# 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. Mar 11, 2016 at 20:32

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? Jun 15, 2019 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.) Jun 15, 2019 at 20:44
• T has to be a bounded operator as nothing guarantees that $Tx_n$ is bounded. Mar 8, 2021 at 17:38

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

Assuming $$T$$ is continuous, you know that if $$||x||=1$$, then $$T(x) \leq M$$ for some $$M \in \mathbb{R}$$. So $$T(B^{X}_{1}(0)$$) (unit ball in $$X$$) is contained in $$\overline{B^{\text{Im}(T)}_{M}(0)}$$, which is compact since the image is finite dimensional and hence isomorphic to $$\mathbb{R}^{n}$$.