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

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

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.


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.)

  • 1
    $\begingroup$ As an aside, if OP is familiar with Fourier series this decomposition should be part of the associated intuition. $\endgroup$ – R.. GitHub STOP HELPING ICE Dec 2 '18 at 14:07
  • $\begingroup$ @R.. Fourier decomposition is way more complicated than necessary. $\endgroup$ – leftaroundabout Dec 2 '18 at 19:05
  • $\begingroup$ @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. $\endgroup$ – R.. GitHub STOP HELPING ICE Dec 2 '18 at 21:18
  • $\begingroup$ @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. $\endgroup$ – leftaroundabout Dec 2 '18 at 21:50
  • 1
    $\begingroup$ 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. $\endgroup$ – copper.hat Dec 2 '18 at 22:36

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