$C^{1}[0,1]$ is not Banach under $\|\cdot\|_{\infty}$ This is a curiosity from a reading a text that offered no proof. Why is $(C^{1}[0,1], \|\cdot\|_{\infty})$ not Banach?

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.
