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.






derivatives epsilon-delta lipschitz-functions gradient-descent







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.






derivatives epsilon-delta lipschitz-functions gradient-descent







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.






derivatives epsilon-delta lipschitz-functions gradient-descent







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




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

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
            Read more

            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

            Image
            0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
            Read more