For
dj=1/j, the
equation Du=f forces uj=jfj. It
has a solution in
ℓ2 exactly when
For example,
fj=1/j defines an element of
ℓ2, but its forced preimage is the constant
sequence uj=1, which is not in
ℓ2. Thus
existence can fail. Since every
dj is nonzero,
D has
trivial kernel and any solution is unique.