This is a curiosity from a reading a text that offered no proof. Why is $(C^{1}[0,1], \|\cdot\|_{\infty})$ not Banach?
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You just need to find a sequence that is Cauchy but not convergent w that metric. – Nap D. Lover Mar 10 '16 at 01:35
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$C^1([0,1])$ is not closed for the $|.|_\infty$ norm. it becomes a Banach space only after you completed $C^1([0,1])$ with the limits of converging sequence (in that case it becomes $L^\infty([0,1])$ ) – reuns Mar 10 '16 at 01:44
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I think it rather becomes $C[0,1]$. – Friedrich Philipp Mar 10 '16 at 01:50
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By Stone—Weierstrass, any continuous function can be uniformly approximated on $[0,1]$ by a sequence of polynomials.
Take any continuous function in $C^0[0,1] \setminus C^1[0,1]$, i.e. $f$ that is continuous but not continuously differentiable on $[0,1]$. Such functions exist.
Now, take a sequence $(P_n)_n$ of polynomials so that $\lVert f - P_n\rVert_\infty\xrightarrow[n\to\infty]{} 0$. Such sequence exists by Stone—Weierstrass, and clearly $P_n \in C^1[0,1]$ for all $n$.
But then you have a Cauchy sequence $(P_n)_n$ (since it converges in $C^0[0,1]$ for the $\lVert\cdot\rVert_\infty$ norm) that does not converge in $C^1[0,1]$ (since $f$ is not $C^1$). So $(C^1[0,1],\lVert\cdot\rVert_\infty)$ is not complete.
Clement C.
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