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Let $C^{(n)}[0,1]=\{f:[0,1]\to \mathbb{K}: f^{(n)} \text{ is continuous}\}$. Define $\|f\|=\sup\limits_{0\leq k\leq n}\{|f^{(k)}(x)|:x\in [0,1]\}$. I have to show that $(C^{(n)}[0,1],\left\|\cdot\right\|)$ is a Banach space.

It can be shown that $(C^{(n)}[0,1],\left\|\cdot\right\|)$ is a normed space. In order to show that it is a Banach space, let $(f_m)$ be a Cauchy sequence in $C^{(n)})[0,1]$. Let $\epsilon >0$. Then there exists $N\in \mathbb N$ such that $\|f_m-f_p\|<\epsilon$ for all $n\geq N$. This implies that $(f^{(k)}_m(x))$ is a Cauchy sequence in $\mathbb K$ for each $x\in [0,1]$ and for all $0\leq k\leq n$. Now how to choose $f$? If I choose $f(x)=\lim\limits_{m\to \infty}f_m^{(k)}(x)$ for all $x$, then does this $f$ belong to $C^{(n)}[0,1]$? Any help is appreciated.

DMcMor
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Anupam
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  • Apply the result of (https://math.stackexchange.com/questions/214218/uniform-convergence-of-derivatives-tao-14-2-7) and induction – user363464 May 27 '17 at 07:05

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