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


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.

  • $\begingroup$ It would be better if you explain the problem and necessary definitions without external links.. $\endgroup$ – Peter Franek Jun 23 '15 at 15:51

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.

  • $\begingroup$ Does the condition (11.2) says ${x \in X : \lambda Tx =x, \lambda \in [0,1]}$ is bounded set? $\endgroup$ – Wojciech Piotrowski Jun 24 '15 at 7:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.