I am reading this paper by J. Appell (if you want, you can download for free by looking for the title in Google), but I have some doubts with example 3.1 on p. 174, which I describe below.
Let the product space $X:=\ell_{2}\times c$, where $\ell_{2}$ is the usual space of square summable sequences and $c$ the space of all convergent sequences. Consider in $X$ the the Euclidean norm $\|(x,y)\|^{2}:=\|x\|^{2}+\|y\|^{2}$. I assume that, really, this norm is $\|(x,y)\|^{2}:=\|x\|_{\ell_{2}}^{2}+\|y\|_{c}^{2}$, where $\|\cdot\|_{\ell_{2}}$ and $\|\cdot\|_{c}$ are the usual norms of $\ell_{2}$ and $c$, respectively. Do you agree?
Then, define the linear operator $L:X\longrightarrow X$ by $L(x,y):=(0,x)$. In the cited paper is stated that $$ L(B(X))=\{(0,y):y\in B(c)\}:=M, $$ where $B(X)$ and $B(c)$ are the closed unit balls of $X$ and $c$, respectively. I am nor sure that this assert be correct: given any $(0,y)\in M$ with $y:=(y_{n})_{n\geq 1}\in B(c)$ such that $\lim_{n}y_{n}=\varepsilon >0$, Can we find $x\in \ell_{2}$ and $y\in c$ such that $L(x,y)=(0,x)\in M$? (recall that as $x\in\ell_{2}$, $\lim_{n} x_{n}^{2}=0$).
On the other hand, in the cited paper is stated the following:
"...for each $(0,y)\in M$ we can find a finite sequence $\tilde{y}$ such that $\|y-\tilde{y}\|\leq 1/2$, and the subset $\tilde{M}:=\{(0,\tilde{y}):y\in M\}$ is a compact $1/\sqrt{2}$-net for $M$ in $X$."
Why is true the above assert? Can we take such $1/\sqrt{2}$-net in $M$?
Many thanks in advance for your suggestions and comments.
If $\|y\|_c=\sup |y_n|$ for each $y=(y_n)\in c$ then for each $0<\varepsilon<1$ there is no set $K$ of $c$ such that $K$ is a compact $\varepsilon$-net for $B(c)$ is $c$ (or, that is the same, a set $\{(0,y):y\in K\}$ is a compact $\varepsilon$-net for $M$ is $X$). Indeed, suppose to the contrary that there exists such a set $K$. Let $A$ be an infinite subset of $B(c)$ such that $\|y-y'\|_c=2$ for each distinct $y,y'\in A$. For instance, we can put $A=\{a^m:m\in\Bbb N\}$, where for each $m$ the first $m$ coordinates of $a^m$ are $1$, the $m+1$-th coordinate is $-1$, and the remaining coordinates are zeroes. Since $K$ is a $\varepsilon$-net for $B(c)$, for each $a^m$ there exists $y^m\in K$ such that $\|y^m-a^m\|_c\le\varepsilon$. The triangle inequality implies that $\|y^m-y^{m'}\|_c\ge 2(1-\varepsilon)$ for each distinct $m$ and $m'$. This follows that $\{y^m\}$ is a sequence of points of a compact metric space, containing no convergent (even no Cauchy) subsequence, which is impossible.