Given normed space $C^1[0,1]$ of differentable functions with continous derivatives on $[0,1]$. The norm is defined as following

$$ ||x|| = \max_{[0,1]} |x(t)| $$

Prove that the given normed space is not a Banach space

In the attempt of solving this problem, I have thought about the possibility to construct a Cauchy sequence in the space which doesn't converge in the space. However until now I have got none of any idea. Then I proceeded to think about constructing an equivalent norm to the given one where the space can be easily proven to be not a Banach space. However, I have got nothing either.

Now I'm stuck without a clue. Please give me a hint to a correct direction. Any help is greatly appreciated. Thank you for reading.

  • 6
    $\begingroup$ Consider any sequence of continuously differentiable functions that converges uniformly to $\lvert x-1/2\rvert$ (or really any function in $C^0([0, 1])\setminus C^1([0, 1])$). This exists by Stone-Weierstrass, for example (although that's a bit heavy-handed for this question). $\endgroup$ – Michael Lee May 5 '18 at 2:58

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 $$ |\sqrt{x+1/N}-\sqrt{x}|<\epsilon $$


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.


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