Compact operators on an infinite dimensional Banach space cannot be surjective

I am reading a book about functional analysis and have a question:

Let $X$ be a infinite-dimensional Banach-space and $A:X \rightarrow X$ a compact operator. How can one show that $A$ can not be surjective?

• What is your preferred definition of "compact operator"? Proofs might look quite different depending on which definition you use. – t.b. Oct 19 '11 at 18:37
• Jonas has given an answer, but you can also remember that in the context of Banach spaces or more generally complete metrizable topological vector spaces, surjective (continuous) linear maps are automatically open. Thus you would have $$c B_X\subset A(B_X)$$ where $c$ is a positive real number and $B_X$ is the unit ball in $X$. The left hand side has compact closure iff $X$ is finite dimensional by a theorem of Riesz, while the right hand side has compact closure by definition of $A$ being compact. So $X$ must be finite dimensional for $A$ to be compact and surjective. – Olivier Bégassat Oct 19 '11 at 18:39
• @t.b. ok, done. – Olivier Bégassat Oct 19 '11 at 18:44
• What about the case when $A\colon X\rightarrow Y$, when $X\neq Y$ in general? – charlestoncrabb Dec 8 '15 at 1:14

Jonas has already given an answer, but here's another one. You can remember that in the context of Banach spaces (or more generally complete metrizable topological vector spaces), surjective (continuous) linear maps are automatically open. Thus you would have $$c B_X\subset A(B_X)$$ where $c$ is a positive real number and $B_X$ is the unit ball in $X$. The left hand side has compact closure iff $X$ is finite dimensional by a theorem of Riesz, while the right hand side has compact closure by definition of $A$ being compact. So $X$ must be finite dimensional for $A$ to be compact and surjective.

• Thanks for moving this to the answer section. Since I had them already prepared: links to the open mapping theorem and Riesz's lemma. – t.b. Oct 19 '11 at 18:45
• @t.b. great additions! – Olivier Bégassat Oct 19 '11 at 18:46
• Why $c B_X \subset A (B_X)$? Thank you! – g.pomegranate May 15 '15 at 17:31
• @r.pomegranate $A(B_X)$ is open by the open mapping theorem, and contains zero, so there is some positive real number $c$ such that $cB_X$ is contained in $A(B_X)$. – Olivier Bégassat May 15 '15 at 18:14
• @r.pomegranate finite rank continuous linear maps are always compact. – Olivier Bégassat May 15 '15 at 20:00

Suppose that $A$ is invertible. Then $I = A^{-1}A$ must also be compact. But $I$ cannot be compact in a infinite-dimensional space.

To prove the last statement note that since $X$ is infinite dimensional, the space contains a sequence of unit vectors $\{x_n\}_n$ in $X$ which does not contain a convergent subsequence (Riesz's lemma). Hence $\{I x_n\} = \{x_n\}$ does not contain a convergent subsequence, hence $I$ cannot be compact.

• But how can you reduce to $A$ invertible? (just saying that you should probably explain this as well) – t.b. Oct 19 '11 at 18:34
• and what if $A$ is not invertible? – Chau Oct 19 '11 at 18:42
• I'm sorry, I'm not fresh enough now to explain why we can reduce to non-invertibility. I'll check it out tomorrow. – Jonas Teuwen Oct 19 '11 at 19:43

I would appeal to Banach open mapping theorem. Suppose $A$ is surjective: then, since $X$ is complete it needs be an open mapping. So $A$ maps the unit ball into some precompact open set, which cannot exist in a infinite-dimensional normed space because of Riesz's lemma - the same Jonas mentions. Indeed, a precompact open set $U$ of $X$ would contain a ball of radius $r$. Call $y_n=\frac{r}{2}x_n$, where $x_n$ is the sequence of unit vectors prescribed by the lemma. Then $y_n \in U$ and so it should have a convergent subsequence, which is a contradiction.

Note [EDIT This example is wrong, see comment by nullUser below.] It is important to assume completeness of $X$. This hypothesis enables us to summon Banach open mapping theorem and without it the claim is false. For example, consider the space $$c_{00}= \{\mathbf{x}=(x_n)_{n \in \mathbf{N}} \mid x_n=0\ \text{for all sufficiently large}\ n\}$$ equipped with the norm $$\lVert \mathbf{x}\rVert^2=\sum_{n=1}^\infty \lvert x_n\rvert^2.$$ This space is not complete and, indeed, the operator $T\colon c_{00} \to c_{00}$ defined by $$T(\mathbf{x})=(x_1, \frac{x_2}{2}, \frac{x_3}{3}, \ldots)$$ is compact and surjective.

• How is that operator compact? Take $T(1,1/2,1/3,\ldots,1/n,0,0,0,\ldots)$. The operands are in $c_{00}$ and $\ell^2$ bounded. But their output tends to $(1,1/4,1/9,\ldots,1/n^2,\ldots)$ which is not an element of $c_{00}$, so there can be no convergent subsequence in $c_{00}$. – nullUser Apr 16 '13 at 22:25
• @nullUser: I guess that you are right. I had hastily classified that operator as 'compact' because indeed it is a compact operator from $\ell^2$ into $\ell^2$. – Giuseppe Negro Apr 17 '13 at 0:16