Functional Analysis – Closedness of Image of Fredholm Operator and Codimension

functional-analysis

Let $X$ and $Y$ be Banach spaces. A bounded operator $T\colon X\to Y$ is called Fredholm iff

  • The dimension of $\ker(T)$ is finite,

  • The codimension of the image $\mathrm{im}(T)$ is finite,

  • The image $\mathrm{im}(T)$ is closed in $Y$.

Question: Is the third condition redundant?

Some lecture notes I'm working through claim that the third condition follows from the second one together with the open mapping theorem. I've checked some books on functional-analysis without finding a proof of this.

Best Answer

Assume that $T$ is injective, because if it is not we know that $\textrm{ker}(T)$ is closed so we can replace $T$ by the induced map from $X/\textrm{ker}(T)$ (which is a Banach space).

Now define $T' := X \oplus C \to Y$ by $T'(x, c) = T(x) + c$ where $C$ is a closed complement for the range of $T$.

This $T'$ is clearly bounded, linear and an isomorphism. So by the open mapping theorem $T'$ is open. Note that $\textrm{im}(T) = T'(X \oplus \{0\})$ which is closed.

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