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Let $X$ be a Banach space with norm $\|\cdot\|_X$. Define $$ \|f\|_p := \left (\sum_{n=1}^\infty \|f(n)\|_X^p \right )^{1/p} $$ for any function $f: \mathbb N \rightarrow X$. Let $\ell^p := \{f : \mathbb N \rightarrow X : \|f\|_p < \infty \}$. I want to show that $\ell^p$ is a Banachspace, too. I have already proven that it is a vectorspace . Further if we have a Cauchysequence $\{f_n\}$ in $\ell^p$ we may define $f:\mathbb N \rightarrow X: n \mapsto \lim_{k \rightarrow \infty} f_k(n)$. This limit exists since every sequence $\{f_n(k)\}_n$ is Cauchy in $X$. But now I get stuck how to prove that $f \in \ell^p$. If I know this I can prove that $f_n \rightarrow f$ in $\ell^p$.

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  • $\begingroup$ Use the fact that your Cauchy sequence is bounded. $\endgroup$ – Etienne Sep 11 '13 at 17:34
  • $\begingroup$ If $\|f_n\|_p \leq M$ for each $n \in \mathbb N$, how can I then take the limit in $n \rightarrow \infty$ ? $\endgroup$ – user42761 Sep 11 '13 at 17:40
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Choose a $k$ such that $\lVert f_i - f_j \rVert_p < 1$ for $i,j \geqslant k$. Then, for every fixed $N \in \mathbb{N}$, you have

$$\left(\sum_{n=1}^N \lVert f_i(n) - f_k(n)\rVert^p \right)^{1/p} \leqslant \lVert f_i - f_k\rVert_p < 1$$

for every $i > k$. Since there are only finitely many terms in the sum, you can take the limit $i \to \infty$, and obtain

$$\left(\sum_{n=1}^N \lVert f(n) - f_k(n)\rVert^p \right)^{1/p} \leqslant 1.$$

Since that inequality holds for all $N$, you have $(f - f_k) \in \ell^p$ and $\lVert f - f_k\rVert_p \leqslant 1$. Since $\ell^p$ is a vector space and $f_k \in \ell^p$, it follows that $f = f_k + (f-f_k) \in \ell^p$. Showing that $f_k \to f$ in $\ell^p$ can be done by the same method.

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