# $T(V)$ is a closed subspace of $V$?

Let $V$ be a normed vector space (not necessarily a Banach space) and let $S$ and $T$ be continuous linear transformations from $V$ to $V$. If we assume that $T=T \circ S \circ T$. Then how to show that $T(V)$ is a closed subspace of $V$? Thank you for your help!

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Hint: Consider $1 - TS$ and its kernel.

We have $T(V) = \ker{(1-TS)}$.

1. Observe that $x \in \ker{(1-TS)}$ means that $x = TSx$, so $x \in T(V)$. Therefore $\ker{(1-TS)} \subset T(V)$.

2. If $x \in T(V)$ then $x = Ty$ for some $y \in V$ and thus $(1-TS)x = (1-TS)Ty = Ty - TSTy = 0$, so $\ker{(1-TS)}\supset T(V)$.

3. It follows from 1. and 2. that $\ker{(1-TS)} = T(V)$.

4. The kernel of a continuous operator between normed spaces is closed since it is the pre-image of $0$. Thus, $T(V) = \ker{(1-TS)}$ is closed.

What is going on is that $T = TST$ implies that $TS$ is a projection: $$(TS)^2 = (TS)(TS) = (TST)S = TS.$$ The range of a projection $P$ is the kernel of its complementary projection $(1-P)$. Apply this to $P = TS$.

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Of $1-TS{}{}{}$? –  Mariano Suárez-Alvarez May 2 '12 at 20:52
Right. Thanks.${}$ –  t.b. May 2 '12 at 20:57