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.

  • $\begingroup$ What is $y$? And what happened to $v$? $\endgroup$ Sep 28, 2011 at 11:48
  • $\begingroup$ Sorry. I'll correct it now. $\endgroup$ Sep 28, 2011 at 11:53
  • 2
    $\begingroup$ 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 $E$. $\endgroup$ Sep 28, 2011 at 12:00

1 Answer 1


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 $E$.

  • $\begingroup$ Thank you. I understand now. :) $\endgroup$ Sep 28, 2011 at 12:07
  • $\begingroup$ Glad to help! $\ $ $\endgroup$
    – user940
    Sep 28, 2011 at 12:08
  • $\begingroup$ @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). $\endgroup$
    – t.b.
    Sep 28, 2011 at 13:54
  • $\begingroup$ @t.b. Quite right. $\endgroup$
    – user940
    Sep 28, 2011 at 13:58

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