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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.

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    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 $E$.2011-09-28

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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 $E$.

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    @t.b. Quite right.2011-09-28