Understanding why the domain of a distribution is defined to be smooth functions with compact support












2














I'm trying to understand the motivation for the definition of a distribution and am probably missing something basic. Why is the domain of distribution / generalized function defined to be the set of smooth functions with compact set support instead of the set of functions with some other property and compact support?



In particular why aren't "competitors" like the following used as the domain




  • the set of differentiable functions from $mathbb{R} to mathbb{R}$ with compact support

  • the set of piecewise linear functions with finitely many pieces with only removable discontinuities with compact support.

  • the set of piecewise constant functions with finitely many pieces with compact support


I'm wondering how you would motivate the actual definition ... and in particular how you would motivate it by considering a couple of simpler ideas and then striking them down as unworkable.





A distribution / generalized function is defined as a map from $T to mathbb{R}$ where $T$, the set of test functions, is defined as follows



$$ T stackrel{def}{=} left{ x in mathbb{R} to mathbb{R} text { where $x$ is smooth and has compact support }right} tag{1} $$



I'm picturing the dirac delta function as a motivating example for a distribution that isn't a function $mathbb{R} to mathbb{R}$ .



Here's a strawman definition of $delta$ as a function $mathbb{R} to mathbb{R} cup left{ infty right}$ . ✱✱ is used here to mark an unacceptable definition.



$$ ✱✱ ;;;delta(x) stackrel{def}{=} text{if } x = 0 text{ then } infty text{ else } 0 tag{2} $$



The definition fails to capture that $int_{-infty}^{infty} delta(x) = 1$ .



But the distributional definition of the Dirac delta is:



$$ leftlangle delta, varphi rightrangle stackrel{def}{=} varphi left(0right) tag{3a} $$
$$ deltaleft(varphiright) stackrel{def}{=} varphi left(0right) tag{3b} $$



where (3a) is conventional notation and (3b) emphasis that $delta$ is a higher order function.



And there is where I get stuck. I don't know what arguments to make to suggest that $T$ as a domain for a generalized function is the natural/correct choice.



The Wikipedia article on distributions offers the following paragraph which briefly describes what you can say about distributions whose domain is a subset of $T$, which suggests that you are doing something nontrivial when you pick $T$ .




Distribution theory reinterprets functions as linear functionals
acting on a space of test functions. Standard functions act by
integration against a test function, but many other linear functionals
do not arise in this way, and these are the "generalized functions".
There are different possible choices for the space of test functions,
leading to different spaces of distributions
. The basic space of test
function consists of smooth functions with compact support, leading to
standard distributions. Use of the space of smooth, rapidly (faster
than any polynomial increases) decreasing test functions (these
functions are called Schwartz functions) gives instead the tempered
distributions, which are important because they have a well-defined
distributional Fourier transform. Every tempered distribution is a
distribution in the normal sense, but the converse is not true: in
general the larger the space of test functions, the more restrictive
the notion of distribution. On the other hand, the use of spaces of
analytic test functions leads to Sato's theory of hyperfunctions; this
theory has a different character from the previous ones because there
are no analytic functions with non-empty compact support.




Emphasis mine.










share|cite|improve this question


















  • 2




    The derivative of a distribution is defined via $langle f',phirangle=-langle f,phi'rangle$ and to make sense of it we need $phi'$ to be in the domain.
    – user8268
    4 hours ago


















2














I'm trying to understand the motivation for the definition of a distribution and am probably missing something basic. Why is the domain of distribution / generalized function defined to be the set of smooth functions with compact set support instead of the set of functions with some other property and compact support?



In particular why aren't "competitors" like the following used as the domain




  • the set of differentiable functions from $mathbb{R} to mathbb{R}$ with compact support

  • the set of piecewise linear functions with finitely many pieces with only removable discontinuities with compact support.

  • the set of piecewise constant functions with finitely many pieces with compact support


I'm wondering how you would motivate the actual definition ... and in particular how you would motivate it by considering a couple of simpler ideas and then striking them down as unworkable.





A distribution / generalized function is defined as a map from $T to mathbb{R}$ where $T$, the set of test functions, is defined as follows



$$ T stackrel{def}{=} left{ x in mathbb{R} to mathbb{R} text { where $x$ is smooth and has compact support }right} tag{1} $$



I'm picturing the dirac delta function as a motivating example for a distribution that isn't a function $mathbb{R} to mathbb{R}$ .



Here's a strawman definition of $delta$ as a function $mathbb{R} to mathbb{R} cup left{ infty right}$ . ✱✱ is used here to mark an unacceptable definition.



$$ ✱✱ ;;;delta(x) stackrel{def}{=} text{if } x = 0 text{ then } infty text{ else } 0 tag{2} $$



The definition fails to capture that $int_{-infty}^{infty} delta(x) = 1$ .



But the distributional definition of the Dirac delta is:



$$ leftlangle delta, varphi rightrangle stackrel{def}{=} varphi left(0right) tag{3a} $$
$$ deltaleft(varphiright) stackrel{def}{=} varphi left(0right) tag{3b} $$



where (3a) is conventional notation and (3b) emphasis that $delta$ is a higher order function.



And there is where I get stuck. I don't know what arguments to make to suggest that $T$ as a domain for a generalized function is the natural/correct choice.



The Wikipedia article on distributions offers the following paragraph which briefly describes what you can say about distributions whose domain is a subset of $T$, which suggests that you are doing something nontrivial when you pick $T$ .




Distribution theory reinterprets functions as linear functionals
acting on a space of test functions. Standard functions act by
integration against a test function, but many other linear functionals
do not arise in this way, and these are the "generalized functions".
There are different possible choices for the space of test functions,
leading to different spaces of distributions
. The basic space of test
function consists of smooth functions with compact support, leading to
standard distributions. Use of the space of smooth, rapidly (faster
than any polynomial increases) decreasing test functions (these
functions are called Schwartz functions) gives instead the tempered
distributions, which are important because they have a well-defined
distributional Fourier transform. Every tempered distribution is a
distribution in the normal sense, but the converse is not true: in
general the larger the space of test functions, the more restrictive
the notion of distribution. On the other hand, the use of spaces of
analytic test functions leads to Sato's theory of hyperfunctions; this
theory has a different character from the previous ones because there
are no analytic functions with non-empty compact support.




Emphasis mine.










share|cite|improve this question


















  • 2




    The derivative of a distribution is defined via $langle f',phirangle=-langle f,phi'rangle$ and to make sense of it we need $phi'$ to be in the domain.
    – user8268
    4 hours ago
















2












2








2


1





I'm trying to understand the motivation for the definition of a distribution and am probably missing something basic. Why is the domain of distribution / generalized function defined to be the set of smooth functions with compact set support instead of the set of functions with some other property and compact support?



In particular why aren't "competitors" like the following used as the domain




  • the set of differentiable functions from $mathbb{R} to mathbb{R}$ with compact support

  • the set of piecewise linear functions with finitely many pieces with only removable discontinuities with compact support.

  • the set of piecewise constant functions with finitely many pieces with compact support


I'm wondering how you would motivate the actual definition ... and in particular how you would motivate it by considering a couple of simpler ideas and then striking them down as unworkable.





A distribution / generalized function is defined as a map from $T to mathbb{R}$ where $T$, the set of test functions, is defined as follows



$$ T stackrel{def}{=} left{ x in mathbb{R} to mathbb{R} text { where $x$ is smooth and has compact support }right} tag{1} $$



I'm picturing the dirac delta function as a motivating example for a distribution that isn't a function $mathbb{R} to mathbb{R}$ .



Here's a strawman definition of $delta$ as a function $mathbb{R} to mathbb{R} cup left{ infty right}$ . ✱✱ is used here to mark an unacceptable definition.



$$ ✱✱ ;;;delta(x) stackrel{def}{=} text{if } x = 0 text{ then } infty text{ else } 0 tag{2} $$



The definition fails to capture that $int_{-infty}^{infty} delta(x) = 1$ .



But the distributional definition of the Dirac delta is:



$$ leftlangle delta, varphi rightrangle stackrel{def}{=} varphi left(0right) tag{3a} $$
$$ deltaleft(varphiright) stackrel{def}{=} varphi left(0right) tag{3b} $$



where (3a) is conventional notation and (3b) emphasis that $delta$ is a higher order function.



And there is where I get stuck. I don't know what arguments to make to suggest that $T$ as a domain for a generalized function is the natural/correct choice.



The Wikipedia article on distributions offers the following paragraph which briefly describes what you can say about distributions whose domain is a subset of $T$, which suggests that you are doing something nontrivial when you pick $T$ .




Distribution theory reinterprets functions as linear functionals
acting on a space of test functions. Standard functions act by
integration against a test function, but many other linear functionals
do not arise in this way, and these are the "generalized functions".
There are different possible choices for the space of test functions,
leading to different spaces of distributions
. The basic space of test
function consists of smooth functions with compact support, leading to
standard distributions. Use of the space of smooth, rapidly (faster
than any polynomial increases) decreasing test functions (these
functions are called Schwartz functions) gives instead the tempered
distributions, which are important because they have a well-defined
distributional Fourier transform. Every tempered distribution is a
distribution in the normal sense, but the converse is not true: in
general the larger the space of test functions, the more restrictive
the notion of distribution. On the other hand, the use of spaces of
analytic test functions leads to Sato's theory of hyperfunctions; this
theory has a different character from the previous ones because there
are no analytic functions with non-empty compact support.




Emphasis mine.










share|cite|improve this question













I'm trying to understand the motivation for the definition of a distribution and am probably missing something basic. Why is the domain of distribution / generalized function defined to be the set of smooth functions with compact set support instead of the set of functions with some other property and compact support?



In particular why aren't "competitors" like the following used as the domain




  • the set of differentiable functions from $mathbb{R} to mathbb{R}$ with compact support

  • the set of piecewise linear functions with finitely many pieces with only removable discontinuities with compact support.

  • the set of piecewise constant functions with finitely many pieces with compact support


I'm wondering how you would motivate the actual definition ... and in particular how you would motivate it by considering a couple of simpler ideas and then striking them down as unworkable.





A distribution / generalized function is defined as a map from $T to mathbb{R}$ where $T$, the set of test functions, is defined as follows



$$ T stackrel{def}{=} left{ x in mathbb{R} to mathbb{R} text { where $x$ is smooth and has compact support }right} tag{1} $$



I'm picturing the dirac delta function as a motivating example for a distribution that isn't a function $mathbb{R} to mathbb{R}$ .



Here's a strawman definition of $delta$ as a function $mathbb{R} to mathbb{R} cup left{ infty right}$ . ✱✱ is used here to mark an unacceptable definition.



$$ ✱✱ ;;;delta(x) stackrel{def}{=} text{if } x = 0 text{ then } infty text{ else } 0 tag{2} $$



The definition fails to capture that $int_{-infty}^{infty} delta(x) = 1$ .



But the distributional definition of the Dirac delta is:



$$ leftlangle delta, varphi rightrangle stackrel{def}{=} varphi left(0right) tag{3a} $$
$$ deltaleft(varphiright) stackrel{def}{=} varphi left(0right) tag{3b} $$



where (3a) is conventional notation and (3b) emphasis that $delta$ is a higher order function.



And there is where I get stuck. I don't know what arguments to make to suggest that $T$ as a domain for a generalized function is the natural/correct choice.



The Wikipedia article on distributions offers the following paragraph which briefly describes what you can say about distributions whose domain is a subset of $T$, which suggests that you are doing something nontrivial when you pick $T$ .




Distribution theory reinterprets functions as linear functionals
acting on a space of test functions. Standard functions act by
integration against a test function, but many other linear functionals
do not arise in this way, and these are the "generalized functions".
There are different possible choices for the space of test functions,
leading to different spaces of distributions
. The basic space of test
function consists of smooth functions with compact support, leading to
standard distributions. Use of the space of smooth, rapidly (faster
than any polynomial increases) decreasing test functions (these
functions are called Schwartz functions) gives instead the tempered
distributions, which are important because they have a well-defined
distributional Fourier transform. Every tempered distribution is a
distribution in the normal sense, but the converse is not true: in
general the larger the space of test functions, the more restrictive
the notion of distribution. On the other hand, the use of spaces of
analytic test functions leads to Sato's theory of hyperfunctions; this
theory has a different character from the previous ones because there
are no analytic functions with non-empty compact support.




Emphasis mine.







distribution-theory






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 5 hours ago









Gregory Nisbet

522312




522312








  • 2




    The derivative of a distribution is defined via $langle f',phirangle=-langle f,phi'rangle$ and to make sense of it we need $phi'$ to be in the domain.
    – user8268
    4 hours ago
















  • 2




    The derivative of a distribution is defined via $langle f',phirangle=-langle f,phi'rangle$ and to make sense of it we need $phi'$ to be in the domain.
    – user8268
    4 hours ago










2




2




The derivative of a distribution is defined via $langle f',phirangle=-langle f,phi'rangle$ and to make sense of it we need $phi'$ to be in the domain.
– user8268
4 hours ago






The derivative of a distribution is defined via $langle f',phirangle=-langle f,phi'rangle$ and to make sense of it we need $phi'$ to be in the domain.
– user8268
4 hours ago












1 Answer
1






active

oldest

votes


















5














As an quick answer, you want the test functions to which you apply your distribution to be defined over a compact interval to ensure that when you integrate this function with the integral definition of your distribution, the integral is guaranteed to be finite. After all, distributions are defined to be a class of bounded linear functionals. As far as being smooth ($c^{infty}$), this is a good requirement to ensure that every derivative of your test function is absolutely continuous, so that you are able to perform integration by parts and thus compute derivatives of any order of you distribution.



It is important to be able to differentiate distributions because this enables people to formulate weak solutions to partial differential equations, where the weak solution satisfies the differential equation in the sense of distributions.






share|cite|improve this answer























    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%2f3056326%2funderstanding-why-the-domain-of-a-distribution-is-defined-to-be-smooth-functions%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









    5














    As an quick answer, you want the test functions to which you apply your distribution to be defined over a compact interval to ensure that when you integrate this function with the integral definition of your distribution, the integral is guaranteed to be finite. After all, distributions are defined to be a class of bounded linear functionals. As far as being smooth ($c^{infty}$), this is a good requirement to ensure that every derivative of your test function is absolutely continuous, so that you are able to perform integration by parts and thus compute derivatives of any order of you distribution.



    It is important to be able to differentiate distributions because this enables people to formulate weak solutions to partial differential equations, where the weak solution satisfies the differential equation in the sense of distributions.






    share|cite|improve this answer




























      5














      As an quick answer, you want the test functions to which you apply your distribution to be defined over a compact interval to ensure that when you integrate this function with the integral definition of your distribution, the integral is guaranteed to be finite. After all, distributions are defined to be a class of bounded linear functionals. As far as being smooth ($c^{infty}$), this is a good requirement to ensure that every derivative of your test function is absolutely continuous, so that you are able to perform integration by parts and thus compute derivatives of any order of you distribution.



      It is important to be able to differentiate distributions because this enables people to formulate weak solutions to partial differential equations, where the weak solution satisfies the differential equation in the sense of distributions.






      share|cite|improve this answer


























        5












        5








        5






        As an quick answer, you want the test functions to which you apply your distribution to be defined over a compact interval to ensure that when you integrate this function with the integral definition of your distribution, the integral is guaranteed to be finite. After all, distributions are defined to be a class of bounded linear functionals. As far as being smooth ($c^{infty}$), this is a good requirement to ensure that every derivative of your test function is absolutely continuous, so that you are able to perform integration by parts and thus compute derivatives of any order of you distribution.



        It is important to be able to differentiate distributions because this enables people to formulate weak solutions to partial differential equations, where the weak solution satisfies the differential equation in the sense of distributions.






        share|cite|improve this answer














        As an quick answer, you want the test functions to which you apply your distribution to be defined over a compact interval to ensure that when you integrate this function with the integral definition of your distribution, the integral is guaranteed to be finite. After all, distributions are defined to be a class of bounded linear functionals. As far as being smooth ($c^{infty}$), this is a good requirement to ensure that every derivative of your test function is absolutely continuous, so that you are able to perform integration by parts and thus compute derivatives of any order of you distribution.



        It is important to be able to differentiate distributions because this enables people to formulate weak solutions to partial differential equations, where the weak solution satisfies the differential equation in the sense of distributions.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 4 hours ago

























        answered 4 hours ago









        D.B.

        9448




        9448






























            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%2f3056326%2funderstanding-why-the-domain-of-a-distribution-is-defined-to-be-smooth-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

            What visual should I use to simply compare current year value vs last year in Power BI desktop

            Alexandru Averescu

            Trompette piccolo