# Application of Brouwer fixed point theorem, why is compactness not required here??

Define a map $K:X_n \to X_n$ where $X=\text{span}(v_1, ..., v_n)$ where $v_i$ are basis functions some Hilbert space $H$. So $X_n$ is finite-dimensional.

$B_r(0)$ denotes the ball of radius $r$ centred at $0$ in $X_n$.

(after some calculations)... Therefore, $K$ is locally Lipschitzian and satisfies $K(\overline{B_R(0)}) \subset \overline{B_R(0)}$, thus Brouwer's fixed point theorem tells us there is a fixed point.

Now, why does Brouwer apply here? For Banach spaces doesn't one require compactness of $K(\overline{B_R(0)})$ in $X_n$? Or do we not need it cause of the finite-dimensional nature of $X_n$?

• Note the bars over $B_R(0)$: This means closed ball :) For Banach spaces, this goes by the name of the Schauder fixed point theorem. – Ted Shifrin Jan 25 '14 at 13:58
• But "bounded + closed = compact" only in $\mathbb{R}^n$ I thought – matt.x Jan 25 '14 at 14:00

$\overline{B_R(0)}$ is compact, since $X_n$ is finite-dimensional. Since $K$ is locally Lipschitz, it is continuous, and hence $K(\overline{B_R(0)})$ is compact too. But for Brouwer's theorem, we only need the continuity, and that $K(\overline{B_R(0)}) \subset \overline{B_R(0)}$.
• Oh I see, I didn't realise closed bounded sets in all finite-dimensional spaces were compact. Thought it was only $\mathbb{R}^n$. Thanks – matt.x Jan 25 '14 at 14:01
• @matt.x: The induced topology on a finite dimensional subspace of a Hilbert space makes it homeomorphic to $\Bbb R^n$, yes. It's not true in general that a closed, bounded subspace of an arbitrary metric space is compact. – Ted Shifrin Jan 25 '14 at 14:03
• @matt.x: If you have a $T_{2}$ topology on $\mathbb{R}^{N}$ or $\mathbb{C}^{N}$ for which vector addition and scalar multiplication are continuous, then the topology is equivalent to the usual Euclidean topology. In particular, all normed topologies and all inner-product topologies on finite-dimensional spaces over the real numbers are equivalent; and, all normed topologies and all inner-product topologies on finite-dimensional spaces over the complex numbers are equivalent. It's a useful result to know about. – Disintegrating By Parts Jan 25 '14 at 14:54
• @matt.x For Brouwer's theorem, you need something homeomorphic to a closed ball in an $\mathbb{R}^n$, so something compact, call it $B$, and a continuous map $f\colon B \to B$. $f(B)$ is then automatically compact as a continuous image of a compact set, you don't need to mention the compactness of $f(B)$ explicitly. That's what I meant. – Daniel Fischer Jan 25 '14 at 18:38