# Equivalent norm in sobolev space H^2

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?

• Only for the subspace $\{w\in H^2(0,a): w(0)=w(a) =w'(0) = w'(a)= 0\}$. – Shuhao Cao Sep 17 '13 at 19:28
• Thank you for answer, but are you sure that only in this subspace? – ela Sep 17 '13 at 19:48
• 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). – Shuhao Cao Sep 17 '13 at 20:17

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.

• 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 $H^{2}$. – ela Sep 17 '13 at 19:30
• From my example, this bound cannot exist, because if $w = 1$ is constant, then $\|w\|_{H^2} \neq 0$ but $\| w'' \|_{L^2} = 0$. – Christopher A. Wong Sep 17 '13 at 23:48
• 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. – ela Sep 18 '13 at 8:46