Discontinuity properties of $f_n$ carries over to the limit function $f$












3












$begingroup$



Suppose that $f_n:[a,b] rightarrow Bbb R$ and $f_n$ converges uniformly to $f$. Which of the following discontinuity properties of the functions $f_n$ carries over to the limit function ?




  • No discontinuities

  • At most ten discontinuities

  • At least ten discontinuities

  • Uncountably many discontinuities

  • Countably many discontinuities

  • No jump discontinuities

  • No oscillating discontinuities





My try :



For first bullet: If each $f_n$ is continuous and convergence is uniform, then by $varepsilon/3$ argument, $f$ is continuous, which means $f$ has no discontinuities.



For fourth bullet: To disprove this , consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; x in Bbb Q cap [0,1]\\0 &text{otherwise}end{cases}$$



Then $f_n$ is discontinuous everywhere on $[0,1]$ whereas the limit $f=0$ is continuous. Of course , the convergence is uniform



For fifth bullet: To disprove this, consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; 0<x<frac{1}{n} \\0 &text{if};x=0 wedge frac{1}{n} leq x leq 1end{cases}$$



Here $f_n$ converges uniformly to $f=0$ and each $f_n$ has discontinuous at $x=0$ and $x=frac{1}{n}$ but $f$ continuous on $[0,1]$.



This link answers the sixth bullet




Is my arguments correct ? Can I have a hint for others ?






real-analysis sequence-of-function







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$endgroup$








  • 1




    $begingroup$
    My reading of the fifth one is that "countably" means the set of discontinuities is either empty, finite, or countably infinite.
    $endgroup$
    – zhw.
    Dec 3 '18 at 18:55










  • $begingroup$
    The first bullet point is exactly about the case where $f_n$ has no discontinuities.
    $endgroup$
    – Ingix
    Dec 4 '18 at 8:35
















3












$begingroup$



Suppose that $f_n:[a,b] rightarrow Bbb R$ and $f_n$ converges uniformly to $f$. Which of the following discontinuity properties of the functions $f_n$ carries over to the limit function ?




  • No discontinuities

  • At most ten discontinuities

  • At least ten discontinuities

  • Uncountably many discontinuities

  • Countably many discontinuities

  • No jump discontinuities

  • No oscillating discontinuities





My try :



For first bullet: If each $f_n$ is continuous and convergence is uniform, then by $varepsilon/3$ argument, $f$ is continuous, which means $f$ has no discontinuities.



For fourth bullet: To disprove this , consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; x in Bbb Q cap [0,1]\\0 &text{otherwise}end{cases}$$



Then $f_n$ is discontinuous everywhere on $[0,1]$ whereas the limit $f=0$ is continuous. Of course , the convergence is uniform



For fifth bullet: To disprove this, consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; 0<x<frac{1}{n} \\0 &text{if};x=0 wedge frac{1}{n} leq x leq 1end{cases}$$



Here $f_n$ converges uniformly to $f=0$ and each $f_n$ has discontinuous at $x=0$ and $x=frac{1}{n}$ but $f$ continuous on $[0,1]$.



This link answers the sixth bullet




Is my arguments correct ? Can I have a hint for others ?






real-analysis sequence-of-function







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    My reading of the fifth one is that "countably" means the set of discontinuities is either empty, finite, or countably infinite.
    $endgroup$
    – zhw.
    Dec 3 '18 at 18:55










  • $begingroup$
    The first bullet point is exactly about the case where $f_n$ has no discontinuities.
    $endgroup$
    – Ingix
    Dec 4 '18 at 8:35














3












3








3





$begingroup$



Suppose that $f_n:[a,b] rightarrow Bbb R$ and $f_n$ converges uniformly to $f$. Which of the following discontinuity properties of the functions $f_n$ carries over to the limit function ?




  • No discontinuities

  • At most ten discontinuities

  • At least ten discontinuities

  • Uncountably many discontinuities

  • Countably many discontinuities

  • No jump discontinuities

  • No oscillating discontinuities





My try :



For first bullet: If each $f_n$ is continuous and convergence is uniform, then by $varepsilon/3$ argument, $f$ is continuous, which means $f$ has no discontinuities.



For fourth bullet: To disprove this , consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; x in Bbb Q cap [0,1]\\0 &text{otherwise}end{cases}$$



Then $f_n$ is discontinuous everywhere on $[0,1]$ whereas the limit $f=0$ is continuous. Of course , the convergence is uniform



For fifth bullet: To disprove this, consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; 0<x<frac{1}{n} \\0 &text{if};x=0 wedge frac{1}{n} leq x leq 1end{cases}$$



Here $f_n$ converges uniformly to $f=0$ and each $f_n$ has discontinuous at $x=0$ and $x=frac{1}{n}$ but $f$ continuous on $[0,1]$.



This link answers the sixth bullet




Is my arguments correct ? Can I have a hint for others ?






real-analysis sequence-of-function







share|cite|improve this question









$endgroup$





Suppose that $f_n:[a,b] rightarrow Bbb R$ and $f_n$ converges uniformly to $f$. Which of the following discontinuity properties of the functions $f_n$ carries over to the limit function ?




  • No discontinuities

  • At most ten discontinuities

  • At least ten discontinuities

  • Uncountably many discontinuities

  • Countably many discontinuities

  • No jump discontinuities

  • No oscillating discontinuities





My try :



For first bullet: If each $f_n$ is continuous and convergence is uniform, then by $varepsilon/3$ argument, $f$ is continuous, which means $f$ has no discontinuities.



For fourth bullet: To disprove this , consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; x in Bbb Q cap [0,1]\\0 &text{otherwise}end{cases}$$



Then $f_n$ is discontinuous everywhere on $[0,1]$ whereas the limit $f=0$ is continuous. Of course , the convergence is uniform



For fifth bullet: To disprove this, consider $$f_n(x)=begin{cases} frac{1}{n} & text{if}; 0<x<frac{1}{n} \\0 &text{if};x=0 wedge frac{1}{n} leq x leq 1end{cases}$$



Here $f_n$ converges uniformly to $f=0$ and each $f_n$ has discontinuous at $x=0$ and $x=frac{1}{n}$ but $f$ continuous on $[0,1]$.



This link answers the sixth bullet




Is my arguments correct ? Can I have a hint for others ?






real-analysis sequence-of-function




real-analysis sequence-of-function






share|cite|improve this question













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asked Dec 3 '18 at 18:20









Chinnapparaj RChinnapparaj R

5,3131828




5,3131828








  • 1




    $begingroup$
    My reading of the fifth one is that "countably" means the set of discontinuities is either empty, finite, or countably infinite.
    $endgroup$
    – zhw.
    Dec 3 '18 at 18:55










  • $begingroup$
    The first bullet point is exactly about the case where $f_n$ has no discontinuities.
    $endgroup$
    – Ingix
    Dec 4 '18 at 8:35














  • 1




    $begingroup$
    My reading of the fifth one is that "countably" means the set of discontinuities is either empty, finite, or countably infinite.
    $endgroup$
    – zhw.
    Dec 3 '18 at 18:55










  • $begingroup$
    The first bullet point is exactly about the case where $f_n$ has no discontinuities.
    $endgroup$
    – Ingix
    Dec 4 '18 at 8:35








1




1




$begingroup$
My reading of the fifth one is that "countably" means the set of discontinuities is either empty, finite, or countably infinite.
$endgroup$
– zhw.
Dec 3 '18 at 18:55




$begingroup$
My reading of the fifth one is that "countably" means the set of discontinuities is either empty, finite, or countably infinite.
$endgroup$
– zhw.
Dec 3 '18 at 18:55












$begingroup$
The first bullet point is exactly about the case where $f_n$ has no discontinuities.
$endgroup$
– Ingix
Dec 4 '18 at 8:35




$begingroup$
The first bullet point is exactly about the case where $f_n$ has no discontinuities.
$endgroup$
– Ingix
Dec 4 '18 at 8:35










1 Answer
1






active

oldest

votes


















1












$begingroup$

First bullet point is correct.



Hint for second: This property also propagates from the $f_n$ to $f$. The proof is similar to the proof for the first bullet point: If $f$ has a discontinuity at some $x_0$, then the $f_n$ must also have a discontinuity at $x_0$ for all $n>N$ for some suitable $N$.



Hint for third: Very similar construction to the 4th, which is correct.



Fifth bullet point is incorrect, as remarked by zhw. Your construction shows a limit $f$ that does have countably many discontinuities (namely $0$). For a hint, prove and then use that if all $f_n$ are continuous at some point $x_0$, $f$ is also continuous at $x_0$.



For the 7th point I'd like to have an exact definition of an oscillation discontinuity. Just as with the proof that non-jump discontinuities can form a jump discontinuity in the limit, by basically masking the jump discontinuity in the $f_n$ by adding a smaller and smaller oscillation, it may be possible to do the same here, depending on the exact definition.






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    $begingroup$

    First bullet point is correct.



    Hint for second: This property also propagates from the $f_n$ to $f$. The proof is similar to the proof for the first bullet point: If $f$ has a discontinuity at some $x_0$, then the $f_n$ must also have a discontinuity at $x_0$ for all $n>N$ for some suitable $N$.



    Hint for third: Very similar construction to the 4th, which is correct.



    Fifth bullet point is incorrect, as remarked by zhw. Your construction shows a limit $f$ that does have countably many discontinuities (namely $0$). For a hint, prove and then use that if all $f_n$ are continuous at some point $x_0$, $f$ is also continuous at $x_0$.



    For the 7th point I'd like to have an exact definition of an oscillation discontinuity. Just as with the proof that non-jump discontinuities can form a jump discontinuity in the limit, by basically masking the jump discontinuity in the $f_n$ by adding a smaller and smaller oscillation, it may be possible to do the same here, depending on the exact definition.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      First bullet point is correct.



      Hint for second: This property also propagates from the $f_n$ to $f$. The proof is similar to the proof for the first bullet point: If $f$ has a discontinuity at some $x_0$, then the $f_n$ must also have a discontinuity at $x_0$ for all $n>N$ for some suitable $N$.



      Hint for third: Very similar construction to the 4th, which is correct.



      Fifth bullet point is incorrect, as remarked by zhw. Your construction shows a limit $f$ that does have countably many discontinuities (namely $0$). For a hint, prove and then use that if all $f_n$ are continuous at some point $x_0$, $f$ is also continuous at $x_0$.



      For the 7th point I'd like to have an exact definition of an oscillation discontinuity. Just as with the proof that non-jump discontinuities can form a jump discontinuity in the limit, by basically masking the jump discontinuity in the $f_n$ by adding a smaller and smaller oscillation, it may be possible to do the same here, depending on the exact definition.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        First bullet point is correct.



        Hint for second: This property also propagates from the $f_n$ to $f$. The proof is similar to the proof for the first bullet point: If $f$ has a discontinuity at some $x_0$, then the $f_n$ must also have a discontinuity at $x_0$ for all $n>N$ for some suitable $N$.



        Hint for third: Very similar construction to the 4th, which is correct.



        Fifth bullet point is incorrect, as remarked by zhw. Your construction shows a limit $f$ that does have countably many discontinuities (namely $0$). For a hint, prove and then use that if all $f_n$ are continuous at some point $x_0$, $f$ is also continuous at $x_0$.



        For the 7th point I'd like to have an exact definition of an oscillation discontinuity. Just as with the proof that non-jump discontinuities can form a jump discontinuity in the limit, by basically masking the jump discontinuity in the $f_n$ by adding a smaller and smaller oscillation, it may be possible to do the same here, depending on the exact definition.






        share|cite|improve this answer









        $endgroup$



        First bullet point is correct.



        Hint for second: This property also propagates from the $f_n$ to $f$. The proof is similar to the proof for the first bullet point: If $f$ has a discontinuity at some $x_0$, then the $f_n$ must also have a discontinuity at $x_0$ for all $n>N$ for some suitable $N$.



        Hint for third: Very similar construction to the 4th, which is correct.



        Fifth bullet point is incorrect, as remarked by zhw. Your construction shows a limit $f$ that does have countably many discontinuities (namely $0$). For a hint, prove and then use that if all $f_n$ are continuous at some point $x_0$, $f$ is also continuous at $x_0$.



        For the 7th point I'd like to have an exact definition of an oscillation discontinuity. Just as with the proof that non-jump discontinuities can form a jump discontinuity in the limit, by basically masking the jump discontinuity in the $f_n$ by adding a smaller and smaller oscillation, it may be possible to do the same here, depending on the exact definition.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 '18 at 8:52









        IngixIngix

        3,409146




        3,409146






























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