Liouville's theorem for harmonic functions












2














I was reading the proof of Liouville's theorem for harmonic functions (in $mathbb{R}^n$) in Wikipedia, but I could not understand where do they use in that proof the assumption that $f$ is bounded.



The proof -



enter image description here



Taken from - https://en.m.wikipedia.org/wiki/Harmonic_function






calculus integration multivariable-calculus harmonic-functions







share|cite|improve this question



























    2














    I was reading the proof of Liouville's theorem for harmonic functions (in $mathbb{R}^n$) in Wikipedia, but I could not understand where do they use in that proof the assumption that $f$ is bounded.



    The proof -



    enter image description here



    Taken from - https://en.m.wikipedia.org/wiki/Harmonic_function






    calculus integration multivariable-calculus harmonic-functions







    share|cite|improve this question

























      2












      2








      2







      I was reading the proof of Liouville's theorem for harmonic functions (in $mathbb{R}^n$) in Wikipedia, but I could not understand where do they use in that proof the assumption that $f$ is bounded.



      The proof -



      enter image description here



      Taken from - https://en.m.wikipedia.org/wiki/Harmonic_function






      calculus integration multivariable-calculus harmonic-functions







      share|cite|improve this question













      I was reading the proof of Liouville's theorem for harmonic functions (in $mathbb{R}^n$) in Wikipedia, but I could not understand where do they use in that proof the assumption that $f$ is bounded.



      The proof -



      enter image description here



      Taken from - https://en.m.wikipedia.org/wiki/Harmonic_function






      calculus integration multivariable-calculus harmonic-functions




      calculus integration multivariable-calculus harmonic-functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 days ago









      ChikChakChikChak

      790418




      790418






















          2 Answers
          2






          active

          oldest

          votes


















          3














          Jose's answer gives details of Nelson's original reasoning, but they aren't quite the same as the details in the proof on Wikipedia.



          The crucial point in the latter is that we assume, without loss of generality, that $f$ is a nonnegative function (we can assume this because we assumed $f$ is bounded from above or below). Then nonnegativity is used in the first displayed inequality to say that an integral over $B_r(y)$ must be at least as large as the integral over $B_R(x)$, since the latter is a subset of the former.






          share|cite|improve this answer





























            2














            The boundeness of $f$ is used in the proof of the fact that, given two points $x$ and $y$, the average value of $f$ on a large disk centered at $x$ and the average value of $f$ on a large disk (with the same radius) centered at $y$ will go to the same value as the radius goes to $infty$. When that happens, the symmetric difference of the two discs gets smaller and smaller in proportion to their intersection. Since $f$ is bounded, the average of the function on one disc is then essentially the average of the function on their intersection. Therefore, as the radius goes to infinity, the average over either disc goes to the same number.






            share|cite|improve this answer





















            • So you're saying that the average value goes to a certain value as the radius goes to $infty$ but the average value on a ball centered at $x$ is in fact fixed $f(x)$, and this average property comes from the fact that $f$ is Harmonic, no boundedness is involved. So I don't get you.
              – mouthetics
              2 days ago








            • 1




              @mouthetics Boundedness is involved when you want to say that the value at $x$ is "essentially" the average over the intersection of the discs (more specifically, that the average over the intersection tends to $f(x)$)
              – Wojowu
              2 days ago











            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%2f3062845%2fliouvilles-theorem-for-harmonic-functions%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









            3














            Jose's answer gives details of Nelson's original reasoning, but they aren't quite the same as the details in the proof on Wikipedia.



            The crucial point in the latter is that we assume, without loss of generality, that $f$ is a nonnegative function (we can assume this because we assumed $f$ is bounded from above or below). Then nonnegativity is used in the first displayed inequality to say that an integral over $B_r(y)$ must be at least as large as the integral over $B_R(x)$, since the latter is a subset of the former.






            share|cite|improve this answer


























              3














              Jose's answer gives details of Nelson's original reasoning, but they aren't quite the same as the details in the proof on Wikipedia.



              The crucial point in the latter is that we assume, without loss of generality, that $f$ is a nonnegative function (we can assume this because we assumed $f$ is bounded from above or below). Then nonnegativity is used in the first displayed inequality to say that an integral over $B_r(y)$ must be at least as large as the integral over $B_R(x)$, since the latter is a subset of the former.






              share|cite|improve this answer
























                3












                3








                3






                Jose's answer gives details of Nelson's original reasoning, but they aren't quite the same as the details in the proof on Wikipedia.



                The crucial point in the latter is that we assume, without loss of generality, that $f$ is a nonnegative function (we can assume this because we assumed $f$ is bounded from above or below). Then nonnegativity is used in the first displayed inequality to say that an integral over $B_r(y)$ must be at least as large as the integral over $B_R(x)$, since the latter is a subset of the former.






                share|cite|improve this answer












                Jose's answer gives details of Nelson's original reasoning, but they aren't quite the same as the details in the proof on Wikipedia.



                The crucial point in the latter is that we assume, without loss of generality, that $f$ is a nonnegative function (we can assume this because we assumed $f$ is bounded from above or below). Then nonnegativity is used in the first displayed inequality to say that an integral over $B_r(y)$ must be at least as large as the integral over $B_R(x)$, since the latter is a subset of the former.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                WojowuWojowu

                17.2k22565




                17.2k22565























                    2














                    The boundeness of $f$ is used in the proof of the fact that, given two points $x$ and $y$, the average value of $f$ on a large disk centered at $x$ and the average value of $f$ on a large disk (with the same radius) centered at $y$ will go to the same value as the radius goes to $infty$. When that happens, the symmetric difference of the two discs gets smaller and smaller in proportion to their intersection. Since $f$ is bounded, the average of the function on one disc is then essentially the average of the function on their intersection. Therefore, as the radius goes to infinity, the average over either disc goes to the same number.






                    share|cite|improve this answer





















                    • So you're saying that the average value goes to a certain value as the radius goes to $infty$ but the average value on a ball centered at $x$ is in fact fixed $f(x)$, and this average property comes from the fact that $f$ is Harmonic, no boundedness is involved. So I don't get you.
                      – mouthetics
                      2 days ago








                    • 1




                      @mouthetics Boundedness is involved when you want to say that the value at $x$ is "essentially" the average over the intersection of the discs (more specifically, that the average over the intersection tends to $f(x)$)
                      – Wojowu
                      2 days ago
















                    2














                    The boundeness of $f$ is used in the proof of the fact that, given two points $x$ and $y$, the average value of $f$ on a large disk centered at $x$ and the average value of $f$ on a large disk (with the same radius) centered at $y$ will go to the same value as the radius goes to $infty$. When that happens, the symmetric difference of the two discs gets smaller and smaller in proportion to their intersection. Since $f$ is bounded, the average of the function on one disc is then essentially the average of the function on their intersection. Therefore, as the radius goes to infinity, the average over either disc goes to the same number.






                    share|cite|improve this answer





















                    • So you're saying that the average value goes to a certain value as the radius goes to $infty$ but the average value on a ball centered at $x$ is in fact fixed $f(x)$, and this average property comes from the fact that $f$ is Harmonic, no boundedness is involved. So I don't get you.
                      – mouthetics
                      2 days ago








                    • 1




                      @mouthetics Boundedness is involved when you want to say that the value at $x$ is "essentially" the average over the intersection of the discs (more specifically, that the average over the intersection tends to $f(x)$)
                      – Wojowu
                      2 days ago














                    2












                    2








                    2






                    The boundeness of $f$ is used in the proof of the fact that, given two points $x$ and $y$, the average value of $f$ on a large disk centered at $x$ and the average value of $f$ on a large disk (with the same radius) centered at $y$ will go to the same value as the radius goes to $infty$. When that happens, the symmetric difference of the two discs gets smaller and smaller in proportion to their intersection. Since $f$ is bounded, the average of the function on one disc is then essentially the average of the function on their intersection. Therefore, as the radius goes to infinity, the average over either disc goes to the same number.






                    share|cite|improve this answer












                    The boundeness of $f$ is used in the proof of the fact that, given two points $x$ and $y$, the average value of $f$ on a large disk centered at $x$ and the average value of $f$ on a large disk (with the same radius) centered at $y$ will go to the same value as the radius goes to $infty$. When that happens, the symmetric difference of the two discs gets smaller and smaller in proportion to their intersection. Since $f$ is bounded, the average of the function on one disc is then essentially the average of the function on their intersection. Therefore, as the radius goes to infinity, the average over either disc goes to the same number.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 2 days ago









                    José Carlos SantosJosé Carlos Santos

                    152k22123226




                    152k22123226












                    • So you're saying that the average value goes to a certain value as the radius goes to $infty$ but the average value on a ball centered at $x$ is in fact fixed $f(x)$, and this average property comes from the fact that $f$ is Harmonic, no boundedness is involved. So I don't get you.
                      – mouthetics
                      2 days ago








                    • 1




                      @mouthetics Boundedness is involved when you want to say that the value at $x$ is "essentially" the average over the intersection of the discs (more specifically, that the average over the intersection tends to $f(x)$)
                      – Wojowu
                      2 days ago


















                    • So you're saying that the average value goes to a certain value as the radius goes to $infty$ but the average value on a ball centered at $x$ is in fact fixed $f(x)$, and this average property comes from the fact that $f$ is Harmonic, no boundedness is involved. So I don't get you.
                      – mouthetics
                      2 days ago








                    • 1




                      @mouthetics Boundedness is involved when you want to say that the value at $x$ is "essentially" the average over the intersection of the discs (more specifically, that the average over the intersection tends to $f(x)$)
                      – Wojowu
                      2 days ago
















                    So you're saying that the average value goes to a certain value as the radius goes to $infty$ but the average value on a ball centered at $x$ is in fact fixed $f(x)$, and this average property comes from the fact that $f$ is Harmonic, no boundedness is involved. So I don't get you.
                    – mouthetics
                    2 days ago






                    So you're saying that the average value goes to a certain value as the radius goes to $infty$ but the average value on a ball centered at $x$ is in fact fixed $f(x)$, and this average property comes from the fact that $f$ is Harmonic, no boundedness is involved. So I don't get you.
                    – mouthetics
                    2 days ago






                    1




                    1




                    @mouthetics Boundedness is involved when you want to say that the value at $x$ is "essentially" the average over the intersection of the discs (more specifically, that the average over the intersection tends to $f(x)$)
                    – Wojowu
                    2 days ago




                    @mouthetics Boundedness is involved when you want to say that the value at $x$ is "essentially" the average over the intersection of the discs (more specifically, that the average over the intersection tends to $f(x)$)
                    – Wojowu
                    2 days ago


















                    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%2f3062845%2fliouvilles-theorem-for-harmonic-functions%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

                    The Binding of Isaac: Rebirth/Afterbirth

                    Image
                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
                    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