Suppose given $f,g\in L^2(\mathbb{R})$ and $f', g' \in L^2(\mathbb{R})$, the linear functional defined by $$F(g):= \int_{\mathbb{R}} f'g' dx $$ is continuous with respect to the derivative, that is $$|F(g)| \leq C \|g'\|_2$$ but not continuous with respect to $g$.

However with the additional assumption $f'' \in L^2(\mathbb{R})$, we have $$|F(g)| = \bigg|\int_{\mathbb{R}} f'g' dx\bigg| = \bigg|\int_{\mathbb{R}} f''g dx\bigg| \leq C\|g\|_2 .$$

Is there a any specific name for this kind of work? I would like to find a reference to study if possible. The reason I asked this question is because the above is a very nice case where $$\int_{\mathbb{R}} f'g' dx = - \int_{\mathbb{R}} f''g dx.$$ In general for $\Omega\subset \mathbb{R}^N$ we have $$\int_\Omega \nabla u\cdot \nabla v dx = \int_{\partial \Omega} v\big(\nabla u \cdot \nu\big) d\mathcal{H}^{N-1} - \int_\Omega v\triangle u dx$$ And I got stuck. With the surface integral over the boundary, what conditions we need for $\int_\Omega \nabla u\cdot \nabla v dx$ to be continuous and linear functional w.r.t.$v$?

  • $\begingroup$ The obvious answer is: You need $\Delta u \in L^2(\Omega)$ and the first integral should vanish (therefore, $\nabla u \cdot \nu$ needs to be well-defined on the boundary). In particular, $u \in H^2(\Omega)$ should be sufficient. $\endgroup$ – gerw Dec 1 '15 at 6:21
  • $\begingroup$ @gerw thank you for the reply. How can we know that first integral vanishes? Or you are imposing this condition together with $\triangle u \in L^2(\Omega)$ $\endgroup$ – Xiao Dec 1 '15 at 6:29
  • $\begingroup$ yes you need additionally $\nabla u \cdot \nu = 0$ on $\partial\Omega$. Note that by $u \in H^2(\Omega)$, you have $\nabla u \in H^1(\Omega)$, so you can take its trace. $\endgroup$ – gerw Dec 1 '15 at 8:11

If $\Omega$ is regular enough ($C^1$ or uniformly Lipschitz) then there exists a continuous linear operator $$\operatorname{Tr}\colon W^{1,p}(\Omega) \to L^p(\partial \Omega,\mathcal{H}^{N-1})$$ such that $\operatorname{Tr}(u) = u$ for every $u \in W^{1,p}(\Omega)\cap C^1(\Omega)$, there is a constant $C = C(\Omega,N,p)$ such that $$\|\operatorname{Tr}u\|_{L^{p}(\partial{\Omega})} \le C\|u\|_{W^{1,p}(\Omega)}$$ and the divergence theorem holds, i.e. for all $\varphi \in C^1_c(\mathbb{R}^N)$, $u \in W^{1,p}(\Omega)$, $i = 1,\dots,N$, $$\int_{\Omega}u\frac{\partial \varphi}{\partial x_i} = -\int_{\Omega}\varphi\frac{\partial u}{\partial x_i} + \int_{\partial \Omega}\varphi\operatorname{Tr}u\,d\mathcal{H}^{N-1}.$$

Now fix $u \in W^{2,2}(\Omega)$ and consider the functional $$\mathcal{F}(v) = \int_{\Omega}\nabla u \cdot \nabla v.$$ Then $\mathcal{F}$ is a linear continuous functional on $W^{1,2}(\Omega)$, indeed by the continuity of the trace operator, we have $$|\mathcal{F}(v)| \le C\|\operatorname{Tr}v\|_{L^2(\partial \Omega)}\|\operatorname{Tr}\nabla u\|_{L^2(\partial \Omega,\mathbb{R}^N)} + \|v\|_{L^2(\Omega)}\|\Delta u\|_{L^2(\Omega)} \le C(\Omega,N,u)\|v\|_{W^{1,2}(\Omega)}.$$

You can read more about how to deal with traces in any introductory book in Sobolev spaces.

EDIT: Since you are interested in continuity with respect to the $L^2$ norm, and since there is no trace operator in $L^2$, you need a condition for which the first integral vanishes, for example $\frac{\partial u}{\partial \nu} = 0$.

  • $\begingroup$ Thank you for the reply, but in my problem I need the linear functional to be continuous with respect to the $L^2$ norm. $\endgroup$ – Xiao Dec 1 '15 at 6:27
  • $\begingroup$ I have edited my answer :) $\endgroup$ – Giovanni Dec 1 '15 at 6:36
  • $\begingroup$ Thank you. I knew trace wouldnt work and wished there might be a weaker condition, but this might be the only way. $\endgroup$ – Xiao Dec 1 '15 at 6:41
  • $\begingroup$ Yeah, I have thought my answer for continuity wrt the $W^{1,2}$ norm, so I understand why is not entirely satisfactory.. I'll follow this question and see what happens :) $\endgroup$ – Giovanni Dec 1 '15 at 6:48

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