Weak topology and convergence . If $X$ and $Y$ are banach spaces and T map from $X$ to $Y$. If every 
sequence $x_n$ in weak topology in $X$ converges to $0$ , and $T(x_n)$ converges to $0$ in weak topology , does that imply that $T$ is bounded ? 
From this question i would like to understand how convergence in weak topology differs from other topology. What would be the case if the topology defined was not weak ? 
Thank you for your kind help . Keenly Looking forward to get some good ideas about weak topologies . 
 A: A Banach space is by definition a normed vector space $X$ which is complete as a metric space, with the distance function induced by the norm. The norm induces a topology on $X$, with respect to which you can consider topological questions such as continuity, compactness, etc.
A linear map $T : (X, ||\;||) \rightarrow (Y, ||\;||)$ between normed vector spaces (not neccesarily Banach) is continuous if and only if it is bounded. It is implicit, that you are talking about continuity with respect to the topologies induced by the norms on $X$ and $Y$.
Now, there are a lot of different useful topologies in functional analysis one can put on a normed vector space $(X, ||\;||)$ other than the topology that is induced from the norm. One such topology is called the weak topology. How is it defined?
The norm structure on $(X, ||\;||)$ allows us to define when a linear functional $\phi : X \rightarrow \mathbb{C}$ is continuous. The collection of all continuous linear functionals on $X$ is denoted by $X^*$, and it by itself is a normed vector space (even Banach), with the norm being the operator norm
$$ ||\phi|| = \sup_{||x|| = 1} |\phi(x)|. $$
The weak topology on $X$ is the weakest topology with respect to which all the maps $\phi \in X^*$ are continuous. Obviously, the original topology, induced by the norm on $X$ is a topology on $X$ under which all maps $\phi \in X^*$ are continuous, but it is quite possible you can throw away some open sets, and still remain with a topology for which all $\phi \in X^*$ are still continuous. Indeed, the weak topology on $X$, which we will denote by $\tau$, is the minimal topology which contains the obvious subsets which must be open for the functionals to be continuous -  $\phi^{-1}(U)$ for $U \subset \mathbb{C}$ open and $\phi \in X^*$. 
Assume that $X$ is infinite dimensional. Some properties of this topology you should get used to and prove:


*

*The weak topology $\tau$ has less open subsets than the norm topology on $X$.

*The weak topology $\tau$ is not first countable. It is not induced by any norm on $X$ nor by any distance function.

*The identity map $\mathrm{id} : (X, ||\;||) \rightarrow (X, \tau)$ is continuous. The inverse of the identity map is not.

*A map $T : X \rightarrow Y$, which is continuous as a map between the normed vector spaces $T : (X, ||\;||_1) \rightarrow (Y, ||\;||_2)$ will also be continuous as a map between the topological vector spaces $T : (X, \tau_1) \rightarrow (Y, \tau_2)$ when $\tau_i$ are the weak topologies on $X$ and $Y$. The other direction is generally false.



Hmm, it seems I misunderstood the question, but this still might be useful...
A: Hint: If $T$ is unbounded then there is a sequence $x_n$ such that $x_n \to 0$ in norm (hence also weakly) but $T x_n$ is unbounded.  However, it follows from the uniform boundedness principle that any weakly convergent sequence is bounded.
