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Let be a subsequence of weakly convergent to, then by Step 7 as.
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Since is weakly compact, there is a subsequence of converging weakly to some.
Since, also converges to Since, and are continuous, we have Thus (ii) Proof follows from (i). (iii) Since is weakly compact, there is a subsequence of converging weakly to some.
Then a subsequence of converges weakly to.
Suppose that a subsequence of converges weakly to.
Now assume that is a weak limit point of and a subsequence of converges weakly to.
The weak compactness of implies that there is a subsequence of converging weakly to as.
So, we assume that a subsequence of converges weakly to We shall show that.
Since is bounded, we can choose a subsequence of converging weakly to.
If follows from reflexivity of and the boundedness of a sequence that there exists which is a subsequence of converging weakly to as.
From Theorem 2.4, for each, there exists such that The weak compactness of implies that there is a subsequence of converging weakly to as.
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