0
$\begingroup$

Leray-Schauder fixed point theorem from Gilbarg and Trudinger book is quoted below. I do not understand remark below this theorem. Could you explain?

picture

Theorem 11.2 from this text is Schauder fixed point theorem says: Every continuous mapping of closed convex subset $X$ of Banach space into itself such that $T(X)$ is precompact has fixed point. Compact operator between $A$ and $B$ is a operator which maps bounded sets in $A$ to precompact sets of $B$. I think I understand proof but I don't understand the remark, especially the second thesis.

$\endgroup$
1
  • $\begingroup$ It would be better if you explain the problem and necessary definitions without external links.. $\endgroup$ Jun 23, 2015 at 15:51

1 Answer 1

1
$\begingroup$

The condition (11.2) says that for all sufficiently large $x$, the image $Tx$ does not lie on the half-line $\{\lambda x : \lambda \ge 1\}$.

If $T$ satisfies this condition, then so does $\sigma T$ for any $\sigma\in [0,1]$. This is what the second sentence of the remark means.

I found the first sentence of the remark a little misleading, because the statement "for some $\sigma\in(0,1]$ the mapping $\sigma T$ possesses a fixed point" does not follow from Theorem 11.3. Rather, it follows from its proof. Indeed, in the course of the proof we found that $T^*$ has a fixed point $x$, without using (11.2). This implies that either $Tx =x$, or $Tx = \lambda x$ with $\lambda >1$. In the former case $T$ has a fixed point; in the latter case $\lambda^{-1}T$ has a fixed point.

$\endgroup$
1
  • $\begingroup$ Does the condition (11.2) says ${x \in X : \lambda Tx =x, \lambda \in [0,1]}$ is bounded set? $\endgroup$ Jun 24, 2015 at 7:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .