# Convex, bounded and closed on the strong topology $\Rightarrow$ compact on the weak topology

The question reads like this:

Let $$X$$ be a reflexive Banach space and $$K \subset X$$ a set.

i) Given $$r > 0$$ define the application $$T_r: X \rightarrow X$$ as $$T(x) = rx$$. Show that $$T_r$$ is continuous considering in $$X$$ the weak topology in the domain and the counter-domain.

ii) Show that if $$K$$ is convex, bounded and closed in the strong topology, then $$K$$ is compact on the weak topology.

For the item i) i used that a linear application between Banach spaces is continuous when both spaces use the weak topology if and only if it's continuous when both use the strong topology. Since $$T_r$$ is bounded on the strong topology, it is continuous on it, and thus continuous on the weak topology.

I don't get how this should help with the item ii).

• It helps maybe a little. The main thing to consider for ii) is Banach-Alaoglu. Aug 21, 2020 at 18:36

Continuous functions map compact sets to compact sets. So i), together with Banach-Alaoglu, shows that all balls centered at $$0$$ are compact.
Being bounded, $$K$$ is inside a ball. Being convex and closed, it is weakly closed. A closed subset of a compact set, in a Hausdorff space, is compact.
Also, you don't really need any theory for i). If $$x_n\to x$$ weakly, this means that $$f(x_n)\to f(x)$$ for all $$f\in X^*$$. Now $$f(T_rx_n)=f(rx_n)=rf(x_n)\to rf(x)=f(rx)=f(T_rx),\qquad f\in X^*.$$ So $$T_rx_n\to T_rx$$ weakly and $$T_r$$ is continuous (note that one is working with nets here, but nothing changes in the argument).