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I read, on an Italian language version of Kolmogorov-Fomin's Introductory Real Analysis, that, for any Banach space $E$, the unit closed sphere $S^{\ast}$ of $E^{\ast}$ is compact in the $\ast$-weak topology for space $E^{\ast}$. The compactness of the image of $S^{\ast}$ (with respect to the metric of $E^{\ast}$) through any compact operator follows.

If it meant that for any compact linear operator $A:E\to E$, the image $A^{\ast}(S^{\ast})$ through the adjoint operator is compact with respect to the strong topology of $E^{\ast}$, that would be clear to me from theorem 4 here. It happens that such theorem 4 immediately precedes the quoted statement in the Italian translation...

Or does it mean that for any compact operator $B:E^{\ast}\to E^{\ast}$ the image $B(S^{\ast})$ is compact? If that is the meaning of the statement, how can we prove it?

$\infty$ thanks!!!

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It must mean "for the transpose of a compact linear operator $A\colon E\to E$".

We can find an easy counterexample to the stronger claim:

Let $E = c_0(\mathbb{N}) = \{x\in \ell^\infty(\mathbb{N}) : x_n \to 0\}$. We know that then $E^\ast$ is isometrically isomorphic to $\ell^1(\mathbb{N})$, so we identify the two spaces. If we take an operator $B$ on $\ell^1(\mathbb{N})$ that is not weak$^\ast$-weak$^\ast$ continuous, we have a chance that $B(S^\ast)$ is not compact. To find such a $B$, we take a continuous linear functional on $\ell^1(\mathbb{N})$ that is not weak$^\ast$ continuous,

$$\lambda \colon x \mapsto \sum_{n=0}^\infty (1-2^{-n-1})x_n$$

serves well. We have $\lVert\lambda\rVert = 1$, and $\lambda(S^\ast) = \mathbb{D} = \{z\in\mathbb{C} : \lvert z\rvert < 1\}$ is open and not compact. Composing $\lambda$ with any linear embedding $\mathbb{C}\to \ell^1(\mathbb{N})$ then yields a compact (since continuous with finite-dimensional range) operator $B\colon \ell^1(\mathbb{N}) \to \ell^1(\mathbb{N})$ such that $B(S^\ast)$ is not compact.

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  • $\begingroup$ I heartily thank you. Very clear. $\endgroup$ – Self-teaching worker Sep 11 '14 at 19:15

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