if $\Lambda = \Lambda ^2$ be a compact operator on a banach space , then $Range(\Lambda)$ is finite dimensional

Question

Let $X$ be a Banach space and let $\Lambda\in\mathfrak L(X,X)$ be compact s. t. $\Lambda=\Lambda ^2$. Namely, $$\Lambda(x)=\Lambda\left(\Lambda(x)\right),\;\forall x\in X.$$ Prove that $\mathcal Im(\Lambda)$ is finite-dimensional.

My attempt:

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Let's take a bounded set $A\subset X$ s.t. $\Lambda(A) = B_1$, so $\Lambda (A)$ is a unit ball in the $\mathcal Im(\Lambda)$ and by the compactness of the operator it is pre-compact, hence closed and bounded.

Applying the operator again,$\Lambda(\Lambda(A))$ is also pre-compact (closed and bounded).

Since $\Lambda\left(\Lambda(A)\right)=\Lambda(A)$ , then $\Lambda(\Lambda(A)) = B_1$ is also a unit ball, so $B_1$ is a pre-compact unit ball. So $\mathcal Im(\Lambda)=\mathcal Im(\Lambda^2)$ is finite dimensional.

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May I ask for verification and advice on improving my proof if necessary?

Answer 1

$\Lambda = I$ on $\mathcal{R}(\Lambda)$ because $\Lambda^2=\Lambda$. And $\mathcal{R}(\Lambda)$ is closed because $\mathcal{R}(\Lambda)=\mathcal{N}(\Lambda-I)$. Therefore, $$ \Lambda : \mathcal{R}(\Lambda)\rightarrow\mathcal{R}(\Lambda) $$ is the identity on $\mathcal{R}(A)$ and this identity operator is compact on $\mathcal{R}(A)$, which forces $\mathcal{R}(\Lambda)$ to be finite dimensional.

Answer 2

Your arguement is not valid because your set $A$ does not exist.

Let $(y_n)$ be a bounded set in the range of $\Lambda$. Let $y_n=\Lambda x_n$. Then $\Lambda (y_n)$ has a convergent subseqeunce (by compactness of $\Lambda)$. But $\Lambda (y_n)=\Lambda (\Lambda (x_n))=\Lambda (x_n)=y_n$ so $(y_n)$ itself has a convergent subsequence. We have proved that the unit ball of $\Lambda (X)$ is compact and this implies that $\Lambda (X)$ is finite dimensional.