How can we prove that the space $V = \{ v \in {H^1}(0,L),\int\limits_0^L {v(x)dx = 0} \} $ is dense in ${L^2}(0,L)$ ? thank you.
Zero average function density
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functional-analysis
pde
1 Answers
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It is not.
Since $v \mapsto \int_0^L v$ is a bounded functional on $L^2(0,L)$, its kernel $K = \{ v \in L^2(0,L): \int_0^L v = 0 \}$ is a closed proper subspace of $L^2(0,L)$. On the other hand $V \subseteq K$ and hence $V$ cannot be dense in $L^2(0,L)$.
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0Can it be dense in $ H^1 $ ? Thank you .... – 2017-02-19
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0No, $V$ is a closed proper subspace of $H^1$. The same argument (with $L^2$ replaced by $H^1$) applies. – 2017-02-19