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






















          1 Answer
          1






          active

          oldest

          votes


















          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











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          StackExchange.ready(function() {
          var channelOptions = {
          tags: "".split(" "),
          id: "69"
          };
          initTagRenderer("".split(" "), "".split(" "), channelOptions);

          StackExchange.using("externalEditor", function() {
          // Have to fire editor after snippets, if snippets enabled
          if (StackExchange.settings.snippets.snippetsEnabled) {
          StackExchange.using("snippets", function() {
          createEditor();
          });
          }
          else {
          createEditor();
          }
          });

          function createEditor() {
          StackExchange.prepareEditor({
          heartbeatType: 'answer',
          autoActivateHeartbeat: false,
          convertImagesToLinks: true,
          noModals: true,
          showLowRepImageUploadWarning: true,
          reputationToPostImages: 10,
          bindNavPrevention: true,
          postfix: "",
          imageUploader: {
          brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
          contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
          allowUrls: true
          },
          noCode: true, onDemand: true,
          discardSelector: ".discard-answer"
          ,immediatelyShowMarkdownHelp:true
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3081628%2fsupnorm-defined-on-set-of-all-uniformly-continuous-functions-is-banach-space%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          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


















          draft saved

          draft discarded




















































          Thanks for contributing an answer to Mathematics Stack Exchange!


          • Please be sure to answer the question. Provide details and share your research!

          But avoid …



          • Asking for help, clarification, or responding to other answers.

          • Making statements based on opinion; back them up with references or personal experience.


          Use MathJax to format equations. MathJax reference.


          To learn more, see our tips on writing great answers.




          draft saved


          draft discarded














          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3081628%2fsupnorm-defined-on-set-of-all-uniformly-continuous-functions-is-banach-space%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown





















































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown

































          Required, but never shown














          Required, but never shown












          Required, but never shown







          Required, but never shown







          -

          Popular posts from this blog

          Mario Kart Wii

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
          Read more

          What does “Dominus providebit” mean?

          Image
          5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
          Read more

          File:Tiny Toon Adventures Wacky Sports JP Title.png

          Image
          Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! File:Tiny Toon Adventures Wacky Sports JP Title.png From The Cutting Room Floor Jump to: navigation, search File File history File usage No higher resolution available. Tiny_Toon_Adventures_Wacky_Sports_JP_Title.png ‎ (160 × 144 pixels, file size: 3 KB, MIME type: image/png ) This image is a screenshot of a(n) Game Boy title screen. File history Click on a date/time to view the file as it appeared at that time. Date/Time Thumbnail Dimensions User Comment current 21:31, 20 August 2016 160 × 144 (3 KB) TheBluegeek (Talk | contribs) 17:25, 29 August 2013 160 × 144 (3 KB) Luigi-San (Talk | contribs) {{titlescreen|Game Boy}} You cannot overwrite this file. File usage The following page links to this file: Tiny Toon Advent
          Read more