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Is set of periodic functions from $\mathbb{R}$ to $\mathbb{R}$ subspace of $\mathbb{R}^{\mathbb{R}}$?

My answer is No. I give the following counterexample:

We consider periodic functions $f(x),g(x)$ so that $f(x)=x$ for $x \in [0,\sqrt{2})$ and $f(x)=f(x+\sqrt{2})$, function $g(x)=x$ for $x \in [0,\sqrt{3})$ and $g(x)=g(x+\sqrt{3})$ then $f+g$ is not periodic.

Indeed, assume that the function $f(x)+g(x)$ is periodic with length $T$. Note that if $T/\sqrt{2}$ is not an integer then $f(x)<f(x+T)$, similar to $T/\sqrt{3}$. Thus, since $T/ \sqrt{2}$ or $T/ \sqrt{3}$ can't be both integers so $f(x)+g(x)<f(x+T)+g(x+T)$, a contradiction. Thus, $f+g$ is non-periodic function. This yields that set of periodic function from $\mathbb{R}$ to $\mathbb{R}$ is not a subspace of $\mathbb{R}^{\mathbb{R}}$.

My questions are

  1. Is this counterexample correct?
  2. Are there any other examples?
Ali
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Tengu
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2 Answers2

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Your argument that $f+g$ is not periodic is not quite correct. For instance, if $T=1$ then $T/\sqrt{2}$ is not an integer, but $f(1)=1>f(1+T)=2-\sqrt{2}$. However, you can fix your argument by specifically looking at $x=0$: then $f(x)+g(x)=0+0=0$, but at least one of $f(x+T)$ and $g(x+T)$ must be positive since $T$ cannot be an integer multiple of both $\sqrt{2}$ and $\sqrt{3}$. So for $x=0$, you must have $f(x)+g(x)<f(x+T)+g(x+T)$.

More generally, a similar argument shows that if $f$ and $g$ are periodic with periods $a$ and $b$ respectively where $a/b$ is irrational, and $f$ and $g$ both achieve their minimum values only at integer multiples of $a$ and $b$ respectively, then $f+g$ is not periodic.

Eric Wofsey
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Edit: The following is not true.

Yes to both: any two periodic functions with incommensurate periods will have a sum that is not periodic.

Ben G.
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  • This is false, see http://math.stackexchange.com/questions/1079/sum-of-two-periodic-functions. – Eric Wofsey Dec 26 '16 at 09:04
  • Correction: Any two periodic functions with incommensurate fundamental periods will have a sum that is not periodic. – Ben G. Dec 26 '16 at 09:26
  • That is still false. Just modify the example in the answer I linked to above to take only $\mathbb{Z}$-linear combinations instead of $\mathbb{Q}$-linear combinations. – Eric Wofsey Dec 26 '16 at 09:29
  • You're right. Now I wonder under what circumstances we can say f+g is not periodic. – Ben G. Dec 26 '16 at 10:39
  • @BenG. If the two functions are continuous, then their sum cannot be periodic unless both are constant. Indeed, the subgroup of $\mathbb{R}^+$ generated by $S$ and $T$ (the incommensurable periods) is dense in $\mathbb{R}$, so $0$ is an accumulation point of it. So if the sum is periodic, it has no minimum positive period, hence it is constant. – egreg Dec 26 '16 at 11:06
  • @egreg. You can also note that with $f,g$ as in the Q, that $ f+g$ is continuous on $[0,1].$ If $f+g$ is periodic then the set of its periods is dense, which means that the closed set ${x\in [0,1]:f(x)+g(x)=f(0)+g(0)}$ is dense in [$0,1]$, making $f+g$ constant on [$0,1],$ a contradiction. – DanielWainfleet Dec 26 '16 at 12:35
  • @egreg: That doesn't work, since there's no reason that $S$ and $T$ need to be periods of the sum, so there is no obvious reason the periods of the sum should be dense. The result happens to be true, but you need a more complicated argument. See http://math.stackexchange.com/questions/1356802/the-sum-of-two-continuous-periodic-functions-is-periodic-if-and-only-if-the-rati. – Eric Wofsey Dec 26 '16 at 20:28