af(x)| =|a| ||f||$



      finally $||f+g||_{C(bar{U})}=sup_{xin U} |f(x)+g(x)|le sup _{xin U} |f(x)|+sup _{xin U} |g(x)|$



      how to prove completeness ..thank you






      functional-analysis sobolev-spaces







      share|cite|improve this question









      $endgroup$




      If we define Supnorm on $C(bar{U})$ is



      $$||f||_{C(bar{U})}:=sup_{xin U} |f(x)|.$$ on $C(bar{U})$ is Banach space



      what I know is :



      I'm proving this normed space



      since $||f||_{C(bar{U})}=sup_{xin U} |f(x)| =0 iff f=0$



      and $||af||_{C(bar{U})}=sup_{xin U} |af(x)| =|a| ||f||$



      finally $||f+g||_{C(bar{U})}=sup_{xin U} |f(x)+g(x)|le sup _{xin U} |f(x)|+sup _{xin U} |g(x)|$



      how to prove completeness ..thank you






      functional-analysis sobolev-spaces




      functional-analysis sobolev-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 21 at 8:16









      learnerlearner

      1188




      1188






















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          $begingroup$

          If $f_n$ is Cauchy and $epsilon >0$ there exists $k$ such that $|f_n(x)-f_m(y)| <epsilon$ for all $x in overline {U}$ $cdots$ (1), for all $n,m geq k$. In particular ${f_n(x)}$ is a Cauchy sequence for each $x$. Hence $f(x)=lim f_n(x)$ exists. Letting $m to infty$ in (1) we get $|f_n(x)-f(y)| leq epsilon$ for all $x in overline {U}$ for all $n geq k$. Hence $f_n to f$ uniformly. Uniform limits of continuous functions are continuous. Hence $f in C(overline {U})$ and $|f_n-f|_{C(overline {U})} leq epsilon $ for $n geq k$. Hence $f_n to f$ in the norm.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I don't understand last lines sir
            $endgroup$
            – learner
            Jan 21 at 12:47










          • $begingroup$
            and how can we say that if sequence is Cauchy then limit exists
            $endgroup$
            – learner
            Jan 21 at 13:03










          • $begingroup$
            and we need to show that f is uniformly continuous on bounded subset of U
            $endgroup$
            – learner
            Jan 21 at 13:06










          • $begingroup$
            $mathbb R$ is complete so $lim f_n(x)$ exists. Uniform limit of uniformly continuous functions is uniformly continuous.
            $endgroup$
            – Kavi Rama Murthy
            Jan 21 at 23:18











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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          1












          $begingroup$

          If $f_n$ is Cauchy and $epsilon >0$ there exists $k$ such that $|f_n(x)-f_m(y)| <epsilon$ for all $x in overline {U}$ $cdots$ (1), for all $n,m geq k$. In particular ${f_n(x)}$ is a Cauchy sequence for each $x$. Hence $f(x)=lim f_n(x)$ exists. Letting $m to infty$ in (1) we get $|f_n(x)-f(y)| leq epsilon$ for all $x in overline {U}$ for all $n geq k$. Hence $f_n to f$ uniformly. Uniform limits of continuous functions are continuous. Hence $f in C(overline {U})$ and $|f_n-f|_{C(overline {U})} leq epsilon $ for $n geq k$. Hence $f_n to f$ in the norm.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I don't understand last lines sir
            $endgroup$
            – learner
            Jan 21 at 12:47










          • $begingroup$
            and how can we say that if sequence is Cauchy then limit exists
            $endgroup$
            – learner
            Jan 21 at 13:03










          • $begingroup$
            and we need to show that f is uniformly continuous on bounded subset of U
            $endgroup$
            – learner
            Jan 21 at 13:06










          • $begingroup$
            $mathbb R$ is complete so $lim f_n(x)$ exists. Uniform limit of uniformly continuous functions is uniformly continuous.
            $endgroup$
            – Kavi Rama Murthy
            Jan 21 at 23:18
















          1












          $begingroup$

          If $f_n$ is Cauchy and $epsilon >0$ there exists $k$ such that $|f_n(x)-f_m(y)| <epsilon$ for all $x in overline {U}$ $cdots$ (1), for all $n,m geq k$. In particular ${f_n(x)}$ is a Cauchy sequence for each $x$. Hence $f(x)=lim f_n(x)$ exists. Letting $m to infty$ in (1) we get $|f_n(x)-f(y)| leq epsilon$ for all $x in overline {U}$ for all $n geq k$. Hence $f_n to f$ uniformly. Uniform limits of continuous functions are continuous. Hence $f in C(overline {U})$ and $|f_n-f|_{C(overline {U})} leq epsilon $ for $n geq k$. Hence $f_n to f$ in the norm.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I don't understand last lines sir
            $endgroup$
            – learner
            Jan 21 at 12:47










          • $begingroup$
            and how can we say that if sequence is Cauchy then limit exists
            $endgroup$
            – learner
            Jan 21 at 13:03










          • $begingroup$
            and we need to show that f is uniformly continuous on bounded subset of U
            $endgroup$
            – learner
            Jan 21 at 13:06










          • $begingroup$
            $mathbb R$ is complete so $lim f_n(x)$ exists. Uniform limit of uniformly continuous functions is uniformly continuous.
            $endgroup$
            – Kavi Rama Murthy
            Jan 21 at 23:18














          1












          1








          1





          $begingroup$

          If $f_n$ is Cauchy and $epsilon >0$ there exists $k$ such that $|f_n(x)-f_m(y)| <epsilon$ for all $x in overline {U}$ $cdots$ (1), for all $n,m geq k$. In particular ${f_n(x)}$ is a Cauchy sequence for each $x$. Hence $f(x)=lim f_n(x)$ exists. Letting $m to infty$ in (1) we get $|f_n(x)-f(y)| leq epsilon$ for all $x in overline {U}$ for all $n geq k$. Hence $f_n to f$ uniformly. Uniform limits of continuous functions are continuous. Hence $f in C(overline {U})$ and $|f_n-f|_{C(overline {U})} leq epsilon $ for $n geq k$. Hence $f_n to f$ in the norm.






          share|cite|improve this answer









          $endgroup$



          If $f_n$ is Cauchy and $epsilon >0$ there exists $k$ such that $|f_n(x)-f_m(y)| <epsilon$ for all $x in overline {U}$ $cdots$ (1), for all $n,m geq k$. In particular ${f_n(x)}$ is a Cauchy sequence for each $x$. Hence $f(x)=lim f_n(x)$ exists. Letting $m to infty$ in (1) we get $|f_n(x)-f(y)| leq epsilon$ for all $x in overline {U}$ for all $n geq k$. Hence $f_n to f$ uniformly. Uniform limits of continuous functions are continuous. Hence $f in C(overline {U})$ and $|f_n-f|_{C(overline {U})} leq epsilon $ for $n geq k$. Hence $f_n to f$ in the norm.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 21 at 8:23









          Kavi Rama MurthyKavi Rama Murthy

          62.2k42262




          62.2k42262












          • $begingroup$
            I don't understand last lines sir
            $endgroup$
            – learner
            Jan 21 at 12:47










          • $begingroup$
            and how can we say that if sequence is Cauchy then limit exists
            $endgroup$
            – learner
            Jan 21 at 13:03










          • $begingroup$
            and we need to show that f is uniformly continuous on bounded subset of U
            $endgroup$
            – learner
            Jan 21 at 13:06










          • $begingroup$
            $mathbb R$ is complete so $lim f_n(x)$ exists. Uniform limit of uniformly continuous functions is uniformly continuous.
            $endgroup$
            – Kavi Rama Murthy
            Jan 21 at 23:18


















          • $begingroup$
            I don't understand last lines sir
            $endgroup$
            – learner
            Jan 21 at 12:47










          • $begingroup$
            and how can we say that if sequence is Cauchy then limit exists
            $endgroup$
            – learner
            Jan 21 at 13:03










          • $begingroup$
            and we need to show that f is uniformly continuous on bounded subset of U
            $endgroup$
            – learner
            Jan 21 at 13:06










          • $begingroup$
            $mathbb R$ is complete so $lim f_n(x)$ exists. Uniform limit of uniformly continuous functions is uniformly continuous.
            $endgroup$
            – Kavi Rama Murthy
            Jan 21 at 23:18
















          $begingroup$
          I don't understand last lines sir
          $endgroup$
          – learner
          Jan 21 at 12:47




          $begingroup$
          I don't understand last lines sir
          $endgroup$
          – learner
          Jan 21 at 12:47












          $begingroup$
          and how can we say that if sequence is Cauchy then limit exists
          $endgroup$
          – learner
          Jan 21 at 13:03




          $begingroup$
          and how can we say that if sequence is Cauchy then limit exists
          $endgroup$
          – learner
          Jan 21 at 13:03












          $begingroup$
          and we need to show that f is uniformly continuous on bounded subset of U
          $endgroup$
          – learner
          Jan 21 at 13:06




          $begingroup$
          and we need to show that f is uniformly continuous on bounded subset of U
          $endgroup$
          – learner
          Jan 21 at 13:06












          $begingroup$
          $mathbb R$ is complete so $lim f_n(x)$ exists. Uniform limit of uniformly continuous functions is uniformly continuous.
          $endgroup$
          – Kavi Rama Murthy
          Jan 21 at 23:18




          $begingroup$
          $mathbb R$ is complete so $lim f_n(x)$ exists. Uniform limit of uniformly continuous functions is uniformly continuous.
          $endgroup$
          – Kavi Rama Murthy
          Jan 21 at 23:18


















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