“…1-3.) If f is a solution of [3] and 5' is an index then, as in (1), there is a lesser 5 for which g = Aj -f is a solution of [2], so that, for an even lesser 6, Aag = g. Thus, by induction on n, (Ab)nf = (A6)Y-'f + g, and hence, for all integers n, ng = (A6)nf -f. [4] Thus, for every semi-norm, IIgI < 1/n2 11f11 for all n, so that jjg = 0. Thusg = 0,orAbf=f.…”