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If $\lim_{k \to \infty} \| u_k - u \|_{L^2(\Bbb R^n)} = 0$ then how can I show that $$ \lim_{k \to \infty} \int_{\Bbb R^n} u_k v = \int_{\Bbb R^n} uv$$ for any $v \in L^2 (\Bbb R^n)$?

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Note that this is a particular case of math.stackexchange.com/questions/235414/… –  Martin Argerami Nov 12 '12 at 15:01
    
Hint: Write the difference as an integral, then use Cauchy-Bunyakovsky-Schwarz inequality. –  Davide Giraudo Nov 12 '12 at 15:02
    
Thank you guys, it was trivial. –  Ann Nov 12 '12 at 15:05

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up vote 3 down vote accepted

$$ \left| \int u_k v - \int u v \right| = \left| \int (u_k -u)v \right| \leqslant \int | u_k - u | |v| \leqslant \| u_k - u \|_2 \| v \|_2 \to 0$$

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Yep, exacttly that. –  Martin Argerami Nov 12 '12 at 15:12
    
And note that it works in any inner-product space. –  Davide Giraudo Nov 12 '12 at 15:17

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