continuous-sup property and equicontinuous set.












0












$begingroup$



Suppose that $Lambda subset C^{0}$ is equicontinuous and bounded.



(a) Prove that $sup{f(x)mid f in Lambda}$ is a continuous function.



(b) Show that (a) fails without equicontinuity.



(c) Show that this continuous-sup property does not imply equicontinuity.



(d) Assume that the continuous-sup property is true for each subset $mathcal{F} subset Lambda$. Is $Lambda$ equicontinuous? Give a proof or counterexample.




My attempt.



(a) We have that
$$sup f(x) = sup (f(x)-f(y) + f(y)) leq sup f(y) + sup(f(x)-f(y)),$$
then
$$sup f(x) - sup f(y) leq sup (f(x)-f(y)) leq Vert f(x) - f(y) Vert$$
and
$$sup f(y) - sup f(x) leq sup (f(y)-f(x)) leq Vert f(x) - f(y) Vert,$$
that is,
$$|sup f(x) - sup f(y)| leq Vert f(x) - f(y) Vert.$$
Since $Lambda$ is equicontinuous, given $epsilon > 0$ there is $delta > 0$ such that $|x - y| < delta$ implies $|f(x) - f(y)| < epsilon$ for each $f in Lambda$. Therefore,
$$|x - y| < delta Longrightarrow |sup f(x) - sup f(y)| < epsilon$$




(b) I can see why without equicontinuity, (a) fails: can be exist a function $f$ that not satisfies $|f(x) - f(y)| < epsilon$. But I cannot find a counterexample.




(c) Consider $C^{0}([0,1],mathbb{R})$ and $Lambda = {f_{n}(x) = x^{n}}_{n}$. Note that
$$sup_{x in [0,1]} f(x) = 1quad forall f in Lambda.$$
Thus, $g = sup f$ is continuous, but $Lambda$ is not equicontinuous.




(d) By Mindlack's observation, the same example of (c) works in (d).




Can someone help me?






real-analysis equicontinuity







share|cite|improve this question











$endgroup$

















    0












    $begingroup$



    Suppose that $Lambda subset C^{0}$ is equicontinuous and bounded.



    (a) Prove that $sup{f(x)mid f in Lambda}$ is a continuous function.



    (b) Show that (a) fails without equicontinuity.



    (c) Show that this continuous-sup property does not imply equicontinuity.



    (d) Assume that the continuous-sup property is true for each subset $mathcal{F} subset Lambda$. Is $Lambda$ equicontinuous? Give a proof or counterexample.




    My attempt.



    (a) We have that
    $$sup f(x) = sup (f(x)-f(y) + f(y)) leq sup f(y) + sup(f(x)-f(y)),$$
    then
    $$sup f(x) - sup f(y) leq sup (f(x)-f(y)) leq Vert f(x) - f(y) Vert$$
    and
    $$sup f(y) - sup f(x) leq sup (f(y)-f(x)) leq Vert f(x) - f(y) Vert,$$
    that is,
    $$|sup f(x) - sup f(y)| leq Vert f(x) - f(y) Vert.$$
    Since $Lambda$ is equicontinuous, given $epsilon > 0$ there is $delta > 0$ such that $|x - y| < delta$ implies $|f(x) - f(y)| < epsilon$ for each $f in Lambda$. Therefore,
    $$|x - y| < delta Longrightarrow |sup f(x) - sup f(y)| < epsilon$$




    (b) I can see why without equicontinuity, (a) fails: can be exist a function $f$ that not satisfies $|f(x) - f(y)| < epsilon$. But I cannot find a counterexample.




    (c) Consider $C^{0}([0,1],mathbb{R})$ and $Lambda = {f_{n}(x) = x^{n}}_{n}$. Note that
    $$sup_{x in [0,1]} f(x) = 1quad forall f in Lambda.$$
    Thus, $g = sup f$ is continuous, but $Lambda$ is not equicontinuous.




    (d) By Mindlack's observation, the same example of (c) works in (d).




    Can someone help me?






    real-analysis equicontinuity







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Suppose that $Lambda subset C^{0}$ is equicontinuous and bounded.



      (a) Prove that $sup{f(x)mid f in Lambda}$ is a continuous function.



      (b) Show that (a) fails without equicontinuity.



      (c) Show that this continuous-sup property does not imply equicontinuity.



      (d) Assume that the continuous-sup property is true for each subset $mathcal{F} subset Lambda$. Is $Lambda$ equicontinuous? Give a proof or counterexample.




      My attempt.



      (a) We have that
      $$sup f(x) = sup (f(x)-f(y) + f(y)) leq sup f(y) + sup(f(x)-f(y)),$$
      then
      $$sup f(x) - sup f(y) leq sup (f(x)-f(y)) leq Vert f(x) - f(y) Vert$$
      and
      $$sup f(y) - sup f(x) leq sup (f(y)-f(x)) leq Vert f(x) - f(y) Vert,$$
      that is,
      $$|sup f(x) - sup f(y)| leq Vert f(x) - f(y) Vert.$$
      Since $Lambda$ is equicontinuous, given $epsilon > 0$ there is $delta > 0$ such that $|x - y| < delta$ implies $|f(x) - f(y)| < epsilon$ for each $f in Lambda$. Therefore,
      $$|x - y| < delta Longrightarrow |sup f(x) - sup f(y)| < epsilon$$




      (b) I can see why without equicontinuity, (a) fails: can be exist a function $f$ that not satisfies $|f(x) - f(y)| < epsilon$. But I cannot find a counterexample.




      (c) Consider $C^{0}([0,1],mathbb{R})$ and $Lambda = {f_{n}(x) = x^{n}}_{n}$. Note that
      $$sup_{x in [0,1]} f(x) = 1quad forall f in Lambda.$$
      Thus, $g = sup f$ is continuous, but $Lambda$ is not equicontinuous.




      (d) By Mindlack's observation, the same example of (c) works in (d).




      Can someone help me?






      real-analysis equicontinuity







      share|cite|improve this question











      $endgroup$





      Suppose that $Lambda subset C^{0}$ is equicontinuous and bounded.



      (a) Prove that $sup{f(x)mid f in Lambda}$ is a continuous function.



      (b) Show that (a) fails without equicontinuity.



      (c) Show that this continuous-sup property does not imply equicontinuity.



      (d) Assume that the continuous-sup property is true for each subset $mathcal{F} subset Lambda$. Is $Lambda$ equicontinuous? Give a proof or counterexample.




      My attempt.



      (a) We have that
      $$sup f(x) = sup (f(x)-f(y) + f(y)) leq sup f(y) + sup(f(x)-f(y)),$$
      then
      $$sup f(x) - sup f(y) leq sup (f(x)-f(y)) leq Vert f(x) - f(y) Vert$$
      and
      $$sup f(y) - sup f(x) leq sup (f(y)-f(x)) leq Vert f(x) - f(y) Vert,$$
      that is,
      $$|sup f(x) - sup f(y)| leq Vert f(x) - f(y) Vert.$$
      Since $Lambda$ is equicontinuous, given $epsilon > 0$ there is $delta > 0$ such that $|x - y| < delta$ implies $|f(x) - f(y)| < epsilon$ for each $f in Lambda$. Therefore,
      $$|x - y| < delta Longrightarrow |sup f(x) - sup f(y)| < epsilon$$




      (b) I can see why without equicontinuity, (a) fails: can be exist a function $f$ that not satisfies $|f(x) - f(y)| < epsilon$. But I cannot find a counterexample.




      (c) Consider $C^{0}([0,1],mathbb{R})$ and $Lambda = {f_{n}(x) = x^{n}}_{n}$. Note that
      $$sup_{x in [0,1]} f(x) = 1quad forall f in Lambda.$$
      Thus, $g = sup f$ is continuous, but $Lambda$ is not equicontinuous.




      (d) By Mindlack's observation, the same example of (c) works in (d).




      Can someone help me?






      real-analysis equicontinuity




      real-analysis equicontinuity






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 10 at 17:28







      Lucas Corrêa

















      asked Jan 10 at 17:11









      Lucas CorrêaLucas Corrêa

      1,6151321




      1,6151321






















          1 Answer
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          1












          $begingroup$

          $b$: try piecewise affine functions $f_n$ that are $0$ on $[0,1/2n]$ and $1$ on $[1/n,1]$.



          $d$: have you tried re-using $c$?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Of course! Seems to me that (c) works in (d), thank you!
            $endgroup$
            – Lucas Corrêa
            Jan 10 at 17:21










          • $begingroup$
            About (a), what would be the problem?
            $endgroup$
            – Lucas Corrêa
            Jan 10 at 17:22






          • 1




            $begingroup$
            Nothing, it works.
            $endgroup$
            – Mindlack
            Jan 10 at 17:24










          • $begingroup$
            All right. I'll try your suggestion for (b). Thanks!
            $endgroup$
            – Lucas Corrêa
            Jan 10 at 17:26











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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          1












          $begingroup$

          $b$: try piecewise affine functions $f_n$ that are $0$ on $[0,1/2n]$ and $1$ on $[1/n,1]$.



          $d$: have you tried re-using $c$?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Of course! Seems to me that (c) works in (d), thank you!
            $endgroup$
            – Lucas Corrêa
            Jan 10 at 17:21










          • $begingroup$
            About (a), what would be the problem?
            $endgroup$
            – Lucas Corrêa
            Jan 10 at 17:22






          • 1




            $begingroup$
            Nothing, it works.
            $endgroup$
            – Mindlack
            Jan 10 at 17:24










          • $begingroup$
            All right. I'll try your suggestion for (b). Thanks!
            $endgroup$
            – Lucas Corrêa
            Jan 10 at 17:26
















          1












          $begingroup$

          $b$: try piecewise affine functions $f_n$ that are $0$ on $[0,1/2n]$ and $1$ on $[1/n,1]$.



          $d$: have you tried re-using $c$?






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Of course! Seems to me that (c) works in (d), thank you!
            $endgroup$
            – Lucas Corrêa
            Jan 10 at 17:21










          • $begingroup$
            About (a), what would be the problem?
            $endgroup$
            – Lucas Corrêa
            Jan 10 at 17:22






          • 1




            $begingroup$
            Nothing, it works.
            $endgroup$
            – Mindlack
            Jan 10 at 17:24










          • $begingroup$
            All right. I'll try your suggestion for (b). Thanks!
            $endgroup$
            – Lucas Corrêa
            Jan 10 at 17:26














          1












          1








          1





          $begingroup$

          $b$: try piecewise affine functions $f_n$ that are $0$ on $[0,1/2n]$ and $1$ on $[1/n,1]$.



          $d$: have you tried re-using $c$?






          share|cite|improve this answer











          $endgroup$



          $b$: try piecewise affine functions $f_n$ that are $0$ on $[0,1/2n]$ and $1$ on $[1/n,1]$.



          $d$: have you tried re-using $c$?







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 10 at 17:25

























          answered Jan 10 at 17:16









          MindlackMindlack

          3,61517




          3,61517












          • $begingroup$
            Of course! Seems to me that (c) works in (d), thank you!
            $endgroup$
            – Lucas Corrêa
            Jan 10 at 17:21










          • $begingroup$
            About (a), what would be the problem?
            $endgroup$
            – Lucas Corrêa
            Jan 10 at 17:22






          • 1




            $begingroup$
            Nothing, it works.
            $endgroup$
            – Mindlack
            Jan 10 at 17:24










          • $begingroup$
            All right. I'll try your suggestion for (b). Thanks!
            $endgroup$
            – Lucas Corrêa
            Jan 10 at 17:26


















          • $begingroup$
            Of course! Seems to me that (c) works in (d), thank you!
            $endgroup$
            – Lucas Corrêa
            Jan 10 at 17:21










          • $begingroup$
            About (a), what would be the problem?
            $endgroup$
            – Lucas Corrêa
            Jan 10 at 17:22






          • 1




            $begingroup$
            Nothing, it works.
            $endgroup$
            – Mindlack
            Jan 10 at 17:24










          • $begingroup$
            All right. I'll try your suggestion for (b). Thanks!
            $endgroup$
            – Lucas Corrêa
            Jan 10 at 17:26
















          $begingroup$
          Of course! Seems to me that (c) works in (d), thank you!
          $endgroup$
          – Lucas Corrêa
          Jan 10 at 17:21




          $begingroup$
          Of course! Seems to me that (c) works in (d), thank you!
          $endgroup$
          – Lucas Corrêa
          Jan 10 at 17:21












          $begingroup$
          About (a), what would be the problem?
          $endgroup$
          – Lucas Corrêa
          Jan 10 at 17:22




          $begingroup$
          About (a), what would be the problem?
          $endgroup$
          – Lucas Corrêa
          Jan 10 at 17:22




          1




          1




          $begingroup$
          Nothing, it works.
          $endgroup$
          – Mindlack
          Jan 10 at 17:24




          $begingroup$
          Nothing, it works.
          $endgroup$
          – Mindlack
          Jan 10 at 17:24












          $begingroup$
          All right. I'll try your suggestion for (b). Thanks!
          $endgroup$
          – Lucas Corrêa
          Jan 10 at 17:26




          $begingroup$
          All right. I'll try your suggestion for (b). Thanks!
          $endgroup$
          – Lucas Corrêa
          Jan 10 at 17:26


















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