Prove that $\|v\|_{H^1(\Omega )}\leq C(\|f\|_{L^2(\Omega )}+\|v\|_{H^{1/2}(\partial \Omega )}+\|\partial _\nu v\|_{H^{1/2}(\partial \Omega )}))$ Let $\Omega $ be a smooth bounded subset of $\mathbb R^d$. Assume that $u=0$ if $u\in H^1(\Omega )$ and $u$ satisfies $$-\Delta u-u=0\ \ in\ \ \Omega \quad \text{and}\quad \partial _\nu u=u=0\ \ on \ \ \partial \Omega .$$
Let $f\in L^2(\Omega )$ and assume that $v$ is solution of $-\Delta v-v=f$ in $\Omega $. Prove that 
$$\|v\|_{H^1(\Omega )}\leq C(\|f\|_{L^2(\Omega )}+\|v\|_{H^{1/2}(\partial \Omega )}+\|\partial _\nu v\|_{H^{-1/2}(\partial \Omega )})),$$
where $C$ is independent of $v$ and $f$.
My attempts
My attempts are unfortunately very few... I'm stuck on this problem since yesterday. I tried to make the substitution $w=v-u$, then $w$ is a solution of $\nabla v-v=f$, but it doesn't look to help a lot.
 A: Assume by contradiction that such inequality doesn't hold for any $C$. Then for every $n \in \mathbb{N}$ one can find $v_n \in H^1(\Omega)$ such that 


*

*$-\Delta v_n - v_n = f_n$, 

*$\| v_n \|_{H^1(\Omega)} = 1$, 

*$\|f_n\|_{L^2(\Omega)}, \|v_n\|_{H^{1/2}(\partial \Omega)}, \|\partial_\nu v_n\|_{H^{-1/2}(\partial \Omega)} < \frac 1n$. 


Since $\Omega$ is bounded, one can choose a subsequence so that $v_n \rightharpoonup v$ in $H^1(\Omega)$ (by Banach-Alaoglu theorem) and $v_n \to v$ in $L^2(\Omega)$ (by Rellich-Kondrashov compactness theorem). It is easy to check that $-\Delta v - v = 0$ and $v = \partial_\nu u = 0$ on $\partial \Omega$. By assumption, the only solution is $v=0$. To summarize, we have that $\| v_n \|_{L^2(\Omega)} \to 0$. 
What is left is to show a contradiction with $\| v_n \|_{H^1(\Omega)} = 1$. To this end, test the definition of $\partial_\nu v_n$ (given at the bottom) with $v_n$ itself: 
\begin{align*}
\int_\Omega |\nabla v_n|^2 
& = \langle \partial_\nu v_n, v_n \rangle - \int_\Omega v_n \Delta v_n \\ 
& = \langle \partial_\nu v_n, v_n \rangle + \int_\Omega v_n (v_n+f_n) \\ 
& \xrightarrow{n \to \infty} 0.\end{align*}
The convergence follows from $\|\partial_\nu v_n\|_{H^{-1/2}(\partial \Omega)}, \|v_n\|_{H^{1/2}(\partial \Omega)} \to 0$ for the first term and from $\|v_n\|_{L^2(\Omega)}, \|f_n\|_{L^2(\Omega)} \to 0$ for the second. Thus we obtain $\|\nabla v_n\|_{L^2(\Omega)} \to 0$, which together with $\| v_n \|_{L^2(\Omega)} \to 0$ gives a contradiction. 
Comment. For regular $v$, the divergence theorem yields the following identity: 
$$ \int_{\partial \Omega} \varphi \partial_\nu v = \int_\Omega \operatorname{div}(\varphi \nabla v) = \int_\Omega \nabla \varphi \nabla v + \varphi \Delta v \quad \text{for any } \varphi \in C^1(\overline{\Omega}). $$
If $v \in H^1(\Omega)$ and $\Delta v \in L^2(\Omega)$ (as it is in our case), the RHS can be used to define $\langle \partial_\nu v, \varphi \rangle$ for any $\varphi \in H^{1/2}(\partial \Omega)$, that is 
$$ \langle \partial_\nu v, \varphi \rangle := \int_\Omega \nabla \varphi \nabla v + \varphi \Delta v \quad \text{for } \varphi \in H^{1/2}(\partial \Omega). $$
It can be checked that this doesn't depend on the choice of the extension $\varphi \in H^{1}(\Omega)$. Look up the discussion here. 
