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Jan 21, 2022 at 17:10 vote accept mnmn1993
Apr 28, 2016 at 16:04 comment added Willie Wong Perhaps it would be easier if you think of the condition in Leray-Schauder as "... there exists some constant $M$ such that if $\|u\| > M$ then $\sigma T(u) \neq u$ for any $\sigma \in [0,1]$." What you are doing is not finding fixed points, but putting constraints on what could potentially be fixed points.
Apr 28, 2016 at 16:03 comment added Willie Wong @mnmn1993: Of course "finding the fixed point of $\sigma T$" is similar to finding those of $T$. But the key thing that you miss is that you can prove the statement "If $u$ solves $\sigma T(u) = u$ then $\|u\| \leq 1$" without knowing that such $u$ actually exist! In other words, in using Leray-Schauder you are never asked to find the fixed points of $\sigma T$.
Apr 28, 2016 at 15:41 comment added mnmn1993 Thank you so much! It is so clear . But for a PDE question , if i choose to apply leray schauder theorem ,can i say finding the fixed of $\sigma T(u)=u$ is very similar to $T(u)=u$ ?
Apr 28, 2016 at 15:15 history answered Willie Wong CC BY-SA 3.0