For n∈N define bounded operators Tₙ,T from ℓ¹ to ℓ∞ by T n(x 1,x 2,)=(x 1,,x n,x n,x n,) and T(x)=x. Which is true?

$\|T_n\|$ does not converge to $\|T\|$.

For every nonzero x there is a continuous linear functional g on ℓ∞ for which g(Tₙx) fails to converge to g(Tx).

GATE 2020 MA – Question 39

Functional Analysis · Bounded Linear Operators and Duality · 2 marks · Multiple choice

For n∈N define bounded operators Tₙ,T from ℓ¹ to ℓ∞ by $T_n(x_1,x_2,\ldots)=(x_1,\ldots,x_n,x_n,x_n,\ldots)$ and $T(x)=x$. Which is true?

$\|T_n\|$ does not converge to $\|T\|$.
$\|T_n-T\|\to0$.
For each x∈ℓ¹, $\|T_nx-Tx\|_\infty\to0$.
For every nonzero x there is a continuous linear functional g on ℓ∞ for which g(Tₙx) fails to converge to g(Tx).

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Correct answer: (C) For each x∈ℓ¹, $\|T_nx-Tx\|_\infty\to0$.

Explanation

For fixed x∈ℓ¹, its entries tend to zero. The sup-norm difference is bounded by $|x_n|+\sup_{j>n}|x_j|\to0$, proving strong convergence, C. Both operator norms are 1, so A fails. Taking x = eₙ gives $\|(T_n-T)e_n\|_\infty=1$, ruling out B. Continuous linear functionals preserve each norm-convergent sequence, ruling out D.

GATE 2020 MA – Question 39

Explanation

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