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



leray schauder fixed point and schauder fixed point
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](https://cmouhot.files.wordpress.com/1900/10/lerayschauder.pdf) (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.)


