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

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

Proposition 1: Let $\mathfrak{A}$ be a normed algebra. Then addition, scalar multiplication, and vector multiplication are all continuous. |

**Proof:**Every normed linear space is a topological vector space and so addition and scalar multiplication on $\mathfrak{A}$ are continuous. All that we need to show is that multiplication is continuous.

- Let $f : \mathfrak{A} \times \mathfrak{A} \to \mathfrak{A}$ be defined for all $(a, b) \in \mathfrak{A} \times \mathfrak{A}$ by $f(a, b) = a \cdot b$. Let $(a_0, b_0) \in \mathfrak{A} \times \mathfrak{A}$. We aim to show that $f$ is continuous at $(a_0, b_0)$ and hence continuous on all of $\mathfrak{A} \times \mathfrak{A}$.

- Let $\epsilon > 0$ be given. If $\| a - a_0 \| < \epsilon$ and $\| b - b_0 \| < \epsilon$ we have that:

\begin{align} \quad \| f(a, b) - f(a_0, b_0) \| &= \| a \cdot b - a_0 \cdot b_0 \| \\ &= \| a \cdot b - a_0 \cdot b + a_0 \cdot b - a_0 \cdot b_0 \| \\ &= \| (a - a_0) \cdot b + a_0(b - b_0) \| \\ &\leq \| a - a_0 \| \| b \| + \| a_0 \| \| b - b_0 \| \\ &< \epsilon (\| b \| + \| a_0 \|) \\ &< \epsilon (\| b - b_0 + b_0 \| + \| a_0 \|) \\ &< \epsilon (\| b - b_0 \| + \| b_0 \| + \| a_0 \|) \\ &< \epsilon ( \epsilon + \| a_0 \| + \| b_0 \|) \\ &< \epsilon^2 + \epsilon (\| a_0 \| + \| b_0 \|) \end{align}

- Since the above equality holds true for all $\epsilon > 0$ we see that $f$ is continuous at $(a_0, b_0)$, and is thus continuous on all of $\mathfrak{A} \times \mathfrak{A}$, i.e., multiplication on $\mathfrak{A}$ is continuous. $\blacksquare$