Show that the following function is continuous on $[0,1].$












2












$begingroup$


Define
$$omega_f(delta) =sup {|f(x)-f(y)|: (x,y)in [0,1]^{2}text{ and } |x-y|leq delta}$$
where $fin mathcal{C}([0,1])$ and $deltageq 0.$ I have proven that for all $delta_1,delta_2geq 0$ we have that:
$$omega_f(delta_1)leq omega_{f}(delta_1+delta_2)leq omega_f(delta_1)+omega_f(delta_2).$$
I now want to show that $omega_f(delta)$ is continuous for all $deltain [0,1].$ I have argued for continuity at $0$ using contradiction and the fact that $omega_f$ is monotonic. Furthermore, for $rin (0,1]$ and $hgeq 0$ we have that
$$omega_f(r) leq omega_f(r+h)leq omega_f(r)+omega_f (h).$$
If we send $hto 0$ we get that $omega_f(r+h)to omega_f(r)$ and so $omega_f$ is right continuous. However, I am not sure how to show $omega_f$ is left-continuous. I tried the following:
$$omega_f(h)leq omega_{f}(h+(r-h))leq omega_f(h)+omega_f(r-h)$$
$$implies omega_{f}(r)-omega_f(h)leq omega_f(r-h)$$
and similarily since,
$$omega_f(r-h)leq omega_{f}((r-h)+h)leq omega_f(r-h)+omega_f(h)$$
$$implies omega_{f}(r-h)leq omega_f(r).$$
Thus we have that
$$omega_f(r)-omega_f(h)leq omega_f(r-h)leq omega_f(r).$$
Sending $hto 0$ should give left continuity by squeeze theorem.



Is this reasoning correct?






real-analysis functional-analysis continuity







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$endgroup$

















    2












    $begingroup$


    Define
    $$omega_f(delta) =sup {|f(x)-f(y)|: (x,y)in [0,1]^{2}text{ and } |x-y|leq delta}$$
    where $fin mathcal{C}([0,1])$ and $deltageq 0.$ I have proven that for all $delta_1,delta_2geq 0$ we have that:
    $$omega_f(delta_1)leq omega_{f}(delta_1+delta_2)leq omega_f(delta_1)+omega_f(delta_2).$$
    I now want to show that $omega_f(delta)$ is continuous for all $deltain [0,1].$ I have argued for continuity at $0$ using contradiction and the fact that $omega_f$ is monotonic. Furthermore, for $rin (0,1]$ and $hgeq 0$ we have that
    $$omega_f(r) leq omega_f(r+h)leq omega_f(r)+omega_f (h).$$
    If we send $hto 0$ we get that $omega_f(r+h)to omega_f(r)$ and so $omega_f$ is right continuous. However, I am not sure how to show $omega_f$ is left-continuous. I tried the following:
    $$omega_f(h)leq omega_{f}(h+(r-h))leq omega_f(h)+omega_f(r-h)$$
    $$implies omega_{f}(r)-omega_f(h)leq omega_f(r-h)$$
    and similarily since,
    $$omega_f(r-h)leq omega_{f}((r-h)+h)leq omega_f(r-h)+omega_f(h)$$
    $$implies omega_{f}(r-h)leq omega_f(r).$$
    Thus we have that
    $$omega_f(r)-omega_f(h)leq omega_f(r-h)leq omega_f(r).$$
    Sending $hto 0$ should give left continuity by squeeze theorem.



    Is this reasoning correct?






    real-analysis functional-analysis continuity







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Define
      $$omega_f(delta) =sup {|f(x)-f(y)|: (x,y)in [0,1]^{2}text{ and } |x-y|leq delta}$$
      where $fin mathcal{C}([0,1])$ and $deltageq 0.$ I have proven that for all $delta_1,delta_2geq 0$ we have that:
      $$omega_f(delta_1)leq omega_{f}(delta_1+delta_2)leq omega_f(delta_1)+omega_f(delta_2).$$
      I now want to show that $omega_f(delta)$ is continuous for all $deltain [0,1].$ I have argued for continuity at $0$ using contradiction and the fact that $omega_f$ is monotonic. Furthermore, for $rin (0,1]$ and $hgeq 0$ we have that
      $$omega_f(r) leq omega_f(r+h)leq omega_f(r)+omega_f (h).$$
      If we send $hto 0$ we get that $omega_f(r+h)to omega_f(r)$ and so $omega_f$ is right continuous. However, I am not sure how to show $omega_f$ is left-continuous. I tried the following:
      $$omega_f(h)leq omega_{f}(h+(r-h))leq omega_f(h)+omega_f(r-h)$$
      $$implies omega_{f}(r)-omega_f(h)leq omega_f(r-h)$$
      and similarily since,
      $$omega_f(r-h)leq omega_{f}((r-h)+h)leq omega_f(r-h)+omega_f(h)$$
      $$implies omega_{f}(r-h)leq omega_f(r).$$
      Thus we have that
      $$omega_f(r)-omega_f(h)leq omega_f(r-h)leq omega_f(r).$$
      Sending $hto 0$ should give left continuity by squeeze theorem.



      Is this reasoning correct?






      real-analysis functional-analysis continuity







      share|cite|improve this question











      $endgroup$




      Define
      $$omega_f(delta) =sup {|f(x)-f(y)|: (x,y)in [0,1]^{2}text{ and } |x-y|leq delta}$$
      where $fin mathcal{C}([0,1])$ and $deltageq 0.$ I have proven that for all $delta_1,delta_2geq 0$ we have that:
      $$omega_f(delta_1)leq omega_{f}(delta_1+delta_2)leq omega_f(delta_1)+omega_f(delta_2).$$
      I now want to show that $omega_f(delta)$ is continuous for all $deltain [0,1].$ I have argued for continuity at $0$ using contradiction and the fact that $omega_f$ is monotonic. Furthermore, for $rin (0,1]$ and $hgeq 0$ we have that
      $$omega_f(r) leq omega_f(r+h)leq omega_f(r)+omega_f (h).$$
      If we send $hto 0$ we get that $omega_f(r+h)to omega_f(r)$ and so $omega_f$ is right continuous. However, I am not sure how to show $omega_f$ is left-continuous. I tried the following:
      $$omega_f(h)leq omega_{f}(h+(r-h))leq omega_f(h)+omega_f(r-h)$$
      $$implies omega_{f}(r)-omega_f(h)leq omega_f(r-h)$$
      and similarily since,
      $$omega_f(r-h)leq omega_{f}((r-h)+h)leq omega_f(r-h)+omega_f(h)$$
      $$implies omega_{f}(r-h)leq omega_f(r).$$
      Thus we have that
      $$omega_f(r)-omega_f(h)leq omega_f(r-h)leq omega_f(r).$$
      Sending $hto 0$ should give left continuity by squeeze theorem.



      Is this reasoning correct?






      real-analysis functional-analysis continuity




      real-analysis functional-analysis continuity






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      edited Jan 7 at 22:59









      Matematleta

      10.2k2918




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      asked Jan 7 at 19:35









      Hello_WorldHello_World

      4,16221631




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          1 Answer
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          $begingroup$

          Monotonicity alone of $omega_f$ is not enough to yield continuity. You have to use that $f$ is (uniformly) continuous.



          A small (maybe pedantic) point of notation: $omega_f(delta)$ is a real number, so it cannot be continuous. What you meant was that you wanted to prove that $omega_f$ was continuous on $[0,1]$, or, if you really want to write $omega_f(delta)$, write that $omega_f(delta)$ is continuous at every $delta in [0,1]$.



          Otherwise, it seems correct to me (however, you that $omega_f(r-h) leq omega_f(r)$ is a straight consequence of its monotonicity).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes I used uniform continuity for showing $omega_f$ is continuous at $0.$ Do you suggest that I need it also for $rin (0,1].$
            $endgroup$
            – Hello_World
            Jan 7 at 19:58










          • $begingroup$
            Not at all, the « triangle inequality » and squeezing theorem work well enough.
            $endgroup$
            – Mindlack
            Jan 7 at 19:59











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          0












          $begingroup$

          Monotonicity alone of $omega_f$ is not enough to yield continuity. You have to use that $f$ is (uniformly) continuous.



          A small (maybe pedantic) point of notation: $omega_f(delta)$ is a real number, so it cannot be continuous. What you meant was that you wanted to prove that $omega_f$ was continuous on $[0,1]$, or, if you really want to write $omega_f(delta)$, write that $omega_f(delta)$ is continuous at every $delta in [0,1]$.



          Otherwise, it seems correct to me (however, you that $omega_f(r-h) leq omega_f(r)$ is a straight consequence of its monotonicity).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes I used uniform continuity for showing $omega_f$ is continuous at $0.$ Do you suggest that I need it also for $rin (0,1].$
            $endgroup$
            – Hello_World
            Jan 7 at 19:58










          • $begingroup$
            Not at all, the « triangle inequality » and squeezing theorem work well enough.
            $endgroup$
            – Mindlack
            Jan 7 at 19:59
















          0












          $begingroup$

          Monotonicity alone of $omega_f$ is not enough to yield continuity. You have to use that $f$ is (uniformly) continuous.



          A small (maybe pedantic) point of notation: $omega_f(delta)$ is a real number, so it cannot be continuous. What you meant was that you wanted to prove that $omega_f$ was continuous on $[0,1]$, or, if you really want to write $omega_f(delta)$, write that $omega_f(delta)$ is continuous at every $delta in [0,1]$.



          Otherwise, it seems correct to me (however, you that $omega_f(r-h) leq omega_f(r)$ is a straight consequence of its monotonicity).






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yes I used uniform continuity for showing $omega_f$ is continuous at $0.$ Do you suggest that I need it also for $rin (0,1].$
            $endgroup$
            – Hello_World
            Jan 7 at 19:58










          • $begingroup$
            Not at all, the « triangle inequality » and squeezing theorem work well enough.
            $endgroup$
            – Mindlack
            Jan 7 at 19:59














          0












          0








          0





          $begingroup$

          Monotonicity alone of $omega_f$ is not enough to yield continuity. You have to use that $f$ is (uniformly) continuous.



          A small (maybe pedantic) point of notation: $omega_f(delta)$ is a real number, so it cannot be continuous. What you meant was that you wanted to prove that $omega_f$ was continuous on $[0,1]$, or, if you really want to write $omega_f(delta)$, write that $omega_f(delta)$ is continuous at every $delta in [0,1]$.



          Otherwise, it seems correct to me (however, you that $omega_f(r-h) leq omega_f(r)$ is a straight consequence of its monotonicity).






          share|cite|improve this answer









          $endgroup$



          Monotonicity alone of $omega_f$ is not enough to yield continuity. You have to use that $f$ is (uniformly) continuous.



          A small (maybe pedantic) point of notation: $omega_f(delta)$ is a real number, so it cannot be continuous. What you meant was that you wanted to prove that $omega_f$ was continuous on $[0,1]$, or, if you really want to write $omega_f(delta)$, write that $omega_f(delta)$ is continuous at every $delta in [0,1]$.



          Otherwise, it seems correct to me (however, you that $omega_f(r-h) leq omega_f(r)$ is a straight consequence of its monotonicity).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 19:52









          MindlackMindlack

          2,72717




          2,72717












          • $begingroup$
            Yes I used uniform continuity for showing $omega_f$ is continuous at $0.$ Do you suggest that I need it also for $rin (0,1].$
            $endgroup$
            – Hello_World
            Jan 7 at 19:58










          • $begingroup$
            Not at all, the « triangle inequality » and squeezing theorem work well enough.
            $endgroup$
            – Mindlack
            Jan 7 at 19:59


















          • $begingroup$
            Yes I used uniform continuity for showing $omega_f$ is continuous at $0.$ Do you suggest that I need it also for $rin (0,1].$
            $endgroup$
            – Hello_World
            Jan 7 at 19:58










          • $begingroup$
            Not at all, the « triangle inequality » and squeezing theorem work well enough.
            $endgroup$
            – Mindlack
            Jan 7 at 19:59
















          $begingroup$
          Yes I used uniform continuity for showing $omega_f$ is continuous at $0.$ Do you suggest that I need it also for $rin (0,1].$
          $endgroup$
          – Hello_World
          Jan 7 at 19:58




          $begingroup$
          Yes I used uniform continuity for showing $omega_f$ is continuous at $0.$ Do you suggest that I need it also for $rin (0,1].$
          $endgroup$
          – Hello_World
          Jan 7 at 19:58












          $begingroup$
          Not at all, the « triangle inequality » and squeezing theorem work well enough.
          $endgroup$
          – Mindlack
          Jan 7 at 19:59




          $begingroup$
          Not at all, the « triangle inequality » and squeezing theorem work well enough.
          $endgroup$
          – Mindlack
          Jan 7 at 19:59


















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