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Prove $\big(C^{(2)}_{[0,1]},||\cdot||_{C^{(2)}_{[0,1]}}\big) $ is complete.

First I derived that $\big(C^{(1)}_{[0,1]},||\cdot||_{C^{(1)}_{[0,1]}}\big) $ is complete. How can I say that $\big(C^{(2)}_{[0,1]},||\cdot||_{C^{(2)}_{[0,1]}}\big) $ is also complete?

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    Is $\|f\|_{C^{(2)}_{[0,1]}} = \|f\|_{\infty}+\|f'\|_{\infty}+\|f''\|_{\infty}?$ – 2017-02-10
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    Yes exactly it is right. – 2017-02-10

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Suppose $f_n$ is a Cauchy sequence in $C^{(2)}_{[0,1]}$. That in particular tells you that $f_n'$ is a Cauchy sequence in $C^{(1)}_{[0,1]}$ (with $\|\cdot\|_{C^{(1)}_{[0,1]}}$ norm) but you know that it is complete, so $f_n'\to h\in C^{(1)}_{[0,1]}$ (with $f_n''\to h'$ in $\|\cdot\|_{\infty}$ norm). On the other hand, $f_n$ is also a Cauchy in $C^{(1)}_{[0,1]}$ so again $f_n\to f\in C^{(1)}_{[0,1]}$.

Moreover (proof of) completeness tells you that $f'=h$, and so $f''=h'\in C[0,1]$ and all this gives you that $f_n\to f$ in $C^{(2)}_{[0,1]}$ norm