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Denote by $C^{\infty}(\mathbb{R})$ the space of infinitely differentiable functions. Prove that $C^{\infty}(\mathbb{R})$ is a complete metric space with respect to the metric $$d(f,g)=\sum_{k=0}^\infty\frac{\|f^{(k)}-g^{(k)}\|_{[-k-1,k+1]}}{1+\|f^{(k)}-g^{(k)}\|_{[-k-1,k+1]}}\frac{1}{2^k}$$ The norm is defined as $\|h\|_K=\sup_{x \in K}{|h(x)|}$

First I define a Cauchy sequence, say $d(f_n,f_m)<\epsilon$. Then I fixed $k$, I obtain $$\frac{\|f^{(k)}-g^{(k)}\|_{[-k-1,k+1]}}{1+\|f^{(k)}-g^{(k)}\|_{[-k-1,k+1]}}\frac{1}{2^k}<\epsilon$$

After some manipulations, I obtain $\|f_n^{(k)}-f_m^{(k)}\|_{[-k-1,k+1]}<\frac{\epsilon2^k}{1-\epsilon}$, which tends to zero as $\epsilon$ can be made arbitrarily small. Hence, $\{f_n^{(k)}\}_{n \geq 1}$ is a Cauchy sequence with the norm $\|.\|_{[-k-1,k+1]}$. Then from here I don't know how to proceed.

My purpose is to find a suitable candidate $f$ such that the sequence converges to $f$ and $f \in C^{\infty}(\mathbb{R})$

Idonknow
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1 Answers1

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I would show that it is complete metric on $C^\infty([-N,N])$ for every $N\in \mathbb{N}$

Use the fact that on $[-N,N]$ the function $f$ can be written as $$ f(x) = P_K(x) + \int_0^x \frac{f^{(K+1)}(t)}{K!} (x-t)^K dt $$ for $K\geq N$. $P_K(x)$ are first $K$ terms in Taylor series.

to find the suitable candidate $f$: Let $f_n$ be Cauchy sequence in your metric. Fix some $K$. Now we will show that $f_n$ is Cauchy in $C[-N,N]$, therefore it has a limit. $$ \|f_n-f_m\|_{[-N,N]} \leq \|P^n_N - P^m_N\|_{[-N,N]} + \left\|\int_0^x \frac{f_n^{(N+1)}(t)-f_m^{(N+1)}(t)}{N!} (x-t)^N dt \right\|_{[-N,N]}\leq $$ $$ \leq \|P^n_N - P^m_N\|_{[-N,N]} + \sup_{x\in[-N,N]}\int_0^x \frac{ \left\|f_n^{(N+1)}-f_m^{(N+1)} \right\|_{[0,x]}}{N!} x^N dt $$ $$ \leq \|P^n_N - P^m_N\|_{[-N,N]} + \left\|f_n^{(N+1)}-f_m^{(N+1)} \right\|_{[-N,N]} \frac{N^{N+1}}{N!} $$

$P^n_N$ contains only derivatives(of order $1$ to $N$) of $f_n$ in zero. Your metric is complete in $C^\infty[-1,1]$ thus $P^n_N$ is Cauchy in $C[-N,N]$. From definition of your metrics follows immediately that $f^{(N+1)}$ is cauchy in $C[-N,N]$. Therefore $f_n$ is cauchy in $C[-N,N]$

To finish proof observe that if you have $f_n$ cauchy in your metrics than $f^{(k)}_n$ is cauchy in $C[-k-1,k+1]$.

We can reformulate it to $f^{(k)}_n$ is cauchy in $C[-l-1,l+1]$ for all $k\geq l$. But you need to show that is cauchy as well for $k<l$. Above I did the proof for $k=0$. The proof is basically the same just use the first formula not for $f$ but for $f^{(k)}$.

tom
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