Continuity of Addition, Scalar Mult., and Mult. on Normed Algebras

# Continuity of Addition, Scalar Multiplication, and Multiplication on Normed Algebras

Proposition 1: Let $(X, \| \cdot \|)$ be a normed algebra. Then additional $+ : X \times X \to X$, scalar multipcation, $\cdot : X \times \mathbf{F} \to X$, and multiplication $\cdot : X \times X \to X$ are continuous with respect to the product topologies on $X \times X$ ($X \times \mathbf{F}$ for scalar multiplication). |

**Proof:**Every normed linear space is a topological vector space and so addition and scalar multiplication are continuous. All that we need to show is that multiplication is continuous.

- Let $f : X \times X \to X$ be defined for all $(x, y) \in X \times X$ by $f(x, y) = x \cdot y$. Let $(a, b) \in X \times X$ and let $\epsilon > 0$ be given. We aim to show that $f$ is continuous at $(a, b)$ and hence continuous on all of $X \times X$.

- Let $\epsilon > 0$ be given with $\epsilon < 1$ as well. Then if $\| x - a \| < \epsilon$ and $\| y - b \| < \epsilon$ we have that:

\begin{align} \quad \| f(x, y) - f(a, b) \| &= \| x \cdot y - a \cdot b \| \\ &= \| x \cdot (y - b) - (a - x) \cdot b \| \\ & \leq \| x \cdot (y - b) \| + \| (a - x) \cdot b \| \\ & \leq \| x \| \| y - b \| + \| a - x \| \| b \| \\ & < \| x \| \epsilon + \| b \| \epsilon \\ & < (\| x \| + \| b \|) \epsilon \\ & < ( \| x - a + a \| + \| b \|) \epsilon \\ & < ( \| x - a \| + \| a \| + \| b \|) \epsilon \\ & < ( \epsilon + \| a \| + \| b \|) \epsilon \\ & < \epsilon^2 + (\| a \| + \| b \|) \epsilon \end{align}

- Since the above inequality holds for all $\epsilon > 0$ with $\epsilon < 1$ we have that $f$ is continuous at $(a, b)$ and hence continuous on all of $X \times X$. $\blacksquare$