Prove that $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ defines an inner product.












0












$begingroup$



Let $V$ be the vector space of all continuous functions $fcolon [-1,1]toBbb R$. Prove that $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ defines an inner product.




I'm fairly new to inner products and further linear algebra and am just trying to wrap my head around this



I understand that for $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ to be an inner product it must satisfy it being symmetric and the positive definiteness.



Where i have;
$(x,y) = (y,x)$ for all $x,y in V$



and



$(x,x) geq 0$ for all $x in V$ and $(x,x)=0$ only if $x$ is the zero vector



I assumed it would simply be plugging in but im not entirely sure.






linear-algebra vector-spaces inner-product-space







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    Let $V$ be the vector space of all continuous functions $fcolon [-1,1]toBbb R$. Prove that $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ defines an inner product.




    I'm fairly new to inner products and further linear algebra and am just trying to wrap my head around this



    I understand that for $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ to be an inner product it must satisfy it being symmetric and the positive definiteness.



    Where i have;
    $(x,y) = (y,x)$ for all $x,y in V$



    and



    $(x,x) geq 0$ for all $x in V$ and $(x,x)=0$ only if $x$ is the zero vector



    I assumed it would simply be plugging in but im not entirely sure.






    linear-algebra vector-spaces inner-product-space







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Let $V$ be the vector space of all continuous functions $fcolon [-1,1]toBbb R$. Prove that $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ defines an inner product.




      I'm fairly new to inner products and further linear algebra and am just trying to wrap my head around this



      I understand that for $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ to be an inner product it must satisfy it being symmetric and the positive definiteness.



      Where i have;
      $(x,y) = (y,x)$ for all $x,y in V$



      and



      $(x,x) geq 0$ for all $x in V$ and $(x,x)=0$ only if $x$ is the zero vector



      I assumed it would simply be plugging in but im not entirely sure.






      linear-algebra vector-spaces inner-product-space







      share|cite|improve this question











      $endgroup$





      Let $V$ be the vector space of all continuous functions $fcolon [-1,1]toBbb R$. Prove that $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ defines an inner product.




      I'm fairly new to inner products and further linear algebra and am just trying to wrap my head around this



      I understand that for $(f, g) = int_{-1}^{1} f(t)g(t) ,mathrm dt$ to be an inner product it must satisfy it being symmetric and the positive definiteness.



      Where i have;
      $(x,y) = (y,x)$ for all $x,y in V$



      and



      $(x,x) geq 0$ for all $x in V$ and $(x,x)=0$ only if $x$ is the zero vector



      I assumed it would simply be plugging in but im not entirely sure.






      linear-algebra vector-spaces inner-product-space




      linear-algebra vector-spaces inner-product-space






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 7 at 14:14









      Christoph

      11.9k1542




      11.9k1542










      asked Jan 7 at 13:18









      L GL G

      248




      248






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          We have
          $$(f,g)=int_{-1}^{1} f(t)g(t) dt=int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$.
          And we have
          $$(f,f)=int_{-1}^{1} (f(t))^2 dtgeq 0$$
          for every f by $(f(t))^2geq 0$ and the monotony of the integral.
          We now have to show that $(f,f)=0$ implies $f=0$.
          It is standard lemma that for a continuous function $ggeq 0$ we have
          $$int_{a}^{b} g(t) dt=0Rightarrow g=0.$$
          We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2geq 0$ is clear) and get
          $$0=(f,f)=int_{-1}^{1} (f(t))^2 dtRightarrow f^2=0Rightarrow f=0.$$
          If you don't know the lemma, I can give you a prove for it.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
            $endgroup$
            – L G
            Jan 7 at 13:26






          • 1




            $begingroup$
            +1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
            $endgroup$
            – Theo Bendit
            Jan 7 at 14:02













          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%2f3064995%2fprove-that-f-g-int-11-ftgt-mathrm-dt-defines-an-inner-produ%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









          2












          $begingroup$

          We have
          $$(f,g)=int_{-1}^{1} f(t)g(t) dt=int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$.
          And we have
          $$(f,f)=int_{-1}^{1} (f(t))^2 dtgeq 0$$
          for every f by $(f(t))^2geq 0$ and the monotony of the integral.
          We now have to show that $(f,f)=0$ implies $f=0$.
          It is standard lemma that for a continuous function $ggeq 0$ we have
          $$int_{a}^{b} g(t) dt=0Rightarrow g=0.$$
          We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2geq 0$ is clear) and get
          $$0=(f,f)=int_{-1}^{1} (f(t))^2 dtRightarrow f^2=0Rightarrow f=0.$$
          If you don't know the lemma, I can give you a prove for it.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
            $endgroup$
            – L G
            Jan 7 at 13:26






          • 1




            $begingroup$
            +1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
            $endgroup$
            – Theo Bendit
            Jan 7 at 14:02


















          2












          $begingroup$

          We have
          $$(f,g)=int_{-1}^{1} f(t)g(t) dt=int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$.
          And we have
          $$(f,f)=int_{-1}^{1} (f(t))^2 dtgeq 0$$
          for every f by $(f(t))^2geq 0$ and the monotony of the integral.
          We now have to show that $(f,f)=0$ implies $f=0$.
          It is standard lemma that for a continuous function $ggeq 0$ we have
          $$int_{a}^{b} g(t) dt=0Rightarrow g=0.$$
          We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2geq 0$ is clear) and get
          $$0=(f,f)=int_{-1}^{1} (f(t))^2 dtRightarrow f^2=0Rightarrow f=0.$$
          If you don't know the lemma, I can give you a prove for it.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
            $endgroup$
            – L G
            Jan 7 at 13:26






          • 1




            $begingroup$
            +1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
            $endgroup$
            – Theo Bendit
            Jan 7 at 14:02
















          2












          2








          2





          $begingroup$

          We have
          $$(f,g)=int_{-1}^{1} f(t)g(t) dt=int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$.
          And we have
          $$(f,f)=int_{-1}^{1} (f(t))^2 dtgeq 0$$
          for every f by $(f(t))^2geq 0$ and the monotony of the integral.
          We now have to show that $(f,f)=0$ implies $f=0$.
          It is standard lemma that for a continuous function $ggeq 0$ we have
          $$int_{a}^{b} g(t) dt=0Rightarrow g=0.$$
          We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2geq 0$ is clear) and get
          $$0=(f,f)=int_{-1}^{1} (f(t))^2 dtRightarrow f^2=0Rightarrow f=0.$$
          If you don't know the lemma, I can give you a prove for it.






          share|cite|improve this answer











          $endgroup$



          We have
          $$(f,g)=int_{-1}^{1} f(t)g(t) dt=int_{-1}^{1} g(t)f(t) dt=(g,f)$$ for every $f,g$.
          And we have
          $$(f,f)=int_{-1}^{1} (f(t))^2 dtgeq 0$$
          for every f by $(f(t))^2geq 0$ and the monotony of the integral.
          We now have to show that $(f,f)=0$ implies $f=0$.
          It is standard lemma that for a continuous function $ggeq 0$ we have
          $$int_{a}^{b} g(t) dt=0Rightarrow g=0.$$
          We can apply this lemma to $f^2$ ($f^2$ is continuous, because $f$ is continuous and $f^2geq 0$ is clear) and get
          $$0=(f,f)=int_{-1}^{1} (f(t))^2 dtRightarrow f^2=0Rightarrow f=0.$$
          If you don't know the lemma, I can give you a prove for it.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 7 at 13:48

























          answered Jan 7 at 13:21









          Student7Student7

          2089




          2089












          • $begingroup$
            ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
            $endgroup$
            – L G
            Jan 7 at 13:26






          • 1




            $begingroup$
            +1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
            $endgroup$
            – Theo Bendit
            Jan 7 at 14:02




















          • $begingroup$
            ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
            $endgroup$
            – L G
            Jan 7 at 13:26






          • 1




            $begingroup$
            +1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
            $endgroup$
            – Theo Bendit
            Jan 7 at 14:02


















          $begingroup$
          ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
          $endgroup$
          – L G
          Jan 7 at 13:26




          $begingroup$
          ok im not entirely sure why but that completely flew over my head, it seems simple enough thank you!
          $endgroup$
          – L G
          Jan 7 at 13:26




          1




          1




          $begingroup$
          +1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
          $endgroup$
          – Theo Bendit
          Jan 7 at 14:02






          $begingroup$
          +1 For completeness, a positive continuous function $g$ with $g(x_0) > 0$, must be at least $g(x_0) / 2 > 0$ on some $delta$-wide neighbourhood, using continuity at $x_0$ with $varepsilon = g(x_0) / 2$. Thus, by the monotonicity of the integral, the integral of $g$ must be at least $delta g(x_0) > 0$.
          $endgroup$
          – Theo Bendit
          Jan 7 at 14:02




















          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%2f3064995%2fprove-that-f-g-int-11-ftgt-mathrm-dt-defines-an-inner-produ%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