# Two measures of noncompactness of operators

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?

• Could you, maybe, add a reference for the fact that $T$ is compact iff $\|T\|_m = 0$? – Jochen Glueck Apr 12 '20 at 14:27
• 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 $T$ – DCM Apr 12 '20 at 14:45

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, $$\|JT\|_e\leq \|T\|_m$$.