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I have seen these 2 fixed point theorem and I think the condition of Leray Schauder fixed point theorem is very strong and we require to consider the fixed point of $u=\sigma Tu$ $\forall \sigma \in[0,1]$.

And I think the result of schauder fixed point theorem is stronger than that of Leray Schauder , Do I have any misunderstanding about these 2 theorem?

Thank you

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  1. Note that Leray-Schauder is usually proven by using the hypotheses to construct a mapping that satisfies the conditions of the Schauder fixed point theorem, and then appealing to the Schauder fixed point theorem. See, e.g. these notes (Theorem 2.2 there is Schauder).

    So in a sense you are right: things that satisfy the hypotheses of Leray-Schauder also (after some work) satisfy the hypotheses of Schauder; and in that way of comparing strength of theorem, you might say that Schauder is stronger.

    (Though in reality to apply Leray-Schauder you first embed your map into a continuous family so the comparison above is sort of comparing apples and oranges.)

  2. However, in terms of applications, Leray-Schauder is often easier to wield. To see this, we can consider the following (slightly weaker) formulation of Schauder's fixed point theorem:

    Let $X$ be a Banach space and let $T:X\to X$ be a continuous, compact map. If there further exists a bounded convex set $K\subseteq X$ such that $T(K) \subseteq K$, then $T$ has a fixed point (in $K$).

    In practice actually locating a convex set $K$ such that $T(K) \subseteq K$ is sometimes difficult (especially, say, in applications in partial differential equations).

    Leray-Schauder, on the other hand, can be formulated as this:

    Let $X$ be a Banach space and let $\mathcal{T}: X\times [0,1]\to X$ be a continuous compact map, such that $\mathcal{T}(X\times \{0\}) = \{0\}$. If we further suppose that all solutions of $\mathcal{T}(x,\sigma) = x$ for $\sigma \in [0,1]$ verify $\|x\| < 1$, then the mapping $T:X \to X$ defined by $\mathcal{T}(x,1) = T(x)$ admits a fixed point.

    What you should think is that with Leray-Schauder, you are off-loading the difficult statement of proving $T(K) \subseteq K$ to the map $\mathcal{T}(\cdot,0)$, for which you can take $K = \{0\}$ by definition. The price you must pay by deforming your original $T = \mathcal{T}(\cdot,1)$ is that you need to check a certain boundedness statement for putative fixed points.

    In the context of PDEs however (where this theorem is most frequently applied), proving stuff about these putative fixed points would be proving a priori estimates, which is often easier (than directly trying to apply Schauder Fixed Point Theorem).

  3. Let's do an example. Suppose our Banach space is $\mathbb{R}^2$. Let our function be $ T(x) = Ax - w $ where $A$ is the matrix $$ A = \begin{pmatrix} 1 & 1 \\ -1 & 1\end{pmatrix}.$$ Notice that we have $\det (\lambda A - I) = (\lambda - 1)^2 + \lambda^2 \neq 0$ for any $\lambda$ so that $\lambda A - I$ is always invertible. We can of course solve the fixed point equation $T(x) = x$ by solving algebraically $$ T(x) = x \iff (A - I)x = w \iff x = (A-I)^{-1} w.$$ This way we see that it has a unique fixed point.

    Suppose however we want to approach using topological methods. With Schauder fixed point theorem, you are out of luck. Notice that since $T$ is affine, it is automatically compact and continuous. But it is not possible to find a non-trivial bounded convex set $K\subseteq \mathbb{R}^2$ such that $T(K) \subseteq K$. This is due to the fact that, letting $x_0$ be the fixed point of $T$, we have $$ \| T(x - x_0)\| = \|A(x - x_0)\| = \sqrt{2}\|x - x_0\| $$ for any $x$. In other words, $T$ is an expanding map. And so the only $K$ to which Schauder's theorem can apply is $K = \{x_0\}$, meaning that to apply Schauder's theorem you would've found the fixed point already.

    Leray-Schauder however is a bit more flexible. Let $T_\lambda(x) = \lambda T(x)$. By definition $T_0$ is the zero map. Now suppose that $x$ is a fixed point of $T_\lambda$. We have formally $$ x = (\lambda A - I)^{-1} \lambda w \tag{*}$$ Now, we know that $\lambda A - I$ are invertible, so we know that there is a universal bound to the operator norm $\| (\lambda A - I)^{-1} \|$ when $\lambda$ belongs to any compact interval. This gives us a universal bound on our fixed point $x$. Note that we got the bound without needing to "explicitly" solve (*).

    (The example is kind of trivial. But you can imagine replacing $A$ by $A_x$ so you have a nonlinear operator, such that $\|A_x y\| \geq 2 \|y\|$ for any $x$ and $y$ [then Schauder won't work], but you have certain uniform bounds on $\lambda A_x - I$ that allows you to still run the Leray-Schauder argument.)

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  • $\begingroup$ 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 $T(u)=u$ ? $\endgroup$ – mnmn1993 Apr 28 '16 at 15:41
  • $\begingroup$ @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 $\sigma T$. $\endgroup$ – Willie Wong Apr 28 '16 at 16:03
  • $\begingroup$ 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. $\endgroup$ – Willie Wong Apr 28 '16 at 16:04

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