If $u \in H^2_0$ and $v \in H^1_0$, I would like to show that $$\forall v \in H^1_0, \mbox{if}~a(u,v) = 0, \forall u \in H^2_0~\mbox{then}~v = 0,$$ where $a(u,v) = \langle Lu,v \rangle$ and $Lu = a(x) u^{(4)} + b(x) u''' + c(x) u'' + d(x) u' +e(x) u$. Also $a(x), b(x), c(x), d(x)$ and $e(x)$ are bounded nonzero functions on the interval $(0,1)$.
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this result is wrong if $a=b=c=d=e=0$. – supinf May 10 '17 at 10:35
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I can put the conditions on these functions. So I will update the post to have nonzero functions. Thanks – Ahmed May 10 '17 at 10:37
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Your notation is unclear. You have a bilinear form $a(.,.)$ and you want to show $\forall u, a(u,v) = 0 \implies v =0$. If $v \ne 0$, you need to construct $u$ such that $a(u,v) \ne 0$ – reuns May 10 '17 at 10:48
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I can restate what I want as follows $$\sup_{u \in H^2_0} a(u,v) > 0~~\forall v \in H^1_0, v\neq 0$$ – Ahmed May 10 '17 at 14:51