# Why this is not a Banach space

When reading about functional analysis I encountered the following example of a Banach space:

$C^1 ([0,1])$ endowed with the norm $\|f\| = \|f\|_\infty + \|f'\|_\infty$.

where $\|\cdot\|_\infty$ denotes the $\sup$-norm.

At first it seemed to me that $C^1 ([0,1])$ endowed with the norm $\|f\| = \|f\|_\infty$ is also a Banach space. The norm $\|f\| = \|f\|_\infty + \|f'\|_\infty$ therefore seemed unnecessarily complicated. But I suspect this is not the case. Hence:

Could anyone provide an example of a Cauchy sequence w.r.t. $\|\cdot\|_\infty$ of $C^1$ functions such that the limit is not $C^1$?

• $$f_n(x) = \sqrt{\left(x-\frac12\right)^2 + \frac{1}{n}}$$ – Daniel Fischer Feb 26 '14 at 14:59
• This isn't an explicit example, but remember that the Weierstrass approximation theorem says that any continuous function $f\colon[0,1]\to \mathbb{R}$ is uniformly approximated by polynomials, and thus every continuous $f\colon [0,1]\to\mathbb{R}$ is a limit of smooth functions w.r.t $\|\cdot\|_\infty$. – froggie Feb 26 '14 at 15:00
• @DanielFischer Thank you this answers my question. The pointwise and uniform limit is $|x-1/2|$ and this is of course not differentiable. – newb Feb 26 '14 at 15:01
• Take any continuous (but not differentiable) function $f$. By Weierstraß' theorem, it is the uniform limit of polynomials. Polynomials are of course $C^1$ (even $C^\infty$, even analytic), so that gives you a sequence of $C^1$ functions converging to $f$ in the sup-norm. – Daniel Fischer Feb 26 '14 at 15:10
• Not here, since that would place the corner on an endpoint of the interval, $\lvert x\rvert$ is continuously differentiable on $[0,1]$. The translation by $\frac12$ serves to place the corner of $\lvert x-c\rvert$ in the interior of the interval. – Daniel Fischer Mar 24 '14 at 10:12