We have the following result:
Let X be a uniformly convex Banach space, $\{x_n\}_{n\in\mathbb{N}} \subset X, x \in X, \|x_n\| \to \|x\|$, $x_n$ converges weakly to $x$.
Then $x_n$ also converges strongly to $x$.
$\textbf{My question:}$
I have a bounded linear operator $T: X\to Y$, where $X$ and $Y$ are reflexive Banach spaces and $Y$ is a uniformly convex spacce. And we know that in reflexive Banach space, a bounded sequence has a weakly convergent subsequence. So if I take a bounded sequence $\{x_n\}\in X$ then we get $\{Tx_n\}$ as bounded in $Y$ [By linearly and boundedness of $T$]. Now since $\{Tx_n\}$ is bounded in a reflexive Banach space $Y$ then it has a weakly convergent subsequence. Also we have for any bounded sequence $\{f_n\} \in X$,$\{Tf_n\}$ is uniformly integrable in $L^{\Psi}$. Now if I can show that $\lVert Tf_n \rVert_{Y}\to ||Tf||_{Y}$ for some $Tf\in Y$ then we can say that $\{Tf_n\}$ converges strongly in $Y$. Is there a condition/assumption that I can add to conclude such convergence?