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












1














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







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  • 1




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






  • 1




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










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


















1














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


















  • 1




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






  • 1




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










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
















1












1








1







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













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











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share|cite|improve this question










asked Dec 4 '18 at 2:41









user13985

3821721




3821721








  • 1




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






  • 1




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










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
















  • 1




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






  • 1




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










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










1




1




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




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




1




1




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




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












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






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












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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.






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    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




























      1














      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


























        1












        1








        1






        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














        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 Mosher

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