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I consider space $H^{2}(0,a)=\{ f\in L^{2}(0,a): f',f''\in L^{2}(0,a) \}$ I define norm $\Vert w \Vert_{H^{2}}:=b\Vert w''\Vert_{L^{2}}$, where b is positive constant.

I couldn't proof that it is norm equivalent to standard norm in $H^{2}$.

Maybe is easier show that $H^{2}$ with this norm is a Hilbert space?

Could you help me?

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    $\begingroup$ Only for the subspace $\{w\in H^2(0,a): w(0)=w(a) =w'(0) = w'(a)= 0\}$. $\endgroup$ – Shuhao Cao Sep 17 '13 at 19:28
  • $\begingroup$ Thank you for answer, but are you sure that only in this subspace? $\endgroup$ – ela Sep 17 '13 at 19:48
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    $\begingroup$ yes, just to rule out the possibility linear functions, if $v$ is linear, then it has a non-zero standard $H^2$-norm, yet $v'' = 0$ and the norm you defined is zero. If you add the boundary condition, if $v$ is linear, it must be zero, so that $\|v''\|_{L^2} = 0$ will imply $v=0$ (a requirement of something being a norm). $\endgroup$ – Shuhao Cao Sep 17 '13 at 20:17
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This is false; consider $w = 1$ (a constant function on the interval). Then clearly $w$ has non-zero $H^2$-norm, but if you just take the $L^2$ norm of its derivative, it'll of course be zero.

However, it is true if you're considering the space $H^2_0(0,a)$, the space of $H^2$ functions with zero trace.

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  • $\begingroup$ Could you explain me how to show the lower bound, i.e. $C\Vert w\Vert_{H^{2}}\leq b\Vert w''\Vert_{L^{2}}$, $C$-constant. Now $\Vert w\Vert_{H^{2}}$ is standard norm in $H^{2}$. $\endgroup$ – ela Sep 17 '13 at 19:30
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    $\begingroup$ From my example, this bound cannot exist, because if $w = 1$ is constant, then $\|w\|_{H^2} \neq 0$ but $\| w'' \|_{L^2} = 0$. $\endgroup$ – Christopher A. Wong Sep 17 '13 at 23:48
  • $\begingroup$ Sorry, I did mistake in remark the space (I forgot the lower index). I mean the lower bound in the space you propose $H_{0}^{2}$. I understand that in your counterexample the lower bound cannot exist. $\endgroup$ – ela Sep 18 '13 at 8:46

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