3
$\begingroup$

From Zeidler's Applied Functional Analysis

Brouwer

The continuous operator $A:M \to M$ has a fixed point provided $M$ is compact, convex, nonempty set in a finite dimensional normed space over $\mathbb{K}$

Schauder

The compact operator $A:M \to M$ has a fixed point provided $M$ is a bounded, closed, convex, nonempty subset of a Banach space $\mathbb{X}$ over $\mathbb{K}$

Claim: if $\dim(\mathbb{X}) < \infty$ then Schauder = Brouwer

Just to clarify:

Why is the operator $A$ assumed to be compact for Schauder but merely continuous for Brouwer? What does finite/infinite dimension have to do with this assumption?

$\endgroup$
3
$\begingroup$

Your claim is correct.

The point is that in finite dimension all continuous operators are compact, while in infinite dimension you can have continuous operators which are not compact, as the following example shows:

In $\ell_2(\mathbb{N})$ consider the operator $T(x)=(\sqrt{1 - \| x\|^2},x_1, x_2, \dots)$ defined for $\|x\| \leq 1$, where $x=(x_1, x_2, \dots)$ and $\|x\|^2= \sum_{i=1}^{\infty} |x_i|^2$. $T$ is continuous, in fact

\begin{align} \|T(x) - T(y)\|^2 & = \left| \sqrt{1 - \|x\|^2} - \sqrt{1 - \|y\|^2}\right|^2 + \|x - y\|^2 \leq \\ & \leq \left| \|x\|^2 - \|y\|^2\right| + \|x - y\|^2 \leq \\ &\leq ( \|x\| + \|y\|)\|x-y\| + \|x-y\|^2 \leq \\ &\leq 2\|x-y\| + \|x-y\|^2. \end{align} Moreover $T$ maps the closed unit ball to its boundary because $\|T(x)\|^2= 1 - \|x\|^2 + \|x\|^2=1$. $T$ does not have fixed points, by contradiction, if we had $T(x)=x$, we would have $\|x\|=1$, but also $(0,x_1,x_2, \dots) = (x_1,x_2, \dots)$, that is $x_i=0$ for every $i$. In that case we would have $\|x\|=0 \neq 1$, which is a contradiction.

In the same book you refer to, you can read a characterization of compact operators as those well approximated by operators with finite range. In some sense that means compact operators are continuous operators with "almost" finite range. Hence, in finite dimension, continuous operators coincide with compact ones and that's the key point (as you can also easily deduce from the definition of compact operators).

Also, in finite dimension being compact for a set is nothing more than being bounded and closed, so to better understand the main difference, I would write Brouwer's theorem as follows:

The continuous operator $A:M \to M$ has a fixed point provided $M$ is bounded, closed, convex, nonempty set in a finite dimensional normed space over $\mathbb{K}$.

$\endgroup$

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.