# weak consecutive connection of straight drivers

Suppose I have a weakly sequentially continual straight driver T in between 2 normed straight rooms X and also Y (i.e. $x_n \stackrel {w}{\rightharpoonup} x$ in $X$ $\Rightarrow$ $T(x_n) \stackrel {w}{\rightharpoonup} T(x)$ in $Y$). Does this indicate that my driver T must be bounded?

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2019-05-13 03:44:53
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In my initial solution I just stated that it benefits $Y$ full, yet as Nate mentioned in a comment, I never ever in fact made use of efficiency of $Y$.
The solution is of course. Weakly convergent series in a normed room are bounded, therefore of the consistent boundedness concept related to the twin room (which is a Banach room) and also the reality that a convergent series of actual (or facility) numbers is bounded. If $T$ is boundless, after that there is a series $x_1,x_2,\ldots$ in $X$ converging in standard (and also therefore weakly) to 0 such that $\|T(x_n)\|\to\infty$, so by the previous sentence this indicates that $T(x_1),T(x_2),\ldots$ does not merge weakly.