# 18.155 LECTURE 17, 2015 Read: Before lecture • A little under prepared.

advertisement ```18.155 LECTURE 17, 2015
RICHARD MELROSE
Abstract. Notes before and after lecture – if you have questions, ask!
Read:
Before lecture
• A little under prepared.
After lecture
The argument I messed up near the end is quite straightforward. Namely if
∆wi = fi where wi is bounded in &lt; x &gt;−w L2 , orthogonal to the null space and
with gi → 0 in &lt; x &gt;−w−2 L2 then wi → 0 in &lt; x &gt;−w L2 . Indeed the boundedness
in this weighted space means that wi is bounded in L2 in the ordinary sense on any
compact subset and then from elliptic regularity it follows that it is bounded in H 2
on any compact subset and so has a weakly convergent subsequence φwi → φw on
any compact subset, and again by elliptic regularity this convergence holds in H 2 .
Now look at the other part
(1)
∆((1 − φ)wi ) = (1 − φ)∆wi + ei
where ei has compact support in a fixed set and involves first derivatives on wi so
converges in L2 . So the right side converges in a space on which ∆ is an isomorphism
and hence so does (1 − φ)wi . Finally then wi → w globally in &lt; x &gt;−w L2 (and
hence in the domain space but this is not really needed) but the limit has to be in
the null space, to which it is orthogonal, so wi → 0.
In the fuller argument this means ui must be bounded in L2 where ∆ui − = ti →
f in &lt; x &gt;−w−2 L2 but then one goes through the same argument to conclude
(maybe along a subsequence) that ui → u in the domain space and so the range is
closed.
Department of Mathematics, Massachusetts Institute of Technology
E-mail address: [email protected]
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