$(C^1[0,1], || . ||_\infty)$ is not Banach Space.

analysisbanach-spacesfunctional-analysis

Recently, I am studying Banach Space.
I know that $(C[0,1], || . ||_\infty)$ is a Banach Space and closed subspace of Banach Space is Banach. This follows from the result that closed subset of complete space is complete. I have shown that $(C^1[0,1], || . ||_\infty)$ is not a closed subspace of $(C[0,1], || . ||_\infty).$ This can be shown using "Stone Weierstrass" theorem. Up to this is fine. Now my problem is from here can I say that $(C^1[0,1], || . ||_\infty)$ is not Banach? Or I have to give a counter-example so that $(C^1[0,1], || . ||_\infty)$ is not complete. Please help me. Thanks in advance.

Best Answer

What about a sequence of $\mathcal C^1(\mathbb R)$ function that converges uniformly to $|x|$ ?

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