Convergence of a series of a given metric..

Question

I'm trouble with a metric defined over a given set: Consider $\mathcal{P}=C_p^\infty([-\pi, \pi])$, that is, $\mathcal{P}$ is the set of all infinitely differentiable functions $f:\mathbb R\rightarrow \mathbb C$ which are $2\pi$-periodic ($f(x+2\pi)=f(x)$, for all $x\in \mathbb R$). We can introduce a metric in this set setting $$\displaystyle d(f, g)=\sum_{k=0}^\infty \frac{1}{2^k}\frac{\|f^{(k)}-g^{(k)}\|_\infty}{1+\|f^{(k)}-g^{(k)}\|_\infty},$$ where $\displaystyle \|f\|_\infty=\sup_{x\in J}|f(x)|$ where $J\subset \mathbb R$ is an interval of length $2\pi$. The problem is: why does the series above converge? Also, how can I prove $f_n\to f$ in $\mathcal{P}$ if and only if $\|f_n^{(k)}-f^{(k)}\|_\infty\to 0$ for all $k\in\mathbb N$?

Answer 1

Each term in the series is bounded by $\frac{1}{2^k}$, since $\frac{x}{1+x}<1$ for all $x\ge 0$. Thus we have $$\sum_{k=0}^\infty \frac{1}{2^k}\frac{\|f^{(k)}-g^{(k)}\|_\infty}{1+\|f^{(k)}-g^{(k)}\|_\infty}\leq \sum_{k=0}^\infty \frac{1}{2^k}=\frac{1}{1-\frac12}=2.$$

I suspect you wish to show that $f_n\to f$ in $\mathcal P$ if and only if $\|f_n^{(k)}-f^{(k)}\|_\infty\to 0$ for all $k$. One direction (that if $f_n\to f$ in $\mathcal P$, $\|f_n^{(k)}-f^{(k)}\|_\infty\to 0$ for all $k$) is easy, and I will leave to you. For the other direction, suppose $\|f_n^{(k)}-f^{(k)}\|_\infty\to 0$ for all $k$ and let $\epsilon>0$. Let $i$ be such that $\frac{1}{2^i}<\frac{\epsilon}{2}$ and $N$ be such that, for all $k\leq i$, $$n\ge N\implies \|f_n^{(i)}-f^{(i)}\|_\infty < \frac{\epsilon}{2(i+1)}.$$ Then for $n\ge N$ we have $$\begin{align} d(f_n,f) &= \sum_{k=0}^\infty \frac{1}{2^k}\frac{\|f_n^{(k)}-f^{(k)}\|_\infty}{1+\|f_n^{(k)}-f^{(k)}\|_\infty}\\ &= \sum_{k=0}^i \frac{1}{2^k}\frac{\|f_n^{(k)}-f^{(k)}\|_\infty}{1+\|f_n^{(k)}-f^{(k)}\|_\infty}+\sum_{k=i+1}^\infty \frac{1}{2^k}\frac{\|f_n^{(k)}-f^{(k)}\|_\infty}{1+\|f_n^{(k)}-f^{(k)}\|_\infty}\\ &\leq \sum_{k=0}^i \frac{1}{2^k}\|f_n^{(k)}-f^{(k)}\|_\infty+\sum_{k=i+1}^\infty \frac{1}{2^k}\\ &\leq \sum_{k=0}^i \frac{1}{2^k}\frac{\epsilon}{2(i+1)}+\frac{1}{2^i}\\ &\leq \sum_{k=0}^i \frac{\epsilon}{2(i+1)}+\frac{\epsilon}{2}=\epsilon\\ \end{align}$$ hence $f_n\to f$ in $\mathcal P$.