Reason for the term “smooth”

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
8
down vote

favorite
2












A normed space $X$ is said to be smooth if for $x in X$ with $||x||=1$ there exists a unique bounded linear functional $f$ such that $||f||=1$ and $f(x)=||x||$. Why the term "smooth" comes?










share|cite|improve this question

























    up vote
    8
    down vote

    favorite
    2












    A normed space $X$ is said to be smooth if for $x in X$ with $||x||=1$ there exists a unique bounded linear functional $f$ such that $||f||=1$ and $f(x)=||x||$. Why the term "smooth" comes?










    share|cite|improve this question























      up vote
      8
      down vote

      favorite
      2









      up vote
      8
      down vote

      favorite
      2






      2





      A normed space $X$ is said to be smooth if for $x in X$ with $||x||=1$ there exists a unique bounded linear functional $f$ such that $||f||=1$ and $f(x)=||x||$. Why the term "smooth" comes?










      share|cite|improve this question













      A normed space $X$ is said to be smooth if for $x in X$ with $||x||=1$ there exists a unique bounded linear functional $f$ such that $||f||=1$ and $f(x)=||x||$. Why the term "smooth" comes?







      functional-analysis banach-spaces normed-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 28 at 8:31









      Infinity

      580313




      580313




















          2 Answers
          2






          active

          oldest

          votes

















          up vote
          10
          down vote



          accepted










          Let us consider the space $X=mathbb R^2$ with the $ell^p$-norm.



          Then for $p=1$ or $p=infty$ we can see that the unit ball has kinks, and does not look smooth.
          It can be shown, that there are points $xin X$ such that there is more than one functional $f$
          with $|f|=1$ and $f(x)=|x|=1$.
          (For example, if $p=1$ consider the point $x=(1,0)$, then $g(x)=x_1$ and $h(x)=x_1+x_2$ are possible choices for $f$.)



          For $p$ with $1<p<infty$ the unit ball looks smooth (its boundary is differentiable).
          And it is possible to show that for each $xin X$ there is exactly one functional $f$
          with $|f|=f(x)=|x|=1$.



          I hope this is sufficient motivation for the term "smooth" for a normed space.






          share|cite|improve this answer



























            up vote
            6
            down vote













            This condition implies that the function $F colon x mapsto |x|$ is (Gâteaux) differentiable in $x$ with $F'(x) = f$.
            Indeed, the set
            $$
            x$$

            coincides with the convex subdifferential of $F$ at $x$.
            If this subdifferential is a singleton, then $F$ is differentiable.






            share|cite|improve this answer






















            • $F'(x)=f$ is a typo?
              – David C. Ullrich
              Nov 28 at 15:41










            • No, the derivative of $F$ at $x$ is the linear function $f$.
              – gerw
              Nov 28 at 18:01










            • The derivative of $F$ at $x$ is $f$? What is $f$??? (Oh: where $f$ is as in the question. Ok, sorry...)
              – David C. Ullrich
              Nov 28 at 18:31











            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',
            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%2f3016897%2freason-for-the-term-smooth%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            10
            down vote



            accepted










            Let us consider the space $X=mathbb R^2$ with the $ell^p$-norm.



            Then for $p=1$ or $p=infty$ we can see that the unit ball has kinks, and does not look smooth.
            It can be shown, that there are points $xin X$ such that there is more than one functional $f$
            with $|f|=1$ and $f(x)=|x|=1$.
            (For example, if $p=1$ consider the point $x=(1,0)$, then $g(x)=x_1$ and $h(x)=x_1+x_2$ are possible choices for $f$.)



            For $p$ with $1<p<infty$ the unit ball looks smooth (its boundary is differentiable).
            And it is possible to show that for each $xin X$ there is exactly one functional $f$
            with $|f|=f(x)=|x|=1$.



            I hope this is sufficient motivation for the term "smooth" for a normed space.






            share|cite|improve this answer
























              up vote
              10
              down vote



              accepted










              Let us consider the space $X=mathbb R^2$ with the $ell^p$-norm.



              Then for $p=1$ or $p=infty$ we can see that the unit ball has kinks, and does not look smooth.
              It can be shown, that there are points $xin X$ such that there is more than one functional $f$
              with $|f|=1$ and $f(x)=|x|=1$.
              (For example, if $p=1$ consider the point $x=(1,0)$, then $g(x)=x_1$ and $h(x)=x_1+x_2$ are possible choices for $f$.)



              For $p$ with $1<p<infty$ the unit ball looks smooth (its boundary is differentiable).
              And it is possible to show that for each $xin X$ there is exactly one functional $f$
              with $|f|=f(x)=|x|=1$.



              I hope this is sufficient motivation for the term "smooth" for a normed space.






              share|cite|improve this answer






















                up vote
                10
                down vote



                accepted







                up vote
                10
                down vote



                accepted






                Let us consider the space $X=mathbb R^2$ with the $ell^p$-norm.



                Then for $p=1$ or $p=infty$ we can see that the unit ball has kinks, and does not look smooth.
                It can be shown, that there are points $xin X$ such that there is more than one functional $f$
                with $|f|=1$ and $f(x)=|x|=1$.
                (For example, if $p=1$ consider the point $x=(1,0)$, then $g(x)=x_1$ and $h(x)=x_1+x_2$ are possible choices for $f$.)



                For $p$ with $1<p<infty$ the unit ball looks smooth (its boundary is differentiable).
                And it is possible to show that for each $xin X$ there is exactly one functional $f$
                with $|f|=f(x)=|x|=1$.



                I hope this is sufficient motivation for the term "smooth" for a normed space.






                share|cite|improve this answer












                Let us consider the space $X=mathbb R^2$ with the $ell^p$-norm.



                Then for $p=1$ or $p=infty$ we can see that the unit ball has kinks, and does not look smooth.
                It can be shown, that there are points $xin X$ such that there is more than one functional $f$
                with $|f|=1$ and $f(x)=|x|=1$.
                (For example, if $p=1$ consider the point $x=(1,0)$, then $g(x)=x_1$ and $h(x)=x_1+x_2$ are possible choices for $f$.)



                For $p$ with $1<p<infty$ the unit ball looks smooth (its boundary is differentiable).
                And it is possible to show that for each $xin X$ there is exactly one functional $f$
                with $|f|=f(x)=|x|=1$.



                I hope this is sufficient motivation for the term "smooth" for a normed space.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 28 at 9:09









                supinf

                5,8181027




                5,8181027




















                    up vote
                    6
                    down vote













                    This condition implies that the function $F colon x mapsto |x|$ is (Gâteaux) differentiable in $x$ with $F'(x) = f$.
                    Indeed, the set
                    $$
                    x$$

                    coincides with the convex subdifferential of $F$ at $x$.
                    If this subdifferential is a singleton, then $F$ is differentiable.






                    share|cite|improve this answer






















                    • $F'(x)=f$ is a typo?
                      – David C. Ullrich
                      Nov 28 at 15:41










                    • No, the derivative of $F$ at $x$ is the linear function $f$.
                      – gerw
                      Nov 28 at 18:01










                    • The derivative of $F$ at $x$ is $f$? What is $f$??? (Oh: where $f$ is as in the question. Ok, sorry...)
                      – David C. Ullrich
                      Nov 28 at 18:31















                    up vote
                    6
                    down vote













                    This condition implies that the function $F colon x mapsto |x|$ is (Gâteaux) differentiable in $x$ with $F'(x) = f$.
                    Indeed, the set
                    $$
                    x$$

                    coincides with the convex subdifferential of $F$ at $x$.
                    If this subdifferential is a singleton, then $F$ is differentiable.






                    share|cite|improve this answer






















                    • $F'(x)=f$ is a typo?
                      – David C. Ullrich
                      Nov 28 at 15:41










                    • No, the derivative of $F$ at $x$ is the linear function $f$.
                      – gerw
                      Nov 28 at 18:01










                    • The derivative of $F$ at $x$ is $f$? What is $f$??? (Oh: where $f$ is as in the question. Ok, sorry...)
                      – David C. Ullrich
                      Nov 28 at 18:31













                    up vote
                    6
                    down vote










                    up vote
                    6
                    down vote









                    This condition implies that the function $F colon x mapsto |x|$ is (Gâteaux) differentiable in $x$ with $F'(x) = f$.
                    Indeed, the set
                    $$
                    x$$

                    coincides with the convex subdifferential of $F$ at $x$.
                    If this subdifferential is a singleton, then $F$ is differentiable.






                    share|cite|improve this answer














                    This condition implies that the function $F colon x mapsto |x|$ is (Gâteaux) differentiable in $x$ with $F'(x) = f$.
                    Indeed, the set
                    $$
                    x$$

                    coincides with the convex subdifferential of $F$ at $x$.
                    If this subdifferential is a singleton, then $F$ is differentiable.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 28 at 21:59

























                    answered Nov 28 at 10:31









                    gerw

                    18.9k11133




                    18.9k11133











                    • $F'(x)=f$ is a typo?
                      – David C. Ullrich
                      Nov 28 at 15:41










                    • No, the derivative of $F$ at $x$ is the linear function $f$.
                      – gerw
                      Nov 28 at 18:01










                    • The derivative of $F$ at $x$ is $f$? What is $f$??? (Oh: where $f$ is as in the question. Ok, sorry...)
                      – David C. Ullrich
                      Nov 28 at 18:31

















                    • $F'(x)=f$ is a typo?
                      – David C. Ullrich
                      Nov 28 at 15:41










                    • No, the derivative of $F$ at $x$ is the linear function $f$.
                      – gerw
                      Nov 28 at 18:01










                    • The derivative of $F$ at $x$ is $f$? What is $f$??? (Oh: where $f$ is as in the question. Ok, sorry...)
                      – David C. Ullrich
                      Nov 28 at 18:31
















                    $F'(x)=f$ is a typo?
                    – David C. Ullrich
                    Nov 28 at 15:41




                    $F'(x)=f$ is a typo?
                    – David C. Ullrich
                    Nov 28 at 15:41












                    No, the derivative of $F$ at $x$ is the linear function $f$.
                    – gerw
                    Nov 28 at 18:01




                    No, the derivative of $F$ at $x$ is the linear function $f$.
                    – gerw
                    Nov 28 at 18:01












                    The derivative of $F$ at $x$ is $f$? What is $f$??? (Oh: where $f$ is as in the question. Ok, sorry...)
                    – David C. Ullrich
                    Nov 28 at 18:31





                    The derivative of $F$ at $x$ is $f$? What is $f$??? (Oh: where $f$ is as in the question. Ok, sorry...)
                    – David C. Ullrich
                    Nov 28 at 18:31


















                    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.





                    Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                    Please pay close attention to the following guidance:


                    • 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.

                    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%2f3016897%2freason-for-the-term-smooth%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

                    How to check contact read email or not when send email to Individual?

                    Bahrain

                    Postfix configuration issue with fips on centos 7; mailgun relay