Question

Why is the set of periodic functions not a vector space?

Answer 1

Answer:

To show that the set of periodic functions $\mathbb R\to\mathbb R$ is not a vector space, you need to show that the sum of two periodic functions might not be periodic. Let $f(x)$ be periodic with period $\alpha$, let $g(x)$ be periodic with period $\beta$, and let $h(x)=f(x)+g(x)$. Suppose $\beta/\alpha$ is rational and can be written as $\beta/\alpha=r/s$ with $r,s\in\mathbb Z$; then $s\beta=r\alpha$. Consequently, $$ h(x+s\beta)=f(x+r\alpha)+g(x+s\beta)=f(x)+g(x)=h(x) $$ and thus $h$ is periodic. This means that if you want $f+g$ not to be periodic, the ratio of the periods of $f$ and $g$ must be irrational, as in your example.

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Answer 2

Answer:

To show that the set of periodic functions $\mathbb R\to\mathbb R$ is not a vector space, you need to show that the sum of two periodic functions might not be periodic. Let $f(x)$ be periodic with period $\alpha$, let $g(x)$ be periodic with period $\beta$, and let $h(x)=f(x)+g(x)$. Suppose $\beta/\alpha$ is rational and can be written as $\beta/\alpha=r/s$ with $r,s\in\mathbb Z$; then $s\beta=r\alpha$. Consequently, $$ h(x+s\beta)=f(x+r\alpha)+g(x+s\beta)=f(x)+g(x)=h(x) $$ and thus $h$ is periodic. This means that if you want $f+g$ not to be periodic, the ratio of the periods of $f$ and $g$ must be irrational, as in your example.

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Explanation: