1

Good day, please a question, is $C^1[a,b]$ with metric $$d(f,g)=\max\left \{ \sup| f-g|,\sup| f'-g'| \right \}$$ complete?

Can I build the limit from this similar expression as with the norm of sup? $| f(x)_m-g_n(x)|< \varepsilon, or | f'(x)_m-g'_n(x)|< \varepsilon $

Thanks.

mathreda
  • 363
  • Bascially the procedure is the same as here. In fact, that metric is easily proved to be strongly equivalent to this one. –  Sep 17 '16 at 23:56
  • show that $C^0([a,b])$ is complete with $|f| = \sup_{x \in [a,b]} |f(x)|$ (uniform convergence of continuous functions), and deduce that your space is complete : if $|g-g_n|+ |g'-g_n'|\to 0$ where $g_n \in C^1([a,b])$, then $g_n' \in C^0([a,b])$ and $|g'-g_n'|\to 0$ so that $g' \in C^0([a,b])$, and $|g-g_n| \to 0 \implies g_n(a) \to g(a)$ i.e. $g(x) = g(a) + \int_a^x g'(t)dt \in C^1([a,b])$ – reuns Sep 18 '16 at 00:28

1 Answers1

0

We say two norms $|| \cdot ||_1$ and $|| \cdot ||_2$ are equivalent if there exist $C_1, C_2>0$ such that \begin{align} C_1 || x ||_1 \leq || x ||_2 \leq C_2|| x||_1. \end{align}

In your case, it's not hard to see $|| f || := \max\{||f||_\infty, ||f'||_\infty\}$ is a norm and

\begin{align} ||f||_\infty + ||f'||_\infty \leq 2\max\{||f||_\infty, ||f'||_{\infty}\} \leq 2(|| f||_\infty + || f'||_\infty) \end{align}

which means the $|| f||$ is equivalent to the $C^1$-norm. Since the $C^1$ norm is complete, then your given norm is also complete.

Note: The norm induces a metric topology.

Edit: Let $\{f_n\}$ be a Cauchy sequence with respect to $|| \cdot || $ then it's also a Cauchy sequence with respect to $|| \cdot ||_{C^1}$ since \begin{align} || f-g|| \leq || f- g||_{C^1}. \end{align} Since $C^1$ is a Banach space with respect to $|| \cdot ||_{C^1}$ then you can find a candidate $f \in C^1[0, 1]$ such that $f_n \rightarrow f$ in $||\cdot||_{C^1}$. But then it follows \begin{align} || f_n - f|| \leq ||f_n - f||_{C^1} \end{align} which means $f_n$ also converges to $f$ with respect to $|| \cdot ||$.

Jacky Chong
  • 26,287