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I'm having a hard time verifying the claim that the Dirichlet energy for the Poisson equation ($\Delta u=f$ in $\Omega$, $u=g$ $\partial \Omega$), $$E(u)=\int_\Omega \frac{1}{2}|Du|^2+\int_\Omega fu$$ is bounded below, where $u\in H^{1,2}(\Omega)$ with the condition that $u-g\in H^{1,2}_0(\Omega)$. Clearly the first term is bounded below by zero, but the $\int fu$ term can get very large (in absolute value). Thus the first term has to be used to control the second. I imagine the boundary condition plays a role then, because for $\int fu$ to get very large, $u$ has to grow a lot in the interior since $u|_{\partial\Omega}$ is prescribed. The first term is also bounded below by the Poincare inequality, but that still doesn't take care of the second term.

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The Hölder inequality gives $$\int_\Omega (-f)u\le|\int_\Omega fu| \le ||f||_2+||u||_2.$$ The negative of this is $$\tag{$*$}\int_\Omega fu\ge -||f||_2-||u||_2,$$ so the first estimate is $$E(u)\ge \int_\Omega \frac{1}{2}|Du|^2-||f||_2-||u||_2.$$ To estimate the first term from below, we use the Poincaré inequality for $H_0^1(\Omega)$. Namely, since $u-g\in H_0^1(\Omega)$, there is a positive constant $C=C(\Omega)$ such that $$||u-g||_2\le C||Du-Dg||_2\le C(||Du||_2+||Dg||_2).$$ If we rearrange and square this, we find $$\tag{$**$}C^2||Du||_2^2\ge ||u-g||_2^2-\underbrace{2C||Dg||_2}_{C_1}||u-g||_2+\underbrace{C^2||Dg||_2^2}_{C_2}$$ The first term is $$ ||u-g||_2^2=\int(u^2+g^2-2ug)\ge||u||_2^2+||g||_2^2+2(-||u||_2-||g||_2)=:||u||_2^2-2||u||_2+C_3 $$ where the inequality is due to $(*)$ with $-g$ in place of $f$. The Minkowski inequality gives $$||u-g||\ge ||u||-||g||.$$ Inserting these estimates into $(**)$ gives $$C^2||Du||_2^2\ge ||u||_2^2+C_4||u||_2+C_5$$ The energy estimate is then $$E(u)\ge C_6(||u||_2^2+C_7||u||_2)+C_8.$$ We note crucially that $C_6$ is positive. Thus the behavior of $E(u)$ is like the polynomial $x^2+C_7x$. This polynomial always has a minimum $m$, no matter the sign of $C_7$. Thus $E$ is bounded below by $C_6m+C_8$.

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Note: for equation $(1)$ and $(2)$ please add a constant in front of it.

By embedding theorem, we have $$ \| u\|^2_{L^2}\leq \|\nabla u\|_{L^2}^2+\|g\|_{H^1}^2 \tag 1 $$ Also, by Holder we have $$ \int_\Omega fu\, dx\leq \|f\|_{L^2}+\|u\|_{L^2}\tag 2 $$ In the view of $(1)$ and $(2)$, we have $$E(u)=\int_\Omega \frac{1}{2}|Du|^2+\int_\Omega fu\geq \|u\|_{L^2}^2-\|g\|_{H^1}^2-\|f\|_{L^2}-\|u\|_{L^2},$$ where $-\|g\|_{H^1}^2-\|f\|_{L^2}$ is only a constant and hence we just need to prove $$ \|u\|_{L^2}^2-\|u\|_{L^2}\tag3 $$ are bounded below. Indeed, we have for positive number $t$, the number $$ \inf\{t^2-t;t>0\} $$ is finite which justifies $(3)$ is bounded below.

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  • $\begingroup$ Which embedding theorem gives (1)? Also in going from (2) to the next one it seems like your signs are wrong. The $fu$ integral should pick up a sign. The second thing can be fixed, but I don't know where your estimate (1) is from. $\endgroup$ – Ryan Unger Jan 31 '17 at 18:33
  • $\begingroup$ for $(1)$, check the trace estimation& embedding for sobolev space. for $(2)$, take absolute value on the both hand side and you should be able to figure it out. $\endgroup$ – spatially Jan 31 '17 at 18:35
  • $\begingroup$ Both of the estimates $(1)$ and $(2)$ are incorrect as stated. The first needs a constant and the second needs both a factor of $1/2$ on the right and squares on both norms. $\endgroup$ – Glitch Feb 1 '17 at 1:16
  • $\begingroup$ @Glitch Sure. I was a bit rush on this answer but I believe I provided the right idea. Also, I think incorrect are bit hush :) $\endgroup$ – spatially Feb 1 '17 at 1:18

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