Is $C^1[a,b]$ with $d(f,g)=\max\left \{ \sup| f-g|,\sup| f'-g'| \right \}$ complete? 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.
 A: 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 $|| \cdot ||$. 
