Sum of continuous functions in non-metrizable space

I found the following problem in Brezis, Functional Analysis. It seems obvious, but I can't find a start point to solve it.

Let $X$ be a topological space and $E$ be a Banach space. Let $u,v : X \to E$ be two continuous maps from $X$ with values in $E$ equipped with the weak topology $\sigma(E,E^*)$.

1. Prove that the map $x \mapsto u(x)+v(x)$ is continuous from $X$ to $E$ equipped with $\sigma(E,E^*)$.

2. Let $a : X \to \Bbb{R}$ be a continuous function. Prove that the map $x \mapsto a(x)u(x)$ is continuous from $X$ into $E$ equipped with $\sigma(E,E^*)$.

Since the weak topology is not metrizable, I cannot use an approach similar to metric spaces case. Please give me a hint. Thank you.

• What is $y$? And what happened to $v$? – Chris Eagle Sep 28 '11 at 11:48
• Sorry. I'll correct it now. – Beni Bogosel Sep 28 '11 at 11:53
• Hint: by definition of the weak topology, a map $f:X \rightarrow E$ is weakly continuous iff $lf$ is continuous whenever $l$ is a bounded linear functional on $E$. – Chris Eagle Sep 28 '11 at 12:00

The $\sigma(E,E^*)$ topology on $E$ is an example of an initial topology; that is, it is the coarsest topology on $E$ that makes $\ell : E\to\mathbb{R}$ continuous for all $\ell\in E^*$.
Consequently, a map $f:X\to E$ from another topological space $X$ is continuous iff every composition $\ell\circ f$ is continuous from $X$ to $\mathbb{R}$.
For instance, if $u,v:X\to E$ are continuous, then so are $\ell\circ u$ and $\ell\circ v$. Therefore, $x\mapsto\ell(u(x)+v(x))=\ell(u(x))+\ell(v(x))$ is continuous into $\mathbb{R}$, and so $x\mapsto u(x)+v(x)$ is continuous into $E$.
Similarly, $x\mapsto\ell(a(x)u(x))=a(x)\ell(u(x))$ is continuous into $\mathbb{R}$, and so $x\mapsto a(x)u(x)$ is continuous into $E$.
• Glad to help! $\$ – user940 Sep 28 '11 at 12:08
• @Beni: I would say the weak topology is a red herring here. To say that $E$ is a topological vector space over $\mathbb{R}$ amounts to saying that addition $E \times E \to E$ and scalar multiplication $\mathbb{R} \times E \to E$ are continuous. By definition of the product topology, the maps $(u,v): X \to E \times E$ and $(a,u) : X \to \mathbb{R} \times E$ are continuous. Now your maps $x \mapsto u(x) + v(x)$ and $x\mapsto a(x) u(x)$ are exhibited as compositions of continuous maps as soon as you check that $(E, \sigma(E,E^{\ast}))$ is a t.v.s. (and that's the heart of Byron's argument). – t.b. Sep 28 '11 at 13:54