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Let $$ C^{\infty} := C^{\infty}([0,1]) = \cap_{n=0}^{\infty} C^n([0,1]) \text{ and } p_n(f) = \|f \|_n, \ f\in C^{\infty}, n \in \mathbb{N}. \\ C^n([0,1]) = \{ f: [0,1] \to \mathbb{R} : f^(k) \in C([0,1]) \ \forall \ 0 \leq k \leq n \} \\ \|f\|_n = \sum_{k=0}^n \|f^{(k)} \|_{\infty} $$ Then $(C^{\infty}, P)$ with $P = \{p_n : n \in \mathbb{N}\}$ is a seminormed vector space.

Define a metric on $C^{\infty}$ by

$$ d(f,g) = \sum_{n=0}^{\infty} \frac{1}{2^n} \frac {p_n(f-g)}{1+p_n(f-g)}. $$ Show that $(C^{\infty}, d)$ is a complete metric space.

> Let $(f_j)_j \subset C^{\infty}$ be Cauchy in $C^{\infty}$, then we have with $\tilde{\epsilon} = 2^n \epsilon_1$ $$ d(f_j,f_i) = \sum_{n=0}^{\infty} \frac{1}{2^n} \frac {p_n(f_j-f_i)}{1+p_n(f_j-f_i)} = \sum_{n=0}^{\infty} \frac{1}{2^n} \frac {\|(f_j-f_i)\|_n}{1+\|(f_j-f_i)\|_n} < \tilde{\epsilon} \Rightarrow \\ \sum_{n=0}^{p} \frac{1}{2^n} \frac {\|(f_j-f_i)\|_n}{1+\|(f_j-f_i)\|_n} < \tilde{\epsilon} \Rightarrow \\ \frac{1}{2^n} \frac {\|(f_j-f_i)\|_n}{1+\|(f_j-f_i)\|_n} < \tilde{\epsilon} \Rightarrow \frac {\|(f_j-f_i)\|_n}{1+\|(f_j-f_i)\|_n} < \epsilon_1 \Rightarrow \|(f_j-f_i)\|_n < \epsilon_1 $$ Thus $(f_j)_j$ is Cauchy in $C^n:=C^n([0,1])$, since $C^n$ is complete $\exists g \in C^n \ : \ f_j \to g$, then $\forall \epsilon >0 \ \exists N\in \mathbb{N}$ s.t $\forall j > N$ we have $\|f_j - g\| < \epsilon$. Now choose $\hat{\epsilon}= \frac12 \epsilon$, then

$$ d(f_j,g) = \sum_{n=0}^{\infty} \frac{1}{2^n} \frac {\|(f_j-g)\|_n}{1+\|(f_j-g)\|_n} < \sum_{n=0}^{\infty} \frac{1}{2^n} \frac{\hat{\epsilon}}{1+\hat{\epsilon}} = \frac{\hat{\epsilon}}{1+\hat{\epsilon}} \sum_{n=0}^{\infty} \frac{1}{2^n} < \\ \frac{\hat{\epsilon}}{1+\hat{\epsilon}} \cdot 2 < \hat{\epsilon}\cdot2 = \epsilon. $$

Is this correct? If not, how can it be improved?

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  • $\begingroup$ When you say $(f_j)$ is Cauchy in $C^n$ and so there exists $g \in C^n$ such that $f_j \to g$, a priori $g$ depends on $n$, doesn't it? $\endgroup$ – Stefano Mar 13 '17 at 9:43
  • $\begingroup$ Yes, but there exists a $g$ for every n. But I couldn't figure out how make it "correct/better", hence I'm asking. @Stefano $\endgroup$ – Olba12 Mar 13 '17 at 9:50

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