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Let $X,Y$ be Banach spaces. We denote by $\mathcal{K}(X,Y)$ the space of all compact operators from $X$ into $Y$. For an operator $T:X\rightarrow Y$, we let $$\|T\|_{e}:=\inf\{\|T-K\|:K\in \mathcal{K}(X,Y)\},$$ and $$\|T\|_{m}:=\inf\{\|T|_{M}\|:codim M<\infty\},$$ where $M$ represents the finite co-dimensional subspace of $X$. It is known that $T$ is compact if and only if $\|T\|_{m}=0$.

Let $X$ be a Banach space. We define $J:X\rightarrow l_{\infty}(B_{X^{*}})$ by $\langle Jx,x^{*}\rangle=\langle x^{*},x\rangle, x\in X, x^{*}\in B_{X^{*}}.$ Let $T:Z\rightarrow X$ be an operator. My question is:

Question. Does $\|JT\|_{e}=\|T\|_{m}$ hold?

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    $\begingroup$ Could you, maybe, add a reference for the fact that $T$ is compact iff $\|T\|_m = 0$? $\endgroup$ – Jochen Glueck Apr 12 '20 at 14:27
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    $\begingroup$ My go-to books for things like this are Pietsch's History of Banach Spaces and Linear Operators and Spectral Theory and Differential Operators by Edmunds and Evans. Both discuss s-numbers at length. If I remember right $\Vert T \Vert_m$ looks like it might be connected with the so-called `Gelfand numbers' of $T$ $\endgroup$ – DCM Apr 12 '20 at 14:45
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The answer is yes. Indeed, let us denote $Y_h=\ell_\infty(B_{Y^*})$. For every $L\in\mathcal{K}(X,Y_h)$,

$\|T\|_m=\|JT\|_m=\|JT-L\|_m\leq \|JT-L\|$. Hence $\|T\|_m\leq \|JT\|_e$.

Conversely, since $Y_h$ is $1$-injective, for every finite codimensional closed subspace $M$ of $X$, then there exists an extension $\hat T:X\to Y_h$ of $JT|_M$ with $\|\hat T\|=\|T|_M\|$. Since $JT-\hat T$ is compact, $\|JT\|_e\leq \|T\|_m$.

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