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

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:

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.

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

• Pre compact does not imply closed. Mar 26, 2020 at 5:15
• Any locally compact tvs. is finite dimensional. Mar 26, 2020 at 5:20

$$\Lambda = I$$ 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.
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.