Finding largest value for which $T$ maps $X$ into $X$












1












$begingroup$


Let $X_l = {f in C([0,l], mathbf{R}), 0 leq f(X) leq 2, forall x in [0,l]}$ and let $T: X rightarrow C([0,l], mathbf{R})$ be defined by $(T(f))(x) = int_0^{x} f(t)dt$



The question is to find the largest number $l_0$ such that $T$ maps $X_{l_0}$ into $X_{l_0}$. My first guess is that $l_0 = 1$ but I'm not sure how to go about showing that.










share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $X_l = {f in C([0,l], mathbf{R}), 0 leq f(X) leq 2, forall x in [0,l]}$ and let $T: X rightarrow C([0,l], mathbf{R})$ be defined by $(T(f))(x) = int_0^{x} f(t)dt$



    The question is to find the largest number $l_0$ such that $T$ maps $X_{l_0}$ into $X_{l_0}$. My first guess is that $l_0 = 1$ but I'm not sure how to go about showing that.










    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Let $X_l = {f in C([0,l], mathbf{R}), 0 leq f(X) leq 2, forall x in [0,l]}$ and let $T: X rightarrow C([0,l], mathbf{R})$ be defined by $(T(f))(x) = int_0^{x} f(t)dt$



      The question is to find the largest number $l_0$ such that $T$ maps $X_{l_0}$ into $X_{l_0}$. My first guess is that $l_0 = 1$ but I'm not sure how to go about showing that.










      share|cite|improve this question











      $endgroup$




      Let $X_l = {f in C([0,l], mathbf{R}), 0 leq f(X) leq 2, forall x in [0,l]}$ and let $T: X rightarrow C([0,l], mathbf{R})$ be defined by $(T(f))(x) = int_0^{x} f(t)dt$



      The question is to find the largest number $l_0$ such that $T$ maps $X_{l_0}$ into $X_{l_0}$. My first guess is that $l_0 = 1$ but I'm not sure how to go about showing that.







      real-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 17 '18 at 4:01









      Tianlalu

      3,09421138




      3,09421138










      asked Dec 17 '18 at 3:25









      hopelessundergradhopelessundergrad

      292




      292






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          If we suppose $T$ maps $X_l$ into $X_l$ then for any function $fin X_l$, its image $T(f)$ must also be an element of $X_l$, i.e. with codomain in the interval $[0,2]$. For the lower bound, observe that $T(f)$ is non-negative because $f$ is non-negative.



          For the upper bound let us consider the maximal function in $X_l$, the constant function defined $f_0(x)=2$ for all $xin[0,l]$. Consider its image under $T$:



          $$Big(T(f_0)Big)(x)=int_0^x 2dt = 2xquadforall xin[0,l]$$



          In particular, for $x=l$, we have $(T(f_0))(l)=2l$. If we insist that $T(f_0)$ is an element of $X_l$, we must have $2lleq 2$, i.e. $lleq 1$.



          Now it suffices to show that $l=1$ is indeed a solution. Let $f$ be any continuous function from $[0,1]$ to $[0,2]$. Then for all $xin[0,1]$ we have the following chain of inequalities:



          $$Big(T(f)Big)(x)=int_0^x f(t)dtleqint_0^1 f(t)dtleqint_0^1 2dt=2$$



          Thus $T(f)in X_l$ and we have that $l_0=1$ is the maximum value of $l$ for which $T(X_l)subseteq X_l$.






          share|cite|improve this answer











          $endgroup$













            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%2f3043497%2ffinding-largest-value-for-which-t-maps-x-into-x%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 we suppose $T$ maps $X_l$ into $X_l$ then for any function $fin X_l$, its image $T(f)$ must also be an element of $X_l$, i.e. with codomain in the interval $[0,2]$. For the lower bound, observe that $T(f)$ is non-negative because $f$ is non-negative.



            For the upper bound let us consider the maximal function in $X_l$, the constant function defined $f_0(x)=2$ for all $xin[0,l]$. Consider its image under $T$:



            $$Big(T(f_0)Big)(x)=int_0^x 2dt = 2xquadforall xin[0,l]$$



            In particular, for $x=l$, we have $(T(f_0))(l)=2l$. If we insist that $T(f_0)$ is an element of $X_l$, we must have $2lleq 2$, i.e. $lleq 1$.



            Now it suffices to show that $l=1$ is indeed a solution. Let $f$ be any continuous function from $[0,1]$ to $[0,2]$. Then for all $xin[0,1]$ we have the following chain of inequalities:



            $$Big(T(f)Big)(x)=int_0^x f(t)dtleqint_0^1 f(t)dtleqint_0^1 2dt=2$$



            Thus $T(f)in X_l$ and we have that $l_0=1$ is the maximum value of $l$ for which $T(X_l)subseteq X_l$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              If we suppose $T$ maps $X_l$ into $X_l$ then for any function $fin X_l$, its image $T(f)$ must also be an element of $X_l$, i.e. with codomain in the interval $[0,2]$. For the lower bound, observe that $T(f)$ is non-negative because $f$ is non-negative.



              For the upper bound let us consider the maximal function in $X_l$, the constant function defined $f_0(x)=2$ for all $xin[0,l]$. Consider its image under $T$:



              $$Big(T(f_0)Big)(x)=int_0^x 2dt = 2xquadforall xin[0,l]$$



              In particular, for $x=l$, we have $(T(f_0))(l)=2l$. If we insist that $T(f_0)$ is an element of $X_l$, we must have $2lleq 2$, i.e. $lleq 1$.



              Now it suffices to show that $l=1$ is indeed a solution. Let $f$ be any continuous function from $[0,1]$ to $[0,2]$. Then for all $xin[0,1]$ we have the following chain of inequalities:



              $$Big(T(f)Big)(x)=int_0^x f(t)dtleqint_0^1 f(t)dtleqint_0^1 2dt=2$$



              Thus $T(f)in X_l$ and we have that $l_0=1$ is the maximum value of $l$ for which $T(X_l)subseteq X_l$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                If we suppose $T$ maps $X_l$ into $X_l$ then for any function $fin X_l$, its image $T(f)$ must also be an element of $X_l$, i.e. with codomain in the interval $[0,2]$. For the lower bound, observe that $T(f)$ is non-negative because $f$ is non-negative.



                For the upper bound let us consider the maximal function in $X_l$, the constant function defined $f_0(x)=2$ for all $xin[0,l]$. Consider its image under $T$:



                $$Big(T(f_0)Big)(x)=int_0^x 2dt = 2xquadforall xin[0,l]$$



                In particular, for $x=l$, we have $(T(f_0))(l)=2l$. If we insist that $T(f_0)$ is an element of $X_l$, we must have $2lleq 2$, i.e. $lleq 1$.



                Now it suffices to show that $l=1$ is indeed a solution. Let $f$ be any continuous function from $[0,1]$ to $[0,2]$. Then for all $xin[0,1]$ we have the following chain of inequalities:



                $$Big(T(f)Big)(x)=int_0^x f(t)dtleqint_0^1 f(t)dtleqint_0^1 2dt=2$$



                Thus $T(f)in X_l$ and we have that $l_0=1$ is the maximum value of $l$ for which $T(X_l)subseteq X_l$.






                share|cite|improve this answer











                $endgroup$



                If we suppose $T$ maps $X_l$ into $X_l$ then for any function $fin X_l$, its image $T(f)$ must also be an element of $X_l$, i.e. with codomain in the interval $[0,2]$. For the lower bound, observe that $T(f)$ is non-negative because $f$ is non-negative.



                For the upper bound let us consider the maximal function in $X_l$, the constant function defined $f_0(x)=2$ for all $xin[0,l]$. Consider its image under $T$:



                $$Big(T(f_0)Big)(x)=int_0^x 2dt = 2xquadforall xin[0,l]$$



                In particular, for $x=l$, we have $(T(f_0))(l)=2l$. If we insist that $T(f_0)$ is an element of $X_l$, we must have $2lleq 2$, i.e. $lleq 1$.



                Now it suffices to show that $l=1$ is indeed a solution. Let $f$ be any continuous function from $[0,1]$ to $[0,2]$. Then for all $xin[0,1]$ we have the following chain of inequalities:



                $$Big(T(f)Big)(x)=int_0^x f(t)dtleqint_0^1 f(t)dtleqint_0^1 2dt=2$$



                Thus $T(f)in X_l$ and we have that $l_0=1$ is the maximum value of $l$ for which $T(X_l)subseteq X_l$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 17 '18 at 5:33

























                answered Dec 17 '18 at 4:30









                M. NestorM. Nestor

                793113




                793113






























                    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%2f3043497%2ffinding-largest-value-for-which-t-maps-x-into-x%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

                    Plaza Victoria

                    Puebla de Zaragoza

                    Musa