# finite dimensional range implies compact operator

Let $X,Y$ be normed spaces over $\mathbb C$. A linear map $T\colon X\to Y$ is compact if $T$ carries bounded sets into relatively compact sets (i.e sets with compact closure). Equivalently if $x_n\in X$ is a bounded sequence, there exist a subsequence $x_{n_k}$ such that $Tx_{n_k}$ is convergent. I want to prove that if $T\colon X\to Y$ has finite dimensional range, then it's compact.

• That's only true if $T$ is continuous. – Daniel Fischer Mar 11 '16 at 20:32

## 4 Answers

Here is a direct proof.

Proof. Since $T$ has finite rank, Im$T$ is a finite-dimensional normed space. Furthermore, for any bounded sequence $\{ x_n \}$ in $X$, the sequence $\{ T x_n \}$ is bounded in Im$T$, so by Bolzano-Weierstrass theorem this sequence must contain a convergent subsequence. Hence $T$ is compact. $\square$

• for any bounded sequence ${x_n}$ in$X$, the sequence ${Tx_n}$ is bounded in $ImT$. Why can we say that? – Lucas Jun 15 at 20:15
• The usual definition of finite rank operator is the following: a linear map $T: X \to Y$ ($X$, $Y$ normed linear spaces) is a finite rank operator if and only if $T$ is bounded and $\mbox{dim} \mbox{Im} T < \infty$. Thus, if $T$ is finite rank, $T$ is bounded. But it is also linear, therefore $T(A)$ is bounded for any bounded $A \subset X$. (Note that this property, as well as boundedness, is equivalent to continuity for linear maps, and continuity is essential, as pointed out by Daniel Fischer just below the OP.) – Federico Jun 15 at 20:44

Hint: Every finite dimensional normed space is isomorphic to $\mathbb R^n$, so any bounded subset is pre-compact (sets with compact closure).

if $T$ is bounded and dim rangeT is finite ,the operator $T$ is compact.(kreyszig p.407)

You require the boundedness of $T$. Fact is that boundedness + finite rank of $T$ gives compactness of $T$. This you can find in any good book in Functional Analysis. I would recommend J.B Conway