Q:

Are weakly compact sets bounded?

Let $X$ be a Hausdorff locally convex topological vector space, and let $X'$ denote its topological dual, that is, the vector space of all continuous linear functionals on $X$.
If $A$ is a weakly compact subset of $X$, that is, if $A$ is $\sigma(X,X')$-compact in $X$, is it (always) true that $A$ is bounded in $X$ ?
If not, can anybody give a counterexample for such a set?   

A:

In a locally convex Hausdorff space all weakly bounded sets are bounded. In particular, weakly compact sets are bounded.
EDIT. That weakly bounded sets are bounded is stated, e.g., in Rudin's Functional Analysis, Theorem 3.18. The proof is a nice application of the uniform boundedness priciple: If $B\subseteq X$ is weakly bounded and $p$ is a continuous seminorm on $X$, then $X^*_p=\{f\in X^*: p^*(f)<\infty \}$ endowed with the dual norm $p^*(f)=\sup\{|f(x)|: p(x)\le 1\}$ is a Banach space. For every $x\in B$ the evaluation $\delta_x(f)=f(x)$ is linear and continuous on $X_p^*$ and $\{\delta_x: x\in B\}$ is pointwise bounded because $B$ is weakly bounded. Hence, by the uniform boundedness principle,  $\sup\{|\delta_x(f)|: x\in B, p^*(f)\le 1\} =c<\infty$ and then Hahn-Banach implies $\sup\{p(x): x\in B\}\le c$, i.e., $B$ is bounded in $X$.

