Doubt in proof of continuity in weak topology

Question

Let $(X,\tau)$ a topological space and $E$ a Banach space. Let $a: (X,\tau)\to (\mathbb{R},\tau_{\mathbb{R}}) $ and $ u: (X,\tau) \to (E,\sigma(E,E^*))$ be continuous maps. I want to prove that the map $ x \mapsto f(x) = a(x)u(x)$ is also continuous. Initially, let $x \in X$, define a neighborhood around $f(x)$ in $(E,\sigma(E,E^*))$: $V = \{y \in E: |\langle\phi_i, y - f(x) \rangle | < \epsilon, \forall i \in \{1,...,m\}\}$ where $\phi_i\in E^*$ But I am with difficulty to define appropriates neighborhoods around $u(x)$ and $a(x)$. Can someone give me a hint?

Answer 1

No need to go to such a level, just use general facts:

First note that the map $(E, \sigma(E,E^*)) \times (\Bbb R, \tau_{\Bbb R}) \to (E, \sigma(E,E^*))$ given by $(u,t) \to t\cdot u$ is continuous (as the weak topology is a topological vector space too).

Also $(X, \tau) \to (E, \sigma(E,E^*)) \times (\Bbb R, \tau_{\Bbb R})$ given by $x \to (u(x), \alpha(x))$ is continuous by general facts on topological products, as $\alpha$ and $u$ are continuous.

Your map is just the composition of two continuous maps (the second product map, followed by the scalar multiplication map) and thus continuous too.