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 $v \in L^2 (\Bbb R^n)$?
Limit and Integral sign in $L^2$.
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real-analysis
functional-analysis
lp-spaces
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1Note that this is a particular case of http://math.stackexchange.com/questions/235414/prove-that-lim-limits-k-rightarrow-infty-int-limits-e-f-k-g-int-limi/235452#235452 – 2012-11-12
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0**Hint:** Write the difference as an integral, then use Cauchy-Bunyakovsky-Schwarz inequality. – 2012-11-12
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0Thank you guys, it was trivial. – 2012-11-12
1 Answers
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$$ \left| \int u_k v - \int u v \right| = \left| \int (u_k -u)v \right| \leqslant \int | u_k - u | |v| \leqslant \| u_k - u \|_2 \| v \|_2 \to 0$$
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1Yep, exacttly that. – 2012-11-12
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0And note that it works in any inner-product space. – 2012-11-12