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 $C^1$?

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    $\begingroup$ $$f_n(x) = \sqrt{\left(x-\frac12\right)^2 + \frac{1}{n}}$$ $\endgroup$ – Daniel Fischer Feb 26 '14 at 14:59
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    $\begingroup$ 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 $\|\cdot\|_\infty$. $\endgroup$ – froggie Feb 26 '14 at 15:00
  • $\begingroup$ @DanielFischer Thank you this answers my question. The pointwise and uniform limit is $|x-1/2|$ and this is of course not differentiable. $\endgroup$ – newb Feb 26 '14 at 15:01
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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Feb 26 '14 at 15:10
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    $\begingroup$ 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. $\endgroup$ – Daniel Fischer Mar 24 '14 at 10:12

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