Theorem 3

Let E be a locally convex space, F a separated locally convex space, $u$ a Fredholm mapping from E into F and $h$ a compact linear mapping from E into F. Then $u+h$ is a Fredholm mapping, and one has ind($u+h$) $=$ ind($u$).

We shall prove the theorem in three steps.

A) Suppose that E and F are Banach spaces. Proposition 2 of III, p. 42 shows that the linear mapping $u$ is a strict morphism with closed image whose kernel and cokernel are of finite dimension. Since E and F are Banach spaces, Theorems 1 and 2 of III, p. 73 imply that, for every $t\in [0,1]$, the linear mapping $u_t=u+th$ has these same properties, and is therefore a Fredholm mapping (III, p. 42, Proposition 2). The mapping $t\mapsto u_t$ of $[0,1]$ into the set $\mathscr{F}(E; F)$ of Fredholm mappings of E into F is continuous. By Theorem 1 of III, p. 58, the mapping $t\mapsto$ ind($u_t$) is locally constant. Since the interval $[0,1]$ of $\mathbf{R}$ is connected, this mapping is constant. Hence ind($u$) $=$ ind($u+h$).

B) Suppose that E = F and $u= 1_E$. By Proposition 5 of III, p. 5, there exist a Banach space G, a continuous linear mapping $p: E\rightarrow G$ and a compact linear mapping $q: G\rightarrow E$ such that $h=q\circ p$. The endomorphism $p\circ q$ of G is compact. By A), $1_G+p\circ q$ is a Fredholm endomorphism of G of index zero. But then $1_E+h= 1_E+q\circ p$ is a Fredholm endomorphism of E of index zero (III, p. 49, Proposition 10, f )).

C) General case. Let $v$ be a quasi-inverse of $u$. The endomorphisms $u\circ v$ and $(u+h)\circ v$ of F differ from $1_F$ by compact linear mappings. By B), they are Fredholm endomorphisms of F of index zero. It follows that $u+h$ is a Fredholm endomorphism (III, p. 40, n$^o2$) of the same index as $u($III, p. 43, n$^o3$, formula (2)).