On finitely supported real sequences C₀₀ define f(x)=Σₙ₌₁∞x_n/√n. I:f extends continuously linearly to ℓ¹. II:f extends continuously linearly to ℓ². Choose truth values.
For ℓ¹, $|f(x)|\le\sum|x_n|=\|x\|_1$, so the same absolutely convergent series defines a continuous extension. For ℓ², the coefficient sequence (1/√n) is not square summable. Explicitly, x_n=1/√n for n≤N gives f(x)=H_N and ‖x‖₂=√H_N, an unbounded ratio. Hence no bounded ℓ² extension. **B**.
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