---
tag: 1A8A1XFG45
kind: definition
name: Leray–Schauder degree
msc: [47H11, 47J05, 35A01]
defines: 3VJC5Q9MYA
uses: [06GZ956657, 09QSZVVM26]
sources:
  - url: https://en.wikipedia.org/wiki/Leray%E2%80%93Schauder_degree
---

## Definition

Let $X$ be a real [[Banach space|06GZ956657]] and let $\Omega \subseteq X$ be a [[bounded|09QSZVVM26]] open set.
Let $K \colon \overline{\Omega} \to X$ be a compact map, that is, continuous with $K(\overline{\Omega})$ relatively compact.
Let $y \in X$ with $y \notin (I - K)(\partial\Omega)$, where $I$ is the identity, and put $r = \inf_{u \in \partial\Omega} \norm{u - K(u) - y}$, which is positive.
Choose a continuous map $K_\varepsilon \colon \overline{\Omega} \to X$ with values in a finite-dimensional subspace $E \subseteq X$ that contains $y$, such that $\norm{K(u) - K_\varepsilon(u)} < r$ for all $u \in \overline{\Omega}$.
Let $\deg_B$ denote the Brouwer degree of continuous maps on closures of bounded open subsets of the finite-dimensional space $E$.
The *Leray–Schauder degree* of $I - K$ on $\Omega$ at $y$ is
$$\deg_{LS}(I - K, \Omega, y) = \deg_B\big((I - K_\varepsilon)|_{\overline{\Omega \cap E}}, \, \Omega \cap E, \, y\big).$$

## 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 $I - K = 2I$.

## Explanations

### L1 human

The Leray–Schauder degree counts solutions of $u - K(u) = y$ with signs, for compact $K$, by approximating $K$ by maps into finite-dimensional spaces and using the finite-dimensional degree there.