Proof. sketch [br97A]

    1. Let $\{ \mathbf{e}_{1}, \dots, \mathbf{e}_{n} \}$ be the standard basis in $\mathbb{R}^{n}$ and suppose $\mathbf{x} = \sum c_{i}\mathbf{e}_{i}$. Now $\lvert \mathbf{x} \rvert \leq 1 \implies \lvert c_{i} \rvert \leq 1$ for all $i$. Then $$\lvert A \mathbf{x} \rvert = \left\lvert \sum c_{i} A \mathbf{e}_{i} \right\rvert \leq \sum \lvert c_{i} \rvert \lvert A \mathbf{e}_{i} \rvert \leq \sum \lvert A \mathbf{e}_{i} \rvert$$ so that $$\lVert A \rVert \leq \sum_{i=1}^{n} \lvert A \mathbf{e}_{i} \rvert < \infty. $$

      Each $A \mathbf{e}_{i}$ is a specific vector in $\mathbb{R}^{m}$, and hence $\lvert A \mathbf{e}_{i} \rvert < \infty$.

      Since $\lvert A \mathbf{x} - A \mathbf{y} \rvert \leq \lVert A \rVert \lvert \mathbf{x} - \mathbf{y} \rvert$ if $\mathbf{x}, \mathbf{y} \in \mathbb{R}^{n}$, we see that $A$ is uniformly continuous.

    2. Omitted.

    3. Omitted.