Consider the Hilbert space ℓ2 of square-summable complex sequences(a1, a2, . . . ).You may assume that the subset S = {e1, e2, . . . } with e1 = (1, 0, 0, . . . ), e2 = (0, 1, 0, . . . ),. . . , is a Schauder basis for ℓ2.Let T ∶ ℓ2 Ð→ ℓ2 be a linear operator.(a) Show that T is a bounded linear operator if and only if the sequence (∥T (en)∥) isbounded.(b) IfT (ej ) =∞∑n=1cjn en,give a condition on the coefficients cjn that is necessary and sufficient for T to beself-adjoint.

(a) We need to show that T is a bounded linear operator if and only if the sequence (∥T (en)∥) is bounded.

First, assume that T is a bounded linear operator. This means that there exists a constant C such that ∥T(x)∥ ≤ C∥x∥ for all x in ℓ2. In particular, for x = en, we have ∥T(en)∥ ≤ C∥en∥ = C. Hence, the sequence (∥T (en)∥) is bounded.

Conversely, assume that the sequence (∥T (en)∥) is bounded. This means that there exists a constant C such that ∥T(en)∥ ≤ C for all n. Given any x = (a1, a2, ...) in ℓ2, we can write x as a linear combination of the en's, i.e., x = ∑ anen. Then, by linearity of T and the triangle inequality, we have ∥T(x)∥ = ∥T(∑ anen)∥ = ∥∑ anT(en)∥ ≤ ∑ |an|∥T(en)∥ ≤ ∑ |an|C = C∥x∥. Hence, T is a bounded linear operator.

(b) For T to be self-adjoint, we need T = T*, where T* is the adjoint of T. This means that for all x, y in ℓ2, we have <Tx, y> = <x, Ty>. In terms of the coefficients cjn, this translates to the condition that cjn = cnj for all j, n. That is, the matrix (cjn) must be Hermitian.