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In the second answer to this post this it is stated:

Since you've already proved that there is a strongly convergent subsequence, let's say $ Tu_{n_k} \to u^* $ for $ k \to \infty $. Then by the weak convergence of $ u_n \rightharpoonup u $ you get immediately that $ Tu_n \rightharpoonup Tu $. Now since strong convergence implies weak convergence and from the uniqueness of the limit of a weak convergent sequence it must be true that $$ u^*= Tu $$

I understand the uniqueness of weak limits to mean if $ Tu_n \rightharpoonup Tu $ and $ Tu_n \rightharpoonup Tu' $ then $Tu = Tu'$. But should that mean the strong limit, $u^*$, should also coincide with $Tu$? This seems like a stronger statement, I'm not quite understanding the logic of how to go back to the strong limit.

yoshi
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  • You dont go back to the strong limit: Since $Tu_{n_k} \to u^$ you get that $Tu_{n_k} \rightharpoonup u^$. But you also have that $Tu_{n_k} \rightharpoonup Tu$ – Evangelopoulos Foivos Jan 07 '23 at 18:37
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    Okay so the idea is that $Tu_{n_k}$ converges to some $u^$. That $u^$ is found to be $Tu$ (because every weak subsequence is converging to $Tu$). So $Tu_{n_k} \rightarrow Tu$ – yoshi Jan 08 '23 at 21:23

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