Is Lipschitz condition on derivatives essentially the same thing as epsilon-delta proof?












1












$begingroup$


I am very confused about Lipchitz gradient/derivative. It's often used in machine learning proofs about gradient descent.



Are Lipschitz and epsilon delta proof essentially the same thing?



Lipchitz: A function $f:Drightarrow mathbb{R}$ is said to be a Lipschitz function, provided if there is a $L>0$ such that,



$$|f(x) - f(y)| le L|x-y| text{ for all } x,y in D$$
Epsilon delta proof: For all $epsilon>0$, there exists a $delta>0$ such that



$$f(x)-f(y) < epsilon text{ if } |x-y| < delta$$



The two ideas are saying the same thing: if $f(x)$ and $f(y)$ differ by this much, then $x$ and $y$ differ by that much. Am I right?



A related question posted here.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The latter is just continuity, the former is Lipschitz continuity. Lipschitz continuity is stronger.
    $endgroup$
    – Ian
    Dec 4 '18 at 2:44






  • 1




    $begingroup$
    The $epsilon$-$delta$ proof is a definition of a uniformly continuous function, which is a weaker condition than Lipschitz continuity.
    $endgroup$
    – 高田航
    Dec 4 '18 at 2:45










  • $begingroup$
    Is weaker or stronger in the sense: Lipschitz is contained in $epsilon$-$delta$ proof, or more precisely, Lipschitz implies $epsilon$-$delta$ proof?
    $endgroup$
    – user13985
    Dec 4 '18 at 3:14


















1












$begingroup$


I am very confused about Lipchitz gradient/derivative. It's often used in machine learning proofs about gradient descent.



Are Lipschitz and epsilon delta proof essentially the same thing?



Lipchitz: A function $f:Drightarrow mathbb{R}$ is said to be a Lipschitz function, provided if there is a $L>0$ such that,



$$|f(x) - f(y)| le L|x-y| text{ for all } x,y in D$$
Epsilon delta proof: For all $epsilon>0$, there exists a $delta>0$ such that



$$f(x)-f(y) < epsilon text{ if } |x-y| < delta$$



The two ideas are saying the same thing: if $f(x)$ and $f(y)$ differ by this much, then $x$ and $y$ differ by that much. Am I right?



A related question posted here.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The latter is just continuity, the former is Lipschitz continuity. Lipschitz continuity is stronger.
    $endgroup$
    – Ian
    Dec 4 '18 at 2:44






  • 1




    $begingroup$
    The $epsilon$-$delta$ proof is a definition of a uniformly continuous function, which is a weaker condition than Lipschitz continuity.
    $endgroup$
    – 高田航
    Dec 4 '18 at 2:45










  • $begingroup$
    Is weaker or stronger in the sense: Lipschitz is contained in $epsilon$-$delta$ proof, or more precisely, Lipschitz implies $epsilon$-$delta$ proof?
    $endgroup$
    – user13985
    Dec 4 '18 at 3:14
















1












1








1





$begingroup$


I am very confused about Lipchitz gradient/derivative. It's often used in machine learning proofs about gradient descent.



Are Lipschitz and epsilon delta proof essentially the same thing?



Lipchitz: A function $f:Drightarrow mathbb{R}$ is said to be a Lipschitz function, provided if there is a $L>0$ such that,



$$|f(x) - f(y)| le L|x-y| text{ for all } x,y in D$$
Epsilon delta proof: For all $epsilon>0$, there exists a $delta>0$ such that



$$f(x)-f(y) < epsilon text{ if } |x-y| < delta$$



The two ideas are saying the same thing: if $f(x)$ and $f(y)$ differ by this much, then $x$ and $y$ differ by that much. Am I right?



A related question posted here.










share|cite|improve this question









$endgroup$




I am very confused about Lipchitz gradient/derivative. It's often used in machine learning proofs about gradient descent.



Are Lipschitz and epsilon delta proof essentially the same thing?



Lipchitz: A function $f:Drightarrow mathbb{R}$ is said to be a Lipschitz function, provided if there is a $L>0$ such that,



$$|f(x) - f(y)| le L|x-y| text{ for all } x,y in D$$
Epsilon delta proof: For all $epsilon>0$, there exists a $delta>0$ such that



$$f(x)-f(y) < epsilon text{ if } |x-y| < delta$$



The two ideas are saying the same thing: if $f(x)$ and $f(y)$ differ by this much, then $x$ and $y$ differ by that much. Am I right?



A related question posted here.







derivatives epsilon-delta lipschitz-functions gradient-descent






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 4 '18 at 2:41









user13985user13985

3821721




3821721








  • 1




    $begingroup$
    The latter is just continuity, the former is Lipschitz continuity. Lipschitz continuity is stronger.
    $endgroup$
    – Ian
    Dec 4 '18 at 2:44






  • 1




    $begingroup$
    The $epsilon$-$delta$ proof is a definition of a uniformly continuous function, which is a weaker condition than Lipschitz continuity.
    $endgroup$
    – 高田航
    Dec 4 '18 at 2:45










  • $begingroup$
    Is weaker or stronger in the sense: Lipschitz is contained in $epsilon$-$delta$ proof, or more precisely, Lipschitz implies $epsilon$-$delta$ proof?
    $endgroup$
    – user13985
    Dec 4 '18 at 3:14
















  • 1




    $begingroup$
    The latter is just continuity, the former is Lipschitz continuity. Lipschitz continuity is stronger.
    $endgroup$
    – Ian
    Dec 4 '18 at 2:44






  • 1




    $begingroup$
    The $epsilon$-$delta$ proof is a definition of a uniformly continuous function, which is a weaker condition than Lipschitz continuity.
    $endgroup$
    – 高田航
    Dec 4 '18 at 2:45










  • $begingroup$
    Is weaker or stronger in the sense: Lipschitz is contained in $epsilon$-$delta$ proof, or more precisely, Lipschitz implies $epsilon$-$delta$ proof?
    $endgroup$
    – user13985
    Dec 4 '18 at 3:14










1




1




$begingroup$
The latter is just continuity, the former is Lipschitz continuity. Lipschitz continuity is stronger.
$endgroup$
– Ian
Dec 4 '18 at 2:44




$begingroup$
The latter is just continuity, the former is Lipschitz continuity. Lipschitz continuity is stronger.
$endgroup$
– Ian
Dec 4 '18 at 2:44




1




1




$begingroup$
The $epsilon$-$delta$ proof is a definition of a uniformly continuous function, which is a weaker condition than Lipschitz continuity.
$endgroup$
– 高田航
Dec 4 '18 at 2:45




$begingroup$
The $epsilon$-$delta$ proof is a definition of a uniformly continuous function, which is a weaker condition than Lipschitz continuity.
$endgroup$
– 高田航
Dec 4 '18 at 2:45












$begingroup$
Is weaker or stronger in the sense: Lipschitz is contained in $epsilon$-$delta$ proof, or more precisely, Lipschitz implies $epsilon$-$delta$ proof?
$endgroup$
– user13985
Dec 4 '18 at 3:14






$begingroup$
Is weaker or stronger in the sense: Lipschitz is contained in $epsilon$-$delta$ proof, or more precisely, Lipschitz implies $epsilon$-$delta$ proof?
$endgroup$
– user13985
Dec 4 '18 at 3:14












1 Answer
1






active

oldest

votes


















1












$begingroup$

First, a counterexample: the function $f : [0,1] to mathbb R$ given by $f(x) = sqrt{x}$ is continuous, and in fact uniformly continuous, but it is not Lipschitz.



Lipschitz continuity means that the difference ratio
$$frac{f(x)-f(y)}{x-y}
$$

is bounded. So, it says something very specific about the comparison between the difference of $f(x)$ and $f(y)$ versus the difference of $x$ and $y$.



One way you can see that the square root function defined above is not Lipschitz is that the limit of difference ratios
$$lim_{x to 0} frac{f(x)-f(0)}{x-0} = lim_{x to 0} frac{sqrt{x}}{x} = lim_{x to 0} frac{1}{sqrt{x}}
$$

is equal to infinity, and so the difference ratio $frac{f(x)-f(y)}{x-y}$ is certainly not bounded.



The $epsilon,delta$ definition says nothing at all about the ratio $frac{f(x)-f(y)}{x-y}$. Granted, though, it does say something about the difference of $f(x)$ and $f(y)$ versus the difference of $x$ and $y$, but what it says is quite a bit more vague than boundedness of the difference ratio.






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%2f3025052%2fis-lipschitz-condition-on-derivatives-essentially-the-same-thing-as-epsilon-delt%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












    $begingroup$

    First, a counterexample: the function $f : [0,1] to mathbb R$ given by $f(x) = sqrt{x}$ is continuous, and in fact uniformly continuous, but it is not Lipschitz.



    Lipschitz continuity means that the difference ratio
    $$frac{f(x)-f(y)}{x-y}
    $$

    is bounded. So, it says something very specific about the comparison between the difference of $f(x)$ and $f(y)$ versus the difference of $x$ and $y$.



    One way you can see that the square root function defined above is not Lipschitz is that the limit of difference ratios
    $$lim_{x to 0} frac{f(x)-f(0)}{x-0} = lim_{x to 0} frac{sqrt{x}}{x} = lim_{x to 0} frac{1}{sqrt{x}}
    $$

    is equal to infinity, and so the difference ratio $frac{f(x)-f(y)}{x-y}$ is certainly not bounded.



    The $epsilon,delta$ definition says nothing at all about the ratio $frac{f(x)-f(y)}{x-y}$. Granted, though, it does say something about the difference of $f(x)$ and $f(y)$ versus the difference of $x$ and $y$, but what it says is quite a bit more vague than boundedness of the difference ratio.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      First, a counterexample: the function $f : [0,1] to mathbb R$ given by $f(x) = sqrt{x}$ is continuous, and in fact uniformly continuous, but it is not Lipschitz.



      Lipschitz continuity means that the difference ratio
      $$frac{f(x)-f(y)}{x-y}
      $$

      is bounded. So, it says something very specific about the comparison between the difference of $f(x)$ and $f(y)$ versus the difference of $x$ and $y$.



      One way you can see that the square root function defined above is not Lipschitz is that the limit of difference ratios
      $$lim_{x to 0} frac{f(x)-f(0)}{x-0} = lim_{x to 0} frac{sqrt{x}}{x} = lim_{x to 0} frac{1}{sqrt{x}}
      $$

      is equal to infinity, and so the difference ratio $frac{f(x)-f(y)}{x-y}$ is certainly not bounded.



      The $epsilon,delta$ definition says nothing at all about the ratio $frac{f(x)-f(y)}{x-y}$. Granted, though, it does say something about the difference of $f(x)$ and $f(y)$ versus the difference of $x$ and $y$, but what it says is quite a bit more vague than boundedness of the difference ratio.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        First, a counterexample: the function $f : [0,1] to mathbb R$ given by $f(x) = sqrt{x}$ is continuous, and in fact uniformly continuous, but it is not Lipschitz.



        Lipschitz continuity means that the difference ratio
        $$frac{f(x)-f(y)}{x-y}
        $$

        is bounded. So, it says something very specific about the comparison between the difference of $f(x)$ and $f(y)$ versus the difference of $x$ and $y$.



        One way you can see that the square root function defined above is not Lipschitz is that the limit of difference ratios
        $$lim_{x to 0} frac{f(x)-f(0)}{x-0} = lim_{x to 0} frac{sqrt{x}}{x} = lim_{x to 0} frac{1}{sqrt{x}}
        $$

        is equal to infinity, and so the difference ratio $frac{f(x)-f(y)}{x-y}$ is certainly not bounded.



        The $epsilon,delta$ definition says nothing at all about the ratio $frac{f(x)-f(y)}{x-y}$. Granted, though, it does say something about the difference of $f(x)$ and $f(y)$ versus the difference of $x$ and $y$, but what it says is quite a bit more vague than boundedness of the difference ratio.






        share|cite|improve this answer











        $endgroup$



        First, a counterexample: the function $f : [0,1] to mathbb R$ given by $f(x) = sqrt{x}$ is continuous, and in fact uniformly continuous, but it is not Lipschitz.



        Lipschitz continuity means that the difference ratio
        $$frac{f(x)-f(y)}{x-y}
        $$

        is bounded. So, it says something very specific about the comparison between the difference of $f(x)$ and $f(y)$ versus the difference of $x$ and $y$.



        One way you can see that the square root function defined above is not Lipschitz is that the limit of difference ratios
        $$lim_{x to 0} frac{f(x)-f(0)}{x-0} = lim_{x to 0} frac{sqrt{x}}{x} = lim_{x to 0} frac{1}{sqrt{x}}
        $$

        is equal to infinity, and so the difference ratio $frac{f(x)-f(y)}{x-y}$ is certainly not bounded.



        The $epsilon,delta$ definition says nothing at all about the ratio $frac{f(x)-f(y)}{x-y}$. Granted, though, it does say something about the difference of $f(x)$ and $f(y)$ versus the difference of $x$ and $y$, but what it says is quite a bit more vague than boundedness of the difference ratio.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 4 '18 at 4:02

























        answered Dec 4 '18 at 3:55









        Lee MosherLee Mosher

        48.5k33681




        48.5k33681






























            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%2f3025052%2fis-lipschitz-condition-on-derivatives-essentially-the-same-thing-as-epsilon-delt%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

            Ellipse (mathématiques)

            Quarter-circle Tiles

            Mont Emei