Leray–Schauder degree

Remarks

Such $K_\varepsilon$ exist because $K(\overline{\Omega})$ is relatively compact, and the right-hand side does not depend on the choice of $K_\varepsilon$ and $E$; this is what makes the definition sound. The degree has these properties.

1. Normalisation: $\deg_{LS}(I, \Omega, y) = 1$ if $y \in \Omega$.
2. Existence: if $\deg_{LS}(I - K, \Omega, y) \ne 0$, then $u - K(u) = y$ for some $u \in \Omega$.
3. Homotopy invariance: if $H \colon \overline{\Omega} \times [0, 1] \to X$ is compact and $u - H(u, t) \ne y$ for all $u \in \partial\Omega$ and $t \in [0, 1]$, then $\deg_{LS}(I - H(\cdot, t), \Omega, y)$ does not depend on $t$.

Example: let $\Omega$ be the open unit ball and let $K$ be compact with $\norm{K(u)} < 1$ on $\overline{\Omega}$. The homotopy $H(u, t) = tK(u)$ has no solution of $u = tK(u)$ on the unit sphere, so $\deg_{LS}(I - K, \Omega, 0) = \deg_{LS}(I, \Omega, 0) = 1$, and $K$ has a fixed point in $\Omega$.

Non-example: if $X$ is infinite-dimensional, the map $K = -I$ on the closed unit ball is not compact, since that ball is not compact. So this construction does not define a degree for $I - K = 2I$.