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According to the book "Applied Functional Analysis" vol. I by Zeidler, the Friedrichs extension of an operator $B\colon D(B)\subset X\to X$, where $X$ is a real Hilbert space, is obtained as follows. One needs:

  1. that $B$ be symmetric, that is $(Bu, v)=(u, Bv)$ for all $u, v\in D(B)$;
  2. that $B$ be strongly monotone (aka strictly positive), that is, $(Bu, u)\ge c\|u\|^2$ for all $u\in D(B)$ for a constant $c>0$.

If those assumptions are satisfies, one defines the energetic space $$X_E=\text{completion of }D(B)\text{ with norm }\|u\|_E^2=(Bu, u).$$ This gives rise to a triple (Hilbert triple as it is sometimes called) $$X_E\subset X\subset X_E^\star, $$ and one defines the energetic extension $$ B_E\colon X_E\to X_E^\star,\qquad {}_{X_E^\star}\langle B_E u, v\rangle_{X_E}=(u, v)_{X_E},$$ that is, $B_E$ equals the Riesz isomorphism of the real Hilbert space $X_E$ with its dual. (It is here that one uses the fact that $X$ is real. In the complex case, this mapping is not linear but conjugate-linear. This is not a big deal nonetheless).

This said, the Friedrichs's extension is the maximal restriction of $B_E$ that does not leave $X$: $$ Au=B_E u\qquad D(A)=\left\{ u\in X_E\ :\ B_E u\in X\right\}.$$

The standard example is obtained on $X=L^2(\Omega)$, where $\Omega$ is a bounded open set in $\mathbb{R}^n$, with $B=-\Delta$ and $D(B)=C^\infty_0(\Omega)$ In this case one has that $X_E=H^1_0(\Omega)$ and the relation $$ Au=f$$ is the abstract reformulation of the boundary value problem $$ \begin{cases} -\Delta u=0 & \Omega \\ u=0 & \text{on }\partial \Omega.\end{cases}$$ In particular, one has obtained Dirichlet boundary conditions.

Question. Can one abstractly obtain Neumann's boundary conditions by means of the same construction? Of course one can assume that $\Omega$ boundary is as smooth as one wishes. If it simplifies things one can even assume that $\Omega$ is the unit ball.

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    $\begingroup$ One problem to address is the additional conditions on $\Omega$ so that the phrase "Neumann boundary conditions" makes sense. After that, the most obvious candidate to try would be $D(B)=$ the infinitely differentiable functions with vanishing normal derivative and with all derivatives bounded. $\endgroup$ Mar 7 '16 at 17:00
  • $\begingroup$ @Justpassingby: That's surely a good start. I forgot to mention that the domain is as smooth as one wishes. The only obstacle is that one does not have $(Bu, u)\ge c(u,u)$, but I think that either considering the operator $B+I$ or quotienting constants out should do the trick. To be continued $\endgroup$ Mar 7 '16 at 19:11
  • $\begingroup$ @Justpassingby: Your idea is the right one. The details are on Davies's book "Spectral theory and differential operators", section 7.2: "Neumann boundary conditions". $\endgroup$ Mar 8 '16 at 11:16
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For smooth functions $u$ with vanishing normal derivative on $\partial\Omega$, \begin{align} (Bu,u) & = \int_{\Omega}(-\Delta u)udx \\ & =\int_{\Omega}\nabla\cdot((-\nabla u)u)+|\nabla u|^2dx \\ & = -\int_{\Omega}u\frac{\partial u}{\partial n}dS+\int_{\Omega}|\nabla u|^2dx \\ & = \int_{\Omega}|\nabla u|^2dx. \end{align} You cannot have $(Bu,u) \ge c\|u\|^2$ for some $c > 0$ because the constant functions $u$ are in the subspace you're considering and $Bu=0$ for a constant function. However, you do have $(Bu,u) \ge 0$. So $(B+\epsilon I)$ is positive definite.

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The answer is affirmative and is based on the idea of the user Justpassingby (see comments to main question). Details are on Davies's book "Spectral theory and differential operators", section 7.2 pag.146. (link)

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