= Solution
Define a <diagonal operator on sequence space> by
$$
\boxed{S:c_0\longrightarrow\ell_2,\qquad Sx=(2^{-n}x_n)_{n\geq1}.}
$$
It is bounded because $\|Sx\|_2^2\leq\|x\|_\infty^2\sum_{n\geq1}4^{-n}=\|x\|_\infty^2/3$. Its truncation $S_mx=(2^{-1}x_1,\ldots,2^{-m}x_m,0,\ldots)$ is a <finite-rank operator>, and
$$
\|S-S_m\|=\left(\sum_{n>m}4^{-n}\right)^{1/2}=\frac{2^{-m}}{\sqrt3}\longrightarrow0.
$$
Equality follows by using <finitely supported sequences> equal to $1$ on longer and longer finite tails. An <operator norm> limit of <compact operators> into the <Banach space> $\ell_2$ is compact; therefore \b[$S$ is a <compact operator>]. The vectors $Se_n=2^{-n}e_n$ are <linearly independent>, so its rank is infinite.
There is no conflict with the preceding result about a compact closed-ball <image>. Indeed, take $x^{(m)}=(1,\ldots,1,0,\ldots)$ with $m$ initial ones. Then $Sx^{(m)}\to(2^{-n})_{n\geq1}$ in $\ell_2$, but that limit has no preimage in $c_0$: its only possible preimage is the constant sequence $1$. Thus $S(B_{c_0})$ is relatively compact but is not closed and hence is not compact.
Back to article page
