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For a Banach space $X$, and Fredholm operators $T_1, T_2: X \rightarrow X$ with index $0$, I want to decide whether the following statement is true:

The operator $S: Sx = T_1x + T_2x$ is either the zero mapping or a Fredholm operator with index zero.

What I know is that the identity mapping has Fredholm index $0$, and in the finite dimensional case, Fredholm index equals the difference between the dimensions $dim(X) - dim(Y)$ for a Fredholm operator $F: X \rightarrow Y$ . So in the finite dimensional case, the statement is likely to be true.

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This is not true. Consider the case where $X=\ell^2$ (or any other classical sequence space), $T_1$ is the identity operator, and $T_2$ is defined by $(T_2x)(n)=x(n)$ if $n=1$ and $(T_2x)(n)=-x(n)$ otherwise. Then $T_1+T_2\neq0$ and $T_1+T_2$ isn't Fredholm.

In fact, if $T_1$ is any Fredholm operator with index $0$, put $T_2=-T_1+K$ for some nonzero compact operator $K$. Then the index of $T_2$ is $0$ as well, but $T_1+T_2=K$ is not Fredholm.

Since you mention the finite-dimensional case, it follows directly from the rank-nullity theorem that any operator on a finite-dimensional space is Fredholm with index zero.

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    $\begingroup$ note your last statement - "any operator on finite-dimensional space is Fredholm with index zeros" - is only true when $dim(X) = dim(Y)$. When this is not the case with $dim(X) = n$ and $dim(Y) = m$ we have $ind(T) = n - m$. $\endgroup$ – lennart Apr 12 at 8:11
  • $\begingroup$ That's true, I was only considering operators from a space to itself. $\endgroup$ – Aweygan Apr 12 at 17:29

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