Prove that $f=1/sqrt{x}$ is continuous on the interval $(0,1]$, but not uniformly continuous.












3












$begingroup$


Prove that $f(x)=1/sqrt{x}$ is continuous on the interval $(0,1]$, but not uniformly continuous.



I believe it follows that $f(x)$ is not uniformly continuous because $f(x)$ is not continuous on the bounded interval including zero. I tried to set up an $epsilon-delta$ proof of the continuity on $(0,1]$ with $delta=frac{p}{(2epsilon+1)^2}$ where $p$ is any point in the interval, but I feel like I am making a leap to say that by that $delta$ we get $|frac{sqrt p-sqrt x}{sqrt psqrt x}|ltepsilon$






real-analysis continuity







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  • 2




    $begingroup$
    Actually, it's because the slope of the tangent line approaches infinity as $xto 0$. Continuous means there is a $delta$ such that blah blah blah, where $delta$ can depend on $x$; uniform continuity means that $delta$ does not depend on $x$.
    $endgroup$
    – Christopher Carl Heckman
    Feb 29 '16 at 1:30






  • 1




    $begingroup$
    Notice that a uniformly continuous function maps Cauchy sequences into Cauchy sequences. In other words, you may disprove the uniform continuity of $f$ by choosing a suitable Cauchy sequence in $(0, 1]$ which is not mapped to a Cauchy sequence.
    $endgroup$
    – Sangchul Lee
    Feb 29 '16 at 1:39










  • $begingroup$
    But can't you also show that, once you have proved continuity on $(0,1]$, that the function is not continuous on the closed interval $[0,1]$, and thus it can not be uniformly continuous on the same interval?
    $endgroup$
    – e2DAeyePi
    Feb 29 '16 at 4:08
















3












$begingroup$


Prove that $f(x)=1/sqrt{x}$ is continuous on the interval $(0,1]$, but not uniformly continuous.



I believe it follows that $f(x)$ is not uniformly continuous because $f(x)$ is not continuous on the bounded interval including zero. I tried to set up an $epsilon-delta$ proof of the continuity on $(0,1]$ with $delta=frac{p}{(2epsilon+1)^2}$ where $p$ is any point in the interval, but I feel like I am making a leap to say that by that $delta$ we get $|frac{sqrt p-sqrt x}{sqrt psqrt x}|ltepsilon$






real-analysis continuity







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Actually, it's because the slope of the tangent line approaches infinity as $xto 0$. Continuous means there is a $delta$ such that blah blah blah, where $delta$ can depend on $x$; uniform continuity means that $delta$ does not depend on $x$.
    $endgroup$
    – Christopher Carl Heckman
    Feb 29 '16 at 1:30






  • 1




    $begingroup$
    Notice that a uniformly continuous function maps Cauchy sequences into Cauchy sequences. In other words, you may disprove the uniform continuity of $f$ by choosing a suitable Cauchy sequence in $(0, 1]$ which is not mapped to a Cauchy sequence.
    $endgroup$
    – Sangchul Lee
    Feb 29 '16 at 1:39










  • $begingroup$
    But can't you also show that, once you have proved continuity on $(0,1]$, that the function is not continuous on the closed interval $[0,1]$, and thus it can not be uniformly continuous on the same interval?
    $endgroup$
    – e2DAeyePi
    Feb 29 '16 at 4:08














3












3








3





$begingroup$


Prove that $f(x)=1/sqrt{x}$ is continuous on the interval $(0,1]$, but not uniformly continuous.



I believe it follows that $f(x)$ is not uniformly continuous because $f(x)$ is not continuous on the bounded interval including zero. I tried to set up an $epsilon-delta$ proof of the continuity on $(0,1]$ with $delta=frac{p}{(2epsilon+1)^2}$ where $p$ is any point in the interval, but I feel like I am making a leap to say that by that $delta$ we get $|frac{sqrt p-sqrt x}{sqrt psqrt x}|ltepsilon$






real-analysis continuity







share|cite|improve this question









$endgroup$




Prove that $f(x)=1/sqrt{x}$ is continuous on the interval $(0,1]$, but not uniformly continuous.



I believe it follows that $f(x)$ is not uniformly continuous because $f(x)$ is not continuous on the bounded interval including zero. I tried to set up an $epsilon-delta$ proof of the continuity on $(0,1]$ with $delta=frac{p}{(2epsilon+1)^2}$ where $p$ is any point in the interval, but I feel like I am making a leap to say that by that $delta$ we get $|frac{sqrt p-sqrt x}{sqrt psqrt x}|ltepsilon$






real-analysis continuity




real-analysis continuity






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Feb 29 '16 at 1:23









e2DAeyePie2DAeyePi

124110




124110








  • 2




    $begingroup$
    Actually, it's because the slope of the tangent line approaches infinity as $xto 0$. Continuous means there is a $delta$ such that blah blah blah, where $delta$ can depend on $x$; uniform continuity means that $delta$ does not depend on $x$.
    $endgroup$
    – Christopher Carl Heckman
    Feb 29 '16 at 1:30






  • 1




    $begingroup$
    Notice that a uniformly continuous function maps Cauchy sequences into Cauchy sequences. In other words, you may disprove the uniform continuity of $f$ by choosing a suitable Cauchy sequence in $(0, 1]$ which is not mapped to a Cauchy sequence.
    $endgroup$
    – Sangchul Lee
    Feb 29 '16 at 1:39










  • $begingroup$
    But can't you also show that, once you have proved continuity on $(0,1]$, that the function is not continuous on the closed interval $[0,1]$, and thus it can not be uniformly continuous on the same interval?
    $endgroup$
    – e2DAeyePi
    Feb 29 '16 at 4:08














  • 2




    $begingroup$
    Actually, it's because the slope of the tangent line approaches infinity as $xto 0$. Continuous means there is a $delta$ such that blah blah blah, where $delta$ can depend on $x$; uniform continuity means that $delta$ does not depend on $x$.
    $endgroup$
    – Christopher Carl Heckman
    Feb 29 '16 at 1:30






  • 1




    $begingroup$
    Notice that a uniformly continuous function maps Cauchy sequences into Cauchy sequences. In other words, you may disprove the uniform continuity of $f$ by choosing a suitable Cauchy sequence in $(0, 1]$ which is not mapped to a Cauchy sequence.
    $endgroup$
    – Sangchul Lee
    Feb 29 '16 at 1:39










  • $begingroup$
    But can't you also show that, once you have proved continuity on $(0,1]$, that the function is not continuous on the closed interval $[0,1]$, and thus it can not be uniformly continuous on the same interval?
    $endgroup$
    – e2DAeyePi
    Feb 29 '16 at 4:08








2




2




$begingroup$
Actually, it's because the slope of the tangent line approaches infinity as $xto 0$. Continuous means there is a $delta$ such that blah blah blah, where $delta$ can depend on $x$; uniform continuity means that $delta$ does not depend on $x$.
$endgroup$
– Christopher Carl Heckman
Feb 29 '16 at 1:30




$begingroup$
Actually, it's because the slope of the tangent line approaches infinity as $xto 0$. Continuous means there is a $delta$ such that blah blah blah, where $delta$ can depend on $x$; uniform continuity means that $delta$ does not depend on $x$.
$endgroup$
– Christopher Carl Heckman
Feb 29 '16 at 1:30




1




1




$begingroup$
Notice that a uniformly continuous function maps Cauchy sequences into Cauchy sequences. In other words, you may disprove the uniform continuity of $f$ by choosing a suitable Cauchy sequence in $(0, 1]$ which is not mapped to a Cauchy sequence.
$endgroup$
– Sangchul Lee
Feb 29 '16 at 1:39




$begingroup$
Notice that a uniformly continuous function maps Cauchy sequences into Cauchy sequences. In other words, you may disprove the uniform continuity of $f$ by choosing a suitable Cauchy sequence in $(0, 1]$ which is not mapped to a Cauchy sequence.
$endgroup$
– Sangchul Lee
Feb 29 '16 at 1:39












$begingroup$
But can't you also show that, once you have proved continuity on $(0,1]$, that the function is not continuous on the closed interval $[0,1]$, and thus it can not be uniformly continuous on the same interval?
$endgroup$
– e2DAeyePi
Feb 29 '16 at 4:08




$begingroup$
But can't you also show that, once you have proved continuity on $(0,1]$, that the function is not continuous on the closed interval $[0,1]$, and thus it can not be uniformly continuous on the same interval?
$endgroup$
– e2DAeyePi
Feb 29 '16 at 4:08










2 Answers
2






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oldest

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0












$begingroup$

Recall that $f$ is uniformly continuous if and only if for each $varepsilon>0$, there exists $delta>0$ such that
$$sup{|f(x)-f(y)| : |x-y|<delta}<varepsilon.$$
Consider the sequence $x_n=frac1{n^2}$. Then for any $N$,



$$
|x_N-x_{N+m}| = frac1{N^2}-frac1{(N+m)^2} stackrel{mtoinfty}longrightarrowfrac1{N^2},
$$
and
$$|f(x_N)-f(x_{N+m})|=left|frac1N-frac1{N+m} right|=frac{m}{N(N+m)}. $$
Choose $varepsilon=frac12$. Then for any $delta>0$, we may choose $N$ such that $frac1{N^2}<delta$, but
$$f(x_N)-f(x_{N+m}) = frac{m}{N(N+m)}> frac m{delta+m}stackrel{mtoinfty}longrightarrow 1,$$
so for $m$ sufficiently large, $$|f(x_N)-f(x_{N+m})|>frac12.$$ It follows that $f$ is not uniformly continuous






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Here we can use the following sequential criterion:




    If $f$ is not uniformly continuous on a given domain $D$, then we can find two sequences $(x_n)$ and $(y_n)$ in $D$ such that $lim_{ntoinfty}(x_n-y_n)=0$ and $|f(x_n)-f(y_n)|geepsilon_0$ for some fixed positive $epsilon_0.$




    Simply choose $x_n=dfrac{1}{n^2}$ and $y_n=dfrac{1}{4n^2}$ for all $ninmathbb{N}.$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      One would need the reverse implication.
      $endgroup$
      – Did
      Apr 27 '16 at 8:19












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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Recall that $f$ is uniformly continuous if and only if for each $varepsilon>0$, there exists $delta>0$ such that
    $$sup{|f(x)-f(y)| : |x-y|<delta}<varepsilon.$$
    Consider the sequence $x_n=frac1{n^2}$. Then for any $N$,



    $$
    |x_N-x_{N+m}| = frac1{N^2}-frac1{(N+m)^2} stackrel{mtoinfty}longrightarrowfrac1{N^2},
    $$
    and
    $$|f(x_N)-f(x_{N+m})|=left|frac1N-frac1{N+m} right|=frac{m}{N(N+m)}. $$
    Choose $varepsilon=frac12$. Then for any $delta>0$, we may choose $N$ such that $frac1{N^2}<delta$, but
    $$f(x_N)-f(x_{N+m}) = frac{m}{N(N+m)}> frac m{delta+m}stackrel{mtoinfty}longrightarrow 1,$$
    so for $m$ sufficiently large, $$|f(x_N)-f(x_{N+m})|>frac12.$$ It follows that $f$ is not uniformly continuous






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Recall that $f$ is uniformly continuous if and only if for each $varepsilon>0$, there exists $delta>0$ such that
      $$sup{|f(x)-f(y)| : |x-y|<delta}<varepsilon.$$
      Consider the sequence $x_n=frac1{n^2}$. Then for any $N$,



      $$
      |x_N-x_{N+m}| = frac1{N^2}-frac1{(N+m)^2} stackrel{mtoinfty}longrightarrowfrac1{N^2},
      $$
      and
      $$|f(x_N)-f(x_{N+m})|=left|frac1N-frac1{N+m} right|=frac{m}{N(N+m)}. $$
      Choose $varepsilon=frac12$. Then for any $delta>0$, we may choose $N$ such that $frac1{N^2}<delta$, but
      $$f(x_N)-f(x_{N+m}) = frac{m}{N(N+m)}> frac m{delta+m}stackrel{mtoinfty}longrightarrow 1,$$
      so for $m$ sufficiently large, $$|f(x_N)-f(x_{N+m})|>frac12.$$ It follows that $f$ is not uniformly continuous






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Recall that $f$ is uniformly continuous if and only if for each $varepsilon>0$, there exists $delta>0$ such that
        $$sup{|f(x)-f(y)| : |x-y|<delta}<varepsilon.$$
        Consider the sequence $x_n=frac1{n^2}$. Then for any $N$,



        $$
        |x_N-x_{N+m}| = frac1{N^2}-frac1{(N+m)^2} stackrel{mtoinfty}longrightarrowfrac1{N^2},
        $$
        and
        $$|f(x_N)-f(x_{N+m})|=left|frac1N-frac1{N+m} right|=frac{m}{N(N+m)}. $$
        Choose $varepsilon=frac12$. Then for any $delta>0$, we may choose $N$ such that $frac1{N^2}<delta$, but
        $$f(x_N)-f(x_{N+m}) = frac{m}{N(N+m)}> frac m{delta+m}stackrel{mtoinfty}longrightarrow 1,$$
        so for $m$ sufficiently large, $$|f(x_N)-f(x_{N+m})|>frac12.$$ It follows that $f$ is not uniformly continuous






        share|cite|improve this answer









        $endgroup$



        Recall that $f$ is uniformly continuous if and only if for each $varepsilon>0$, there exists $delta>0$ such that
        $$sup{|f(x)-f(y)| : |x-y|<delta}<varepsilon.$$
        Consider the sequence $x_n=frac1{n^2}$. Then for any $N$,



        $$
        |x_N-x_{N+m}| = frac1{N^2}-frac1{(N+m)^2} stackrel{mtoinfty}longrightarrowfrac1{N^2},
        $$
        and
        $$|f(x_N)-f(x_{N+m})|=left|frac1N-frac1{N+m} right|=frac{m}{N(N+m)}. $$
        Choose $varepsilon=frac12$. Then for any $delta>0$, we may choose $N$ such that $frac1{N^2}<delta$, but
        $$f(x_N)-f(x_{N+m}) = frac{m}{N(N+m)}> frac m{delta+m}stackrel{mtoinfty}longrightarrow 1,$$
        so for $m$ sufficiently large, $$|f(x_N)-f(x_{N+m})|>frac12.$$ It follows that $f$ is not uniformly continuous







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 29 '16 at 2:10









        Math1000Math1000

        19.4k31746




        19.4k31746























            0












            $begingroup$

            Here we can use the following sequential criterion:




            If $f$ is not uniformly continuous on a given domain $D$, then we can find two sequences $(x_n)$ and $(y_n)$ in $D$ such that $lim_{ntoinfty}(x_n-y_n)=0$ and $|f(x_n)-f(y_n)|geepsilon_0$ for some fixed positive $epsilon_0.$




            Simply choose $x_n=dfrac{1}{n^2}$ and $y_n=dfrac{1}{4n^2}$ for all $ninmathbb{N}.$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              One would need the reverse implication.
              $endgroup$
              – Did
              Apr 27 '16 at 8:19
















            0












            $begingroup$

            Here we can use the following sequential criterion:




            If $f$ is not uniformly continuous on a given domain $D$, then we can find two sequences $(x_n)$ and $(y_n)$ in $D$ such that $lim_{ntoinfty}(x_n-y_n)=0$ and $|f(x_n)-f(y_n)|geepsilon_0$ for some fixed positive $epsilon_0.$




            Simply choose $x_n=dfrac{1}{n^2}$ and $y_n=dfrac{1}{4n^2}$ for all $ninmathbb{N}.$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              One would need the reverse implication.
              $endgroup$
              – Did
              Apr 27 '16 at 8:19














            0












            0








            0





            $begingroup$

            Here we can use the following sequential criterion:




            If $f$ is not uniformly continuous on a given domain $D$, then we can find two sequences $(x_n)$ and $(y_n)$ in $D$ such that $lim_{ntoinfty}(x_n-y_n)=0$ and $|f(x_n)-f(y_n)|geepsilon_0$ for some fixed positive $epsilon_0.$




            Simply choose $x_n=dfrac{1}{n^2}$ and $y_n=dfrac{1}{4n^2}$ for all $ninmathbb{N}.$






            share|cite|improve this answer











            $endgroup$



            Here we can use the following sequential criterion:




            If $f$ is not uniformly continuous on a given domain $D$, then we can find two sequences $(x_n)$ and $(y_n)$ in $D$ such that $lim_{ntoinfty}(x_n-y_n)=0$ and $|f(x_n)-f(y_n)|geepsilon_0$ for some fixed positive $epsilon_0.$




            Simply choose $x_n=dfrac{1}{n^2}$ and $y_n=dfrac{1}{4n^2}$ for all $ninmathbb{N}.$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 27 '16 at 10:52

























            answered Apr 27 '16 at 7:52









            BumblebeeBumblebee

            9,74912551




            9,74912551












            • $begingroup$
              One would need the reverse implication.
              $endgroup$
              – Did
              Apr 27 '16 at 8:19


















            • $begingroup$
              One would need the reverse implication.
              $endgroup$
              – Did
              Apr 27 '16 at 8:19
















            $begingroup$
            One would need the reverse implication.
            $endgroup$
            – Did
            Apr 27 '16 at 8:19




            $begingroup$
            One would need the reverse implication.
            $endgroup$
            – Did
            Apr 27 '16 at 8:19


















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