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I'm reading M.E.Taylor's PDE, Vol I, Chapter 5 Linear elliptic equations. I have some problem on Proposition 1.7. I will quote it here:

Consider the following boundary problem for $u$: $$ \Delta u=0 \text{ on } \Omega, \quad u|_{\partial \Omega}=f, $$ where $f\in C^{\infty}(\partial\Omega)$ is given. We denote the solution to it by $$ u=\mathrm{PI} \,\,f. \tag{1} $$

Proposition 1.7. The map $(1)$ has a unique continuous extension $$ \mathrm{PI}: H^s(\partial\Omega)\rightarrow H^{s+1/2}(\Omega),\quad s\geqslant \frac{1}{2}. $$

Proof. It suffices to prove this for $s=k+1/2,\,k=0,1,2,\cdots$, by interpolation. Given $f\in H^{k+1/2}(\partial \Omega)$, there exists $F\in H^{k+1}(\Omega)$ such that $F|_{\partial\Omega}=f$, by the property of trace operator. Then $\mathrm{PI}\,\,f=F+v$, where $v$ is defined by $$ \Delta v=-\Delta F\in H^{k-1}(\Omega),\quad v\in H_0^1(\Omega). $$ The regularity result gives $v\in H^{k+1}(\Omega)$, which establishes the result for $s=k+1/2$. $\blacksquare$

Now, I know there is an extension, but where is the "continuous"?

I know $$ \Vert u\Vert_{H^{k+1}(\Omega)}\leqslant \Vert F\Vert_{H^{k+1}(\Omega)}+\Vert v\Vert_{H^{k+1}(\Omega)} $$ $$ \leqslant \Vert F\Vert_{H^{k+1}(\Omega)}+C\Vert \Delta F\Vert_{H^{k-1}(\Omega)}+C\Vert v\Vert_{H^k(\Omega)} $$ $$ \leqslant C\Vert F\Vert_{H^{k+1}(\Omega)}+C\Vert v\Vert_{H^k(\Omega)} $$ $$ \leqslant C\Vert f\Vert_{H^{k+1/2}(\partial\Omega)}+\Vert v\Vert_{H^k(\Omega)}. $$

Without the second term $\Vert v\Vert_{H^k(\Omega)}$ I can see the "continuous". But I can't remove it! So, why is it continuous?

Any advice will be appreciated!


I should assume $\Omega$ is bounded with smooth boundary before. I think $$ \Vert u\Vert_{H^{k+1}(\Omega)}\leqslant \Vert F\Vert_{H^{k+1}(\Omega)}+\Vert v\Vert_{H^{k+1}(\Omega)}\leqslant \Vert F\Vert_{H^{k+1}(\Omega)}+C\Vert \Delta F\Vert_{H^{k-1}(\Omega)}+C\Vert v\Vert_{H^k(\Omega)} $$ Can be rewrite as $$ \Vert u\Vert_{H^{k+1}(\Omega)}\leqslant \Vert F\Vert_{H^{k+1}(\Omega)}+\Vert v\Vert_{H^{k+1}(\Omega)}\leqslant \Vert F\Vert_{H^{k+1}(\Omega)}+C(k)\Vert \Delta F\Vert_{H^{-1}(\Omega)}. $$ Since $$ \Vert v\Vert_{H^{k+1}(\Omega)}\leqslant C(k)\Vert v\Vert_{H^1(\Omega)}\leqslant \underbrace{C(k) \Vert \Delta F\Vert_{H^{-1}(\Omega)}\leqslant C(k) \Vert F\Vert_{H^{k+1}(\Omega)}.}_{\text{I'm not sure about this}} $$ Hence we have $\Vert u\Vert_{H^{k+1}(\Omega)}\leqslant C(k)\Vert f\Vert_{H^{k+1/2}(\partial\Omega)}$, which implies the "continuous"! But Here we have a constant depends on $k$(and also the diameter of $\Omega$). Can we have a uniform constant?

Edit(about $\Vert \Delta F\Vert_{H^{-1}(\Omega)}\leqslant \Vert F\Vert_{H^{k+1}(\Omega)}$)

I think this follows from $$ \Vert \Delta F\Vert_{H^{-1}(\Omega)}=\sup_{\Vert v\Vert_{H_0^1(\Omega)}=1} \int_{\Omega}v\Delta F \leqslant \Vert \nabla F \Vert_{L^2(\Omega)}\leqslant \Vert F\Vert_{H^{k+1}(\Omega)} $$

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I guess it depends on elliptic estimates for solutions. –  Siminore May 7 '12 at 16:55

1 Answer 1

You said any advice will be appreciated, so I'll try to give you a hint, not more.

If you have a solution of a linear elliptic PDE $Lu = f$ with $u$ in $H^1_0$ and $f\in L^2$ then $$||u||_{H^1} \le C \left(||f||_{L^2} + ||u||_{L^2} \right)$$ If, however, $u$ is the unique solution of that equation, then the last term on the right hand side (the $L^2$ -norm of $u$) can be omitted, because in that case there is an estimate $$||u||_{L^2} \le C || f ||_{L^2}$$

If you don't know this here is an idea how to see that: if this were not true, there would exist sequences $u_k\in H^1_0, f_k \in L^2$ such that $L u_k = f_k$ but $$||u_k||_{L^2} \ge k || f_k ||_{L^2} $$

Wlog $||u_k||_{L^2}=1$. Now show that:

i) $f_k \rightarrow 0 $ in $L^2$

ii) a subsequence of $u_k$ converges weakly in $H^1_0$, strongly in $L^2$ (show it's bounded in $H^1_0$)

iii) the limit of $u_k$ satisfies $Lu=0$ weakly.

Conclude there is a contradiction to $||u_k||_{L2}=1$ and uniqueness.

Now check what this means for your solution of $\Delta v = - \Delta F$ with $v\in H^1_0$

(note: this is a special case of the boundedness of the inverse of $I-T$ for compact T in case $I-T$ is invertible. See e.g. Gilbarg and Trudinger's Elliptic Partial Differential Equations of Second Order, Thm 5.3)

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Thanks very much! You show me the case $k=0$. I see $\Delta$ is a bijection form $H_0^1(\Omega)$ to $H^{-1}(\Omega)$, and $\Vert u\Vert_{H^1(\Omega)}\leqslant\Vert \Delta u\Vert_{H^{-1}(\Omega)}$. I should assume $\Omega$ is bounded with smooth boundary before. I add an edit to my question because it is a little long for the comment space. –  Y.Z May 8 '12 at 5:51

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