Consider the function g ∶ ℓ2 Ð→ F given byg(x) =∞∑n=1xnn2 .(a) Find y ∈ ℓ2 such thatg(x) = ⟨x, y⟩ for all x ∈ ℓ2.(b) Deduce that g is linear and bounded and find its norm ∥g∥.[Hint: You may use without proof the fact that∞∑n=11n4 = π490 .]

(a) The function g is defined as an infinite series, and it's given that g(x) = ∑ (from n=1 to ∞) x_n/n^2. We want to find a y in ℓ2 such that g(x) = ⟨x, y⟩ for all x in ℓ2. The inner product ⟨x, y⟩ in ℓ2 is defined as ∑ (from n=1 to ∞) x_n * y_n. So, we want to find a y such that ∑ (from n=1 to ∞) x_n/n^2 = ∑ (from n=1 to ∞) x_n * y_n. This equality holds if y_n = 1/n^2 for all n. So, y = (1, 1/4, 1/9, 1/16, ...) is in ℓ2 and satisfies g(x) = ⟨x, y⟩ for all x in ℓ2.

(b) The linearity of g follows directly from the linearity of the inner product. For any x, z in ℓ2 and scalar a, we have g(ax + z) = ⟨ax + z, y⟩ = a⟨x, y⟩ + ⟨z, y⟩ = ag(x) + g(z).

The boundedness of g and its norm can be found using the Cauchy-Schwarz inequality, which states that |⟨x, y⟩| ≤ ||x|| * ||y|| for all x, y in ℓ2. So, |g(x)| = |⟨x, y⟩| ≤ ||x|| * ||y||. This shows that g is bounded and its norm is less than or equal to ||y||.

The norm of y in ℓ2 is ||y|| = sqrt(∑ (from n=1 to ∞) |y_n|^2) = sqrt(∑ (from n=1 to ∞) 1/n^4). Using the hint, we know that ∑ (from n=1 to ∞) 1/n^4 = π^4/90. So, ||y|| = sqrt(π^4/90), and therefore ||g|| ≤ sqrt(π^4/90). Since ||g|| is the supremum of |g(x)|/||x|| over all non-zero x in ℓ2, and we have shown that |g(x)|/||x|| ≤ sqrt(π^4/90) for all x, we conclude that ||g|| = sqrt(π^4/90).