Prove that $(C^1[0,1], \|\cdot\|)$ is not a Banach space

Solution 1:

Take $\sqrt{x+1/n}$, which is $C^1[0,1]$ for each $n$ thanks to our shifting of the discontinuity in the derivative.

To see this converges uniformly, note that $\sqrt{x}$ is uniformly continuous on compact sets. On say $[0,2]$ we may use uniform continuity to meet any epsilon challenge with an $N$ not dependent on $x$ with $$ |\sqrt{x+1/N}-\sqrt{x}|<\epsilon $$

Solution 2:

The idea is to find a sequence of functions $\{f_n\}$ which does converge in the space $C^0$ of continuous functions with the max norm, but whose limit is not differentiable. For example, $$f_n(x) = \left\{\begin{matrix} \left|x - \frac{1}{2}\right| &\text{if } \left|x - \frac{1}{2}\right|\ge\frac{1}{n} \\ \frac{1}{2n}+\frac{n}{2}\left(x-\frac{1}{2}\right)^2 & \text{if }\left|x - \frac{1}{2}\right|<\frac{1}{n}\end{matrix}\right. $$ converges in $C^0$ under the max norm to $f(x) = \left|x-\frac{1}{2}\right|$ (in fact $\max|f-f_n| = \frac{1}{2n}$), and hence is Cauchy under the max norm in $C^1$; however if were to converge in $C^1$ its limit would have to equal $f$ in $C^0$, which is impossible.