Correct answer: (C) T is bounded and its inverse on its range exists but is unbounded.
Explanation
‖Tx‖∞ ≤ ‖x‖∞, and nonzero scaling in every coordinate makes T injective. On the range, inverse multiplies coordinate n by n. For the unit coordinate vector e_n, ‖Te_n‖∞ = 1/n while ‖e_n‖∞ = 1, so inverse norms have ratios n without bound. **C**.
More Bounded Linear Operators and Duality questions
GATE 2026 MA Q53– For a normed linear space X and its continuous dual X′, which statements hold?
GATE 2025 MA Q17– X=f∈C[0,1]:f(0)=f(1)=0 with sup norm; Y=C[0,1] with L² norm. On their closed unit balls U…
GATE 2025 MA Q50– For 0<α<1, Cα[0,1] is the Hölder Banach space with norm f(0) +sup (s≠t) f(t)−f(s) / t−s α.…
GATE 2024 MA Q47– On finitely supported real sequences C₀₀ define f(x)=Σₙ₌₁∞x n/√n. I:f extends continuously…
GATE 2024 MA Q49– X=ℝ² with norm x + y . On S=(x,0), f(x,0)=2x. A Hahn–Banach extension F agrees with f and…
GATE 2023 MA Q23– In real C[0,1] with supremum norm let X=f:f(0)=f(1)=0, S X=f∈X:‖f‖∞=1 and F(f)=∫₀¹f(t)dt.…