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The problem can be formulated as following

Let $L$ be a second order differential operator $L = > a^{ij}D_{ij}+b^iD_i+c$ where Einstein summation is imposed. And the underlying field for function spaces is $\mathbb{R}^d$.

Assume $\exists \lambda, N_0\in (0,\infty) $ such that for $\forall > u\in C^2_0$ and $t\in[0,1]$, we have $$\|u\|_{W^{2,2}}\leq N_0 \|L_t u\|_2,$$ where $L_t = (1-t)(\lambda-\Delta)+tL$ and $W^{2,2}$ is Sobolev space.

I want to extend the inequality to $u\in W^{2,2}.$ It is easy to show that $Lt$ is bounded.

And since $C^2_0$ dense in $W^{2,2}$, take $u^n \in C^2_0$ s.t. $u^n \rightarrow u$ ( in the sense of $W^{2,2}$).

we have $$\|u\|_{W^{2,2}} \leq \|u-u^n\|_{W^{2,2}}+\|u^n\|_{W^{2,2}} \leq |u-u^n\|_{W^{2,2}} + N_0\|L_t u^n\|_2.$$

And here is my question:

How to show $\lim_{n\rightarrow \infty}\|L_t u^n\|_2 \leq \|\lim_{n\rightarrow \infty}L_t u^n\|_2 $?

Reverse Fatou lemma seems to be a good option, but I don't know how to show it is valid in this case.

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up vote 1 down vote accepted

You said that $L_t$ is bounded. In what sense? I guess what you mean is that it is bounded as an operator $L_t:W^{2,2}\to L^2$. This is the same as continuity of $L_t:W^{2,2}\to L^2$, which implies the desired convergence.

To provide the details, let $A:X\to Y$ be a bounded linear operator between the normed spaces $X$ and $Y$, and let $x_n\to x$ in $X$. Then $Ax_n\to Ax$ in $Y$, since $$ \|Ax_n-Ax\|_Y=\|A(x_n-x)\|_Y\leq C\|x_n-x\|_X\to0, $$ as $n\to\infty$. This implies also $\|Ax_n\|_Y\to\|Ax\|_Y$, because $$ \|Ax_n\|_Y\leq \|Ax_n-Ax\|_Y+\|Ax\|_Y $$ and $$ \|Ax\|_Y\leq \|Ax-Ax_n\|_Y+\|Ax_n\|_Y $$ together imply $$ \big|\|Ax_n\|_Y-\|Ax\|_Y\big|\leq \|Ax_n-Ax\|_Y. $$ As a side note, observe that the latter inequality is general, meaning that $$ \big|\|y\|-\|z\|\big|\leq \|y-z\|, $$ holds in any normed space.

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Yes, I do mean bounded operator. How can I show the implication in details? – newbie May 11 '12 at 23:24
@newbie: I updated the answer. – timur May 12 '12 at 0:34
Thanks. I made the problem way to complicated. – newbie May 12 '12 at 0:51
Sometimes a little abstraction is useful. – timur May 12 '12 at 0:54

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