$B \subseteq X$ is bounded $\Leftrightarrow$ $\forall U \in \mathcal{U}(0) ~ \exists \lambda_U: ~B \subseteq \lambda_u U $

Question

I have some troubles with the following problem and hope some of you can help me.

Let $X$ be a vector space, equipped with the $\sigma$-weak topology $\sigma(X,Y)$, where $Y$ is a subspace of the algebraic dual-space (of $X$) $X^*$. Let $||.||$ be a norm, such that $(X,||.||)$ is a Banach space and in addition to the former definitions let be $Y = X'= L_b(X, \mathbb{C})$, where $X'$ is the topological dual space and $L_b(X,\mathbb{C})$ stands for all bounded Maps from $X$ to $\mathbb{C}$.

I have to show, that:

$B \subseteq X$ is bounded (relating to $||.||$) $\Leftrightarrow$ $\forall U \in \mathcal{U}(0) ~ \exists \lambda_U: ~B \subseteq \lambda_u U $

.

Some additional information about my notation:

The topology $\sigma(X,Y)$ is the initial-topology referred with the familiy of all $f\in Y$, so all $f \in Y$ are continous.

I guess the easier direction might be "=>", so I tried to start with this one, but due to I hardly can imagine the $\mathcal{U}(0)$ in the given toplology I fail finding a connex from bounded by norm to bounded in topological sense...

I would be very grateful, if some of you can help me! :)

Answer 1

B is topologically bounded when exists compact K with B subset K.
B is bounded by a metric or norm when B is a subset of a ball.
Equivalently, the diameter of B is finite.

Assume B is bounded by a norm n(x) = ||x||.
Thus exists s > 0, a in X with B subset B(a,s) = { x : n(x - a) < s }
B subset B(0,s + n(a))
since for all x in B, n(x) <= n(a) + n(x - a) < n(a) + s = k.

Let U be a nhood of 0. 
So exists r > 0 with B(0,r) = { x : n(x) < r } subset U.
Consequently, B subset B(0,k) = (k/r)B(0,r) subset (k/r)U.

Conversely, assume for all U nhood 0, exists r > 0 with B subset rU.
U = B(0,1) = { x : n(x) < 1 } is a nhood of 0.
Thus exists r > 0 with B subset r.B(0,1) = B(0,r).
So B is bounded.