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\begin{equation} P = \{f \in\ C ( \mathbb{R}, \mathbb{R}) \mid f(x+2 \pi ) = f(x)\} \end{equation} be the set of $2\pi$-periodic function.

1) Show that $P$ is a subspace of $C( \mathbb{R}, \mathbb{R})$

2) Let $\|F\| = \sup_{x\in[0,2\pi]} |f(x)| $ for all $ f\in P$. Show that $ (P, \| \cdot \|)$ is a normed space.

3) Let $d(f,g) = \| f - g \| $ for all $f,g\in P$. Show that $d$ is a metric and $(P,d)$ is a complete metric space.

I have an exam soon and this was one of the practice questions.

For 1) part: I know that for $P$ to be subspace of $C( \mathbb{R}, \mathbb{R})$, I have to show that it is closed under vector addition and scalar multiplication, right? But how?

For 2) part: I can prove it using the axioms for a normed space.

For 3) part: I proved that $d$ is a metric by using axioms, but then how do I prove that $(P,d)$ is a complete metric space? I know that a metric space is called complete if every Cauchy sequence in that space is convergent, but how do I prove it for this part?

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(1) You have to show that $P$ is closed under vector addition and scalar multiplication and that $P\ne\varnothing$. That last is trivial, though: the constant $0$ function is in $P$, as are $\sin x$ and $\cos x$. You can show closure under vector addition and scalar multiplication simultaneously by showing that for any $f,g\in P$ and any scalars $\alpha,\beta\in\Bbb R$, $\alpha f+\beta g\in P$. To show that $\alpha f+\beta g$, you must show that it’s continuous and $2\pi$-periodic. You should have available some general facts about continuity that make that part very easy. To show that $\alpha f+\beta g$ is $2\pi$-periodic, you must show that $$(\alpha f+\beta g)(x+2\pi)=(\alpha f+\beta g)(x)\tag{1}$$ for all $x\in\Bbb R$. This is also very easy: just expand the lefthand side of $(1)$ and use the fact that $f$ and $g$ are $2\pi$-periodic.

(3) Let $\langle f_n:n\in\Bbb N\rangle$ by a Cauchy sequence in $C(\Bbb R,\Bbb R)$. This means that for each $\epsilon>0$ there is an $n_\epsilon\in\Bbb N$ such that

$$\|f_m-f_n\|=\sup_{x\in[0,2\pi]}|f_m(x)-f_n(x)|<\epsilon\quad\text{whenever}\quad m,n\ge n_\epsilon\;.\tag{2}$$

Now let $x\in[0,2\pi]$; $(2)$ clearly implies that $$|f_m(x)-f_n(x)|<\epsilon\quad\text{whenever}\quad m,n\ge n_\epsilon\;,$$ so $\langle f_n(x):n\in\Bbb N\rangle$ is a Cauchy sequence in $\Bbb R$. $\Bbb R$ (with the usual metric) is complete, so the sequence $\langle f_n(x):n\in\Bbb N\rangle$ converges to some real number; call that number $g(x)$. This defines a function $g:[0,2\pi]\to\Bbb R$. Now try to carry out the following program:

  1. Extend $g$ to a $2\pi$-periodic function $f:\Bbb R\to\Bbb R$.
  2. Prove that $f$ is continuous and therefore in $P$.
  3. Prove that $\langle f_n:n\in\Bbb N\rangle$ converges to $f$ in $\left\langle P,\|\cdot\|\right\rangle$.
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For part $1)$ you need to show that for any $f, g\in P$ and $ \lambda,\mu \in F$, where $F$ is the field you are working over (presumably $\mathbb{R}$) $\lambda f + \mu g \in P$. That is to say that $\lambda f + \mu g \in C(\mathbb{R},\mathbb{R})$ and $[\lambda f + \mu g](x+2\pi) = [\lambda f + \mu g](x)$. (Technically you also need to show that $0 \in P$)

For part $3)$ recall that a closed subset of a complete metric space is also complete.

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