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Let $X,Y$ be Banach spaces and $A\in\mathcal L(X,Y)$ . The task is to prove the following:

$A$ is compact if and only if the image of the closed unit ball in $X$ is compact in $Y$.

I have proven this when $X$ is a reflexive space.

Proof. Let $X$ be a reflexive space, $\bar B$ the closed unit ball in $X$, and $A$ a compact operator. Let further $y_n=Ax_n$ be a sequence in $A(\bar B)$.

In reflexive spaces $\bar B$ is weakly compact, so there exists a subsequence $x_{n_j} \to x$ weakly.

Because $A$ is compact, $Ax_{n_j}\to Ax$ strongly.

On the other side, $A(\bar B)$ is relatively compact, so there exists $z_k=Ax_{n_{j_k}}$ that converges strongly to $y\in Y$.

But $z_k\to Ax$ strongly. So by unicity of the limit $y=Ax$ and the image is compact.

It's easily proved that if the image is compact, the operator is also compact.

But I don't know what to do in case of nonreflexive spaces. Is there any counterexample or proof in such case?

MorganeMaPh
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Тимофей Ломоносов
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  • You might find this interesting: http://math.stackexchange.com/questions/270862/compact-operators-why-is-the-image-of-the-unit-ball-only-assumed-to-be-relative – kahen May 28 '13 at 06:32
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    What is your definition of a compact operator? – Vishal Gupta May 28 '13 at 06:51
  • Forgive me: I'm trying to understand why Because A is compact, $A_{{x_n}_j}→Ax$ strongly. I only know that $\forall f\in X^{\ast} f(x_{n_j})\to f(x)$ strongly. Could you be so kind to explain that to me? $\infty$ thanks! – Self-teaching worker Sep 11 '14 at 23:03
  • @Self-teachingworker See https://math.stackexchange.com/q/70410/491450 – Smiley1000 May 21 '25 at 15:54

2 Answers2

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It is false in general that the image of the closed unit ball under a compact operator is closed (and hence compact). Here is an easy example:

Consider $X = C[0,1]$ with the uniform norm, and the compact operator $A \in B(X)$ defined by the formula:

$\displaystyle\qquad Af(x) = \int_0^x f(t)\,dt$.

Compactness of $A$ is easily proven using Arzelà–Ascoli. Our operator $A$ produces an anti-derivative of any input given to it, and the image of the closed unit ball of $X$ under $A$ is the set

$\displaystyle\qquad \{f \in C^1[0,1] \mathrel: f(0)=0,\ \lVert f'\rVert \leq 1\}$

which certainly is not a closed subset of $X$.

kahen
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  • Thank you @Kahen. Do we necessarily need $X$ and $Y$ to be Banach? Is it because we want more than closedness: Compactness? - Because I know that following statement holds: If $X$ is reflexive normed vector space, $A\in \mathcal{L}(X,Y)$ with $Y$ a normed vector space a compact operator. Then the image of the closed unit ball under A is a closed subset of $Y$. – MorganeMaPh Jul 23 '16 at 09:09
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This is much easier. Use the definition of compact operator. An operator is compact if and only if the image of a bounded set is precompact. The unit ball is bounded and any other bounded set is inside some $\lambda$ times the unit ball.

Vishal Gupta
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  • There are many definitions of compact operators and the one used by the OP is probably the one given in the course/book. Using a different equivalent characterization means that one need to prove that the definitions are equivalent. // "The unit ball is compact"? Do you mean weakly compact? Then you need to argue why a compact operator is continuous with respect to the weak topology on the unit ball and the norm topology in the target. – Martin May 28 '13 at 06:45
  • sorry, I meant bounded! – Vishal Gupta May 28 '13 at 06:50