I need to show that the space $C^1([0,1])$ is incomplete in the norm $\|f\|_{\infty}=\sup _{[0,1]}|f|$ but complete in the norm $\|f\|_{\infty}+\left\|f^{\prime}\right\|_{\infty}$.
My attempt: To show that the space $C^1([0,1])$ is incomplete in the norm $|f|{\infty}=\sup{[0,1]}|f|$, we can construct a Cauchy sequence in $C^1([0,1])$ that does not converge in this norm. Consider the sequence $(f_n)_{n=1}^{\infty}$ defined by $f_n(x) = \sqrt{x+\frac{1}{n}}$.
To show that this sequence is Cauchy, suppose $m>n$ and let $\epsilon>0$ be given. Then for $x \in [0,1]$, we have \begin{align*} |f_m(x) - f_n(x)| &= \left| \sqrt{x+\frac{1}{m}} - \sqrt{x+\frac{1}{n}} \right| \ &\leq \left| \sqrt{x+\frac{1}{m}} - \sqrt{x+\frac{1}{n}} \right| \frac{\sqrt{m}-\sqrt{n}}{\sqrt{m}-\sqrt{n}} \ &= \frac{1}{\sqrt{m}+\sqrt{n}} \left| \frac{1}{\sqrt{x+\frac{1}{m}} + \sqrt{x+\frac{1}{n}}} \right| \ &\leq \frac{1}{\sqrt{m}+\sqrt{n}} \leq \epsilon, \end{align*} where we have used the reverse triangle inequality and the fact that $m>n$.
Thus, $(f_n){n=1}^{\infty}$ is Cauchy in $C^1([0,1])$. However, this sequence does not converge in the norm $|f|{\infty}$, since for each $n$, we have $|f_n|{\infty} = \sup{[0,1]} |f_n| = \sqrt{1+\frac{1}{n}}$, which goes to 1 as $n$ goes to infinity, while the pointwise limit of $f_n$ as $n\to \infty$ is the function $f(x) = \sqrt{x}$, which is not in $C^1([0,1])$.
To show that $C^1([0,1])$ is complete in the norm $|f|{\infty}+\left|f^{\prime}\right|{\infty}$, let $(f_n){n=1}^{\infty}$ be a Cauchy sequence in $C^1([0,1])$ with respect to this norm. Then, for any $\epsilon>0$, there exists an $N$ such that for all $n,m\geq N$, we have \begin{align*} |f_n-f_m|{\infty} &\leq \frac{\epsilon}{2}, \ |f_n'-f_m'|_{\infty} &\leq \frac{\epsilon}{2}. \end{align*} Since each $f_n$ is continuous on $[0,1]$, it follows that $f_n(x)$ is uniformly Cauchy on $[0,1]$, and hence converges uniformly to a function $f(x)$. Moreover, since $f_n'(x)$ is also uniformly Cauchy, it follows that $f_n'(x)$ converges uniformly to a function $g(x)$.
