Compact operator with closed range has finite dimensional range

Question

I have to prove that:

If $X,Y$ be Banach Spaces, and $T\in B(X,Y)$ is a compact operator, then $T(X)$ is closed in $Y$ if and only if $\dim T(X)<\infty$.

Can anybody help me with this proof, please? There is surely some property I haven't thought about, but I'm getting really weird right now... Thank you!

Answer 1

Let $Z=T(X)$. Then $Z$ is also a Banach space, as a closed subspace of a Banach space, and $T:X\to Z$ is onto, and hence open, due to Open Mapping Theorem. If $Z$ were infinite dimensional, then $T$ would not be compact, as open sets in infinite dimensional spaces are not pre-compact.

Answer 2

Hint: Try the open mapping theorem.

Answer 3

$\textbf{Lemma:}$ Let $X$ and $Y$ be Banach spaces, and let $T \in \mathscr{B}(X, Y)$ such that $\operatorname{ran} T$ is closed. Then there exists a constant $M > 0$ such that for any $y \in \operatorname{ran} T$, there exists $x \in X$ with $y = Tx$ and $\|x\| \leq M\|y\|$.

$\textbf{Proof:}$ Only by the unique invertible operator $\hat{T} \in \mathscr{B}(X/\operatorname{ker} T, \operatorname{ran} T)$ such that $T = \hat{T} \circ \pi$.

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$\textbf{Proof of the question:}$ Assume that $\operatorname{ran} T$ is both closed and of infinite dimension. Then there exists a constant $M > 0$ such that for any $y \in \operatorname{ran} T$, there exists $x \in X$ satisfying $y = Tx$ and $\|x\| \leq M\|y\|$. Since $\operatorname{ran} T$ is of infinite dimension, by Riesz's lemma, there exists a sequence of unit vectors $\{y_n\}_{n=1}^\infty \subseteq \operatorname{ran} T$ such that $\|y_i - y_j\| > \frac{1}{2}$ for $i \neq j$. Correspondingly, there is a sequence $\{x_n\}_{n=1}^\infty \subseteq X$ such that $y_n = Tx_n$ and $\|x_n\| \leq M\|y_n\| = M$. This implies that $\{y_n\}$ should have a convergent subsequence, leading to a contradiction.

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$\textbf{Corollary1:}$ If $X$ and $Y$ are Banach spaces and $Y$ is of infinite dimension, and if $T \in \mathscr{B}_0(X,Y)$, then $\operatorname{ran} T \neq Y$.

$\textbf{Corollary2:}$ If there exist $A\in \mathscr{B}_0(X)$ such that $A$ is invertible, then $\operatorname{dim} X< \infty$.

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We can also conclude it when $X=Y$ is a Hilbert space

$\textbf{Lemma:}$ If $\mathcal{H}$ is a Hilbert space and $K\in\mathscr{B}_0(\mathcal{H})$, then each closed subspace contained in $\operatorname{ran}K$ is finite-dimensional.

$\textbf{Proof:}$If $M$ is a closed subspace contained in $\operatorname{ran}K$, let $P_M$ be the projection onto $M$ and by the Douglas's lemma, there exist $A\in \mathscr{B}(\mathcal{H})$ on $\mathcal{H}$ such that $P_M=KA$, so $P_M$ is compact and therefore $M$ is of finite dimension.

Answer 4

Hint: an infinite dimensional Banach space is never $\sigma$-compact (by Baire category theorem).