Intereting Posts

how to prove that every low-dimensional topological manifold has a unique smooth structure up to diffeomorphism?
Why do the endpoints of the Maclaurin series for arcsin converge?
algebraic manipulation of differential form
Complex number trigonometry problem
Proving the convergence of $\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2}$
Number theory question with automorphic numbers
$G\times H\cong G$ with $H$ non-trivial
Triangle Formula for alternative Points
A uniform bound on $u_n$ in $L^\infty(0,T;L^\infty(\Omega))$
Sum of eigenvalues and singular values
Show that a connected graph on $n$ vertices is a tree if and only if it has $n-1$ edges.
Examples proving that the tensor product does not commute with direct products
proof of functional completeness of logical operators
Trigonometry problem on product of trig functions.
Scheme glued out of finitely many spectra of local rings

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?

- Dense subsets in tensor products of Banach spaces
- Is it true that $\dim(X) \leq \dim(X^{\ast})$ for every infinite dimentional Banach space $X$?
- How to proof that a finite-dimensional linear subspace is a closed set
- Every separable Banach space is isomorphic to $\ell_1/A$ for some closed $A\subset \ell_1$
- $C ( \times \to \mathbb R)$ dense in $C ( \rightarrow L^{2} ( \to \mathbb R))$?
- A few questions about the Hilbert triple/Gelfand triple

- construction of a linear functional in $\mathcal{C}()$
- Every closed subspace of ${\scr C}^0$ of continuously differentiable funcions must have finite dimension.
- The space of Riemannian metrics on a given manifold.
- Bounded linear operator maps norm-bounded, closed sets to closed sets. Implies closed range?
- A Banach space that is not a Hilbert space
- Compact multiplication operators
- Coordinate functions of Schauder basis
- Spectra of restrictions of bounded operators
- On the weak closure
- Weakly compact implies bounded in norm

By popular request ^^ this comment is made into an answer.

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.

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.

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.

- Can this standard calculus result be explained “intuitively”
- Showing that if the initial ideal of I is radical, then I is radical.
- least common multiple $\lim\sqrt{}=e$
- Topological space definition in terms of open-sets
- Difference between metric and norm made concrete: The case of Euclid
- Problem on the number of generators of some ideals in $k$
- Well-ordering theorem and second-order logic
- Warm start of simplex algorithm after update of constraint matrix
- Number of well-ordering relations on a well-orderable infinite set $A$?
- Intuition for idempotents, orthogonal idempotents?
- Elementary Number Theory.. If a divides..
- What is the isomorphism function in $M_m(M_n(\mathbb R))\cong M_{mn}(\mathbb R)$?
- Closed form for the integral $\int_0^\infty t^s/(1+t^2)$
- A sequence of continuous functions on $$ which converge pointwise a.e. but does not converge uniformly on any interval
- Is this GCD statement true?