Is the Set of Continuous Functions that are the Sum of Even and Odd Functions Meager?












6














Consider $X = mathcal{C}([−1,1])$ with the usual norm $|f|_{infty} = sup_{tin [−1,1]}|f(t)|.$



Define
$$mathcal{A}_{+}={ f in X : f(t)=f(−t) space forall tin [−1,1]},$$
$$mathcal{A}_{−}={ f in X : f(t)=−f(−t) space forall t in [−1,1]}. $$



Is $mathcal{A}_{+} +mathcal{A}_{−} = {f +g : f in mathcal{A}_{+},g in mathcal{A}_{−}}$ meager?




I know this set is dense by the Stone-Weierstrass Theorem. However, that doesn't really help. I also know that if the set is closed, then it is meager, but I have difficulties deciding whether it is closed or not. I know the exponential function is a limit of a sequence of a sum of even and odd functions, however one could define it to be that, in which case it doesn't help.




Any hints on how to get going on this problem, and on whether the set $mathcal{A}_{+}+{A}_{-} $ is closed or not? Thank you in advance.






real-analysis general-topology functional-analysis metric-spaces baire-category







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    6














    Consider $X = mathcal{C}([−1,1])$ with the usual norm $|f|_{infty} = sup_{tin [−1,1]}|f(t)|.$



    Define
    $$mathcal{A}_{+}={ f in X : f(t)=f(−t) space forall tin [−1,1]},$$
    $$mathcal{A}_{−}={ f in X : f(t)=−f(−t) space forall t in [−1,1]}. $$



    Is $mathcal{A}_{+} +mathcal{A}_{−} = {f +g : f in mathcal{A}_{+},g in mathcal{A}_{−}}$ meager?




    I know this set is dense by the Stone-Weierstrass Theorem. However, that doesn't really help. I also know that if the set is closed, then it is meager, but I have difficulties deciding whether it is closed or not. I know the exponential function is a limit of a sequence of a sum of even and odd functions, however one could define it to be that, in which case it doesn't help.




    Any hints on how to get going on this problem, and on whether the set $mathcal{A}_{+}+{A}_{-} $ is closed or not? Thank you in advance.






    real-analysis general-topology functional-analysis metric-spaces baire-category







    share|cite|improve this question

























      6












      6








      6


      2





      Consider $X = mathcal{C}([−1,1])$ with the usual norm $|f|_{infty} = sup_{tin [−1,1]}|f(t)|.$



      Define
      $$mathcal{A}_{+}={ f in X : f(t)=f(−t) space forall tin [−1,1]},$$
      $$mathcal{A}_{−}={ f in X : f(t)=−f(−t) space forall t in [−1,1]}. $$



      Is $mathcal{A}_{+} +mathcal{A}_{−} = {f +g : f in mathcal{A}_{+},g in mathcal{A}_{−}}$ meager?




      I know this set is dense by the Stone-Weierstrass Theorem. However, that doesn't really help. I also know that if the set is closed, then it is meager, but I have difficulties deciding whether it is closed or not. I know the exponential function is a limit of a sequence of a sum of even and odd functions, however one could define it to be that, in which case it doesn't help.




      Any hints on how to get going on this problem, and on whether the set $mathcal{A}_{+}+{A}_{-} $ is closed or not? Thank you in advance.






      real-analysis general-topology functional-analysis metric-spaces baire-category







      share|cite|improve this question













      Consider $X = mathcal{C}([−1,1])$ with the usual norm $|f|_{infty} = sup_{tin [−1,1]}|f(t)|.$



      Define
      $$mathcal{A}_{+}={ f in X : f(t)=f(−t) space forall tin [−1,1]},$$
      $$mathcal{A}_{−}={ f in X : f(t)=−f(−t) space forall t in [−1,1]}. $$



      Is $mathcal{A}_{+} +mathcal{A}_{−} = {f +g : f in mathcal{A}_{+},g in mathcal{A}_{−}}$ meager?




      I know this set is dense by the Stone-Weierstrass Theorem. However, that doesn't really help. I also know that if the set is closed, then it is meager, but I have difficulties deciding whether it is closed or not. I know the exponential function is a limit of a sequence of a sum of even and odd functions, however one could define it to be that, in which case it doesn't help.




      Any hints on how to get going on this problem, and on whether the set $mathcal{A}_{+}+{A}_{-} $ is closed or not? Thank you in advance.






      real-analysis general-topology functional-analysis metric-spaces baire-category




      real-analysis general-topology functional-analysis metric-spaces baire-category






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




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      asked Dec 2 at 7:14









      Gaby Alfonso

      682315




      682315






















          1 Answer
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          16














          Note that any function can be written as
          $f(x) = {1 over 2} (f(x) + f(-x)) + {1 over 2} (f(x) - f(-x)) $, so
          $mathcal{A}_{+} +mathcal{A}_{−} = X$, which is not meagre.



          (It is not meagre because $C[-1,1]$ is a complete metric space.)






          share|cite|improve this answer



















          • 1




            As an aside, if OP is familiar with Fourier series this decomposition should be part of the associated intuition.
            – R..
            Dec 2 at 14:07










          • @R.. Fourier decomposition is way more complicated than necessary.
            – leftaroundabout
            Dec 2 at 19:05










          • @leftaroundabout: Absolutely. copper.hat's answer is simpler and more direct, but having studied Fourier series should leave one with an intuition that there is a natural decomposition into even and odd parts, whereas otherwise this may seem like a surprising and/or obscure result.
            – R..
            Dec 2 at 21:18










          • @R.. hm, I rather see it the other way around – this decomposition is one of the foundations that make Fourier analysis possible. But, I suppose this is somewhat a matter of taste.
            – leftaroundabout
            Dec 2 at 21:50






          • 1




            I guess I would see the decomposition of square matrices into symmetric and skew symmetric parts a more natural intuition. Fourier series requires a lot more baggage.
            – copper.hat
            Dec 2 at 22:36











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          16














          Note that any function can be written as
          $f(x) = {1 over 2} (f(x) + f(-x)) + {1 over 2} (f(x) - f(-x)) $, so
          $mathcal{A}_{+} +mathcal{A}_{−} = X$, which is not meagre.



          (It is not meagre because $C[-1,1]$ is a complete metric space.)






          share|cite|improve this answer



















          • 1




            As an aside, if OP is familiar with Fourier series this decomposition should be part of the associated intuition.
            – R..
            Dec 2 at 14:07










          • @R.. Fourier decomposition is way more complicated than necessary.
            – leftaroundabout
            Dec 2 at 19:05










          • @leftaroundabout: Absolutely. copper.hat's answer is simpler and more direct, but having studied Fourier series should leave one with an intuition that there is a natural decomposition into even and odd parts, whereas otherwise this may seem like a surprising and/or obscure result.
            – R..
            Dec 2 at 21:18










          • @R.. hm, I rather see it the other way around – this decomposition is one of the foundations that make Fourier analysis possible. But, I suppose this is somewhat a matter of taste.
            – leftaroundabout
            Dec 2 at 21:50






          • 1




            I guess I would see the decomposition of square matrices into symmetric and skew symmetric parts a more natural intuition. Fourier series requires a lot more baggage.
            – copper.hat
            Dec 2 at 22:36
















          16














          Note that any function can be written as
          $f(x) = {1 over 2} (f(x) + f(-x)) + {1 over 2} (f(x) - f(-x)) $, so
          $mathcal{A}_{+} +mathcal{A}_{−} = X$, which is not meagre.



          (It is not meagre because $C[-1,1]$ is a complete metric space.)






          share|cite|improve this answer



















          • 1




            As an aside, if OP is familiar with Fourier series this decomposition should be part of the associated intuition.
            – R..
            Dec 2 at 14:07










          • @R.. Fourier decomposition is way more complicated than necessary.
            – leftaroundabout
            Dec 2 at 19:05










          • @leftaroundabout: Absolutely. copper.hat's answer is simpler and more direct, but having studied Fourier series should leave one with an intuition that there is a natural decomposition into even and odd parts, whereas otherwise this may seem like a surprising and/or obscure result.
            – R..
            Dec 2 at 21:18










          • @R.. hm, I rather see it the other way around – this decomposition is one of the foundations that make Fourier analysis possible. But, I suppose this is somewhat a matter of taste.
            – leftaroundabout
            Dec 2 at 21:50






          • 1




            I guess I would see the decomposition of square matrices into symmetric and skew symmetric parts a more natural intuition. Fourier series requires a lot more baggage.
            – copper.hat
            Dec 2 at 22:36














          16












          16








          16






          Note that any function can be written as
          $f(x) = {1 over 2} (f(x) + f(-x)) + {1 over 2} (f(x) - f(-x)) $, so
          $mathcal{A}_{+} +mathcal{A}_{−} = X$, which is not meagre.



          (It is not meagre because $C[-1,1]$ is a complete metric space.)






          share|cite|improve this answer














          Note that any function can be written as
          $f(x) = {1 over 2} (f(x) + f(-x)) + {1 over 2} (f(x) - f(-x)) $, so
          $mathcal{A}_{+} +mathcal{A}_{−} = X$, which is not meagre.



          (It is not meagre because $C[-1,1]$ is a complete metric space.)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 2 at 7:23

























          answered Dec 2 at 7:17









          copper.hat

          126k559159




          126k559159








          • 1




            As an aside, if OP is familiar with Fourier series this decomposition should be part of the associated intuition.
            – R..
            Dec 2 at 14:07










          • @R.. Fourier decomposition is way more complicated than necessary.
            – leftaroundabout
            Dec 2 at 19:05










          • @leftaroundabout: Absolutely. copper.hat's answer is simpler and more direct, but having studied Fourier series should leave one with an intuition that there is a natural decomposition into even and odd parts, whereas otherwise this may seem like a surprising and/or obscure result.
            – R..
            Dec 2 at 21:18










          • @R.. hm, I rather see it the other way around – this decomposition is one of the foundations that make Fourier analysis possible. But, I suppose this is somewhat a matter of taste.
            – leftaroundabout
            Dec 2 at 21:50






          • 1




            I guess I would see the decomposition of square matrices into symmetric and skew symmetric parts a more natural intuition. Fourier series requires a lot more baggage.
            – copper.hat
            Dec 2 at 22:36














          • 1




            As an aside, if OP is familiar with Fourier series this decomposition should be part of the associated intuition.
            – R..
            Dec 2 at 14:07










          • @R.. Fourier decomposition is way more complicated than necessary.
            – leftaroundabout
            Dec 2 at 19:05










          • @leftaroundabout: Absolutely. copper.hat's answer is simpler and more direct, but having studied Fourier series should leave one with an intuition that there is a natural decomposition into even and odd parts, whereas otherwise this may seem like a surprising and/or obscure result.
            – R..
            Dec 2 at 21:18










          • @R.. hm, I rather see it the other way around – this decomposition is one of the foundations that make Fourier analysis possible. But, I suppose this is somewhat a matter of taste.
            – leftaroundabout
            Dec 2 at 21:50






          • 1




            I guess I would see the decomposition of square matrices into symmetric and skew symmetric parts a more natural intuition. Fourier series requires a lot more baggage.
            – copper.hat
            Dec 2 at 22:36








          1




          1




          As an aside, if OP is familiar with Fourier series this decomposition should be part of the associated intuition.
          – R..
          Dec 2 at 14:07




          As an aside, if OP is familiar with Fourier series this decomposition should be part of the associated intuition.
          – R..
          Dec 2 at 14:07












          @R.. Fourier decomposition is way more complicated than necessary.
          – leftaroundabout
          Dec 2 at 19:05




          @R.. Fourier decomposition is way more complicated than necessary.
          – leftaroundabout
          Dec 2 at 19:05












          @leftaroundabout: Absolutely. copper.hat's answer is simpler and more direct, but having studied Fourier series should leave one with an intuition that there is a natural decomposition into even and odd parts, whereas otherwise this may seem like a surprising and/or obscure result.
          – R..
          Dec 2 at 21:18




          @leftaroundabout: Absolutely. copper.hat's answer is simpler and more direct, but having studied Fourier series should leave one with an intuition that there is a natural decomposition into even and odd parts, whereas otherwise this may seem like a surprising and/or obscure result.
          – R..
          Dec 2 at 21:18












          @R.. hm, I rather see it the other way around – this decomposition is one of the foundations that make Fourier analysis possible. But, I suppose this is somewhat a matter of taste.
          – leftaroundabout
          Dec 2 at 21:50




          @R.. hm, I rather see it the other way around – this decomposition is one of the foundations that make Fourier analysis possible. But, I suppose this is somewhat a matter of taste.
          – leftaroundabout
          Dec 2 at 21:50




          1




          1




          I guess I would see the decomposition of square matrices into symmetric and skew symmetric parts a more natural intuition. Fourier series requires a lot more baggage.
          – copper.hat
          Dec 2 at 22:36




          I guess I would see the decomposition of square matrices into symmetric and skew symmetric parts a more natural intuition. Fourier series requires a lot more baggage.
          – copper.hat
          Dec 2 at 22:36


















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