Proof. Suppose $(V,\Vert\cdot\Vert)$ is the normed vector space, and $(e_i)_{i=1}^N$ is a basis for $V$ . For $x=\sum_{j=1}^N \lambda_j e_j$ , we can then define $$ \Vert x \Vert' = \sqrt{\sum_{j=1}^N |\lambda_j|^2} $$ whence $\Vert\cdot\Vert'\colon V\to \sR$ is a norm for $V$ . Since all norms on a finite dimensional vector space are equivalent, there is a constant $C>0$ such that $$ \frac{1}{C} \Vert x \Vert' \le \Vert x \Vert \le C \Vert x \Vert', \quad x\in V. $$ To prove that $V$ is a Banach space, let $x_1,x_2,\ldots$ be a Cauchy sequence in $(V,\Vert\cdot \Vert)$ . That is, for all $\varepsilon>0$ there is an $M\ge 1$ such that $$ \Vert x_j-x_k \Vert <\varepsilon, \ \ \mbox{for all} j,k\ge M. $$ Let us write each $x_k$ in this sequence in the basis $(e_j)$ as $x_k=\sum_{j=1}^N \lambda_{k,j} e_j$ for some constants . For $k,l\ge 1$ we then have \begin{eqnarray*} \Vert x_k-x_l\Vert &\ge& \frac{1}{C} \Vert x_k-x_l \Vert' \\ &\ge& \frac{1}{C} \sqrt{\sum_{j=1}^N |\lambda_{k,j}-\lambda_{l,j}|^2} \\ &\ge& \frac{1}{C} |\lambda_{k,j}-\lambda_{l,j}| \end{eqnarray*}for all $j=1,\ldots, N$ . It follows that $(\lambda_{k,1})_{k=1}^\infty, \ldots, (\lambda_{k,N})_{k=1}^\infty$ are Cauchy sequences in . As is complete, these converge to some complex numbers $\lambda_1, \ldots, \lambda_N$ . Let $x=\sum_{j=1}^N \lambda_j e_j$ .

For each $k=1,2,\ldots$ , we then have \begin{eqnarray*} \Vert x-x_k\Vert &\le& C \Vert x-x_k\Vert' \\ &\le& C \sqrt{\sum_{j=1}^N |\lambda_{j}-\lambda_{k,j}|^2}. \end{eqnarray*}By taking $k\to \infty$ it follows that $(x_j)$ converges to $x\in V$ .