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Define the norm as: $$\vert\vert f \vert\vert=\{\int_a^b [f^2+(f')^2]dx\}^{\frac{1}{2}}$$

I have shown $\vert\vert \cdot\vert\vert$ is actually a norm.

I know a corollary of the open mapping theorem, which says that if a vector space $X$ is complete with respect to two norms, and one norm is greater than the other one, then these two norm are equivalent.

I want to use this corollary, but I can't find a suitable norm($X$ needs to be complete with respect to this norm and this norm can be compared with $\vert\vert \cdot\vert\vert$).

If our space is $L^p[a,b]$, I can use the $L^p$-norm to argue with this corollary. But I have no idea about $C^1[a,b]$.

The suggested answer gives a sequence of functions $\{f_n\}$ in $C^1[0,1]$ which is Cauchy but not converges in $C^1[0,1]$, where $$f_n(x)=\sqrt{x^2+\frac{1}{n^2}}, x\in [0,1]. $$

But I think its estimate which intends to show $\{f_n\}$ is Cauchy has something wrong.

Can someone help me show $C^1[a,b]$ with this norm is not complete?

Thanks a lot.

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I found a suitable norm occasionally just now, in Stackexchange.

The suitable norm is $||f||_1 = \sup_{a\leq x\leq b}|f(x)|+\sup_{a\leq x\leq b}|f'(x)|.$

And with the corollary what I said in the problem, we can prove the problem above.

More details about this suitable norm please refer to: Prove that $C^1([a,b])$ with the $C^1$- norm is a Banach Space

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