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In this Wikipedia link, it's stated that the condition of $\operatorname{im} T$ being closed in the definition of Fredholm Operator is redundant. Why is that?

Edit: Most proofs use the existence of a closed $N$ such that $T:X\rightarrow Y$, $Y=imT\oplus N$. However, why does such a $N$ exist?

An old man in the sea.
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  • @kkc the answer to that question is useful, but it still left me with a doubt. I wrote a comment asking for more informations, but I doubt I'll get an answer(was written in 2010) – An old man in the sea. Jul 29 '19 at 23:29
  • As an additional remark in the case that $T:D(T)\to Y$ is a densely defined unbounded operator, I believe that closedness of the range of $T$ does not follow from $T$ having finite dimensional co-kernel. Thus this condition is not redundant in such setting. – s.harp Jul 30 '19 at 17:55

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The reference given on the wiki page is to [Abramovich/Aliprantis, Lemma 4.38]:

Any [linear] operator $T : X \to Y$ between Banach spaces with a finite defect has a closed range and satisfies [a dimension formula].

The proof refers to Corollary 2.17:

Let $T : X \to Y$ be a bounded [linear] operator between two Banach spaces. If the quotient vector space $Y / R(T)$ is finite dimensional, then the range of $T$ is closed.

This is a corollary to Theorem 2.16:

A bounded [linear] operator $T : X \to Y$ between two Banach spaces has a closed range if and only if there exists a closed subspace $Z \subset Y$ such that $R(T) \cap Z = \{ 0 \}$ and the vector subspace $V := R(T) \oplus Z$ is closed in $Y$.

Proof of the "if" part:

  • The vector space $V / Z$ with the quotient norm is a Banach space.
  • $J : R(T) \to V / Z$, $J(y) := y + Z$ is a surjective linear isomorphism.
  • With the notation $\|y\|_T := \|y\| + \|T^{-1} y\|$ for $y \in R(T)$, the latter norm being the quotient norm on $T / N(T)$, we have $\| J(y) \| \leq \| y \| \leq \| y \|_T$, so the operator $J$ is continuous, and, being one-to-one and onto, Theorem 2.5 below gives that $J$ is bounded below.
  • Thus the norms $\|\cdot\|$ and $\|\cdot\|_T$ are equivalent on $R(T)$. Since $R(T)$ is closed for $\|\cdot\|_T$, it is also closed for $\|\cdot\|$.

Theorem 2.5:

A continuous operator $T : X \to Y$ between Banach spaces is bounded below if and only if $T$ is one-to-one and has a closed range.

The proof is not difficult and uses the Open Mapping Theorem at one point.

[Probably there is a more concise way to answer you question, but that, it seems to me, is how the book proceeds.]

user66081
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