Question

Let \(\ell^{\infty} = \{x = (x_n)_{n \geq 1} \mid x_n \in \mathbb{R},\ \sup\{|x_n| : n = 1, 2, \ldots\} < \infty\}\) with the supremum norm. Let \(T : \ell^{\infty} \to \ell^{\infty}\) be given by \(T(x_1, x_2, x_3, \ldots) = \left(x_1, \dfrac{x_2}{2}, \dfrac{x_3}{3}, \ldots \right)\). Which one of the following is TRUE?

• A. \(T\) is bounded but not one-to-one.
• B. \(T\) is one-to-one but not bounded.
• C. \(T\) is bounded and the inverse (from the range of \(T\)) exists but not bounded.
• D. \(T\) is bounded and the inverse (from the range of \(T\)) is bounded.

Hint:

Check boundedness using the multiplier 1/n, injectivity by solving Tx=0, then test the inverse (multiply back by n) on a sequence like y_n = 1/sqrt(n) truncated at N and let N grow.

Answer 1

The Correct Option is C

$T$ is a diagonal operator on $\ell^{\infty}$: it multiplies the $n$-th coordinate of $x$ by the scalar $a_n = 1/n$. For a diagonal operator, three standard facts about the multipliers $a_n$ settle everything.

1. Boundedness of $T$: a diagonal operator with multipliers $a_n$ is bounded on $\ell^{\infty}$ exactly when $(a_n)$ is a bounded sequence, and then $\|T\| = \sup_n |a_n|$. Here $a_n = 1/n \leq 1$ for all $n \geq 1$, so $T$ is bounded with $\|T\| = 1$, attained at $n=1$.
2. Injectivity of $T$: a diagonal operator is one-to-one exactly when no multiplier is zero. Since $a_n = 1/n \neq 0$ for every $n$, $T$ never sends a non zero coordinate to zero, so $T$ is injective.
3. Boundedness of the inverse: on the range of $T$, the inverse is again diagonal, with multipliers $1/a_n = n$. A diagonal operator is bounded exactly when its multipliers form a bounded sequence. Here the multipliers are $n = 1,2,3,\ldots$, an unbounded sequence, so $T^{-1}$ cannot be bounded.

A concrete check on point 3: take $y = (1,1,\ldots,1,0,0,\ldots)$ with exactly $N$ ones, so $\|y\|_{\infty}=1$. Its preimage under $T$ has $n$-th entry $n \cdot 1 = n$ for $n \leq N$, giving $x = (1,2,3,\ldots,N,0,0,\ldots)$ with $\|x\|_{\infty} = N$. The ratio $\|T^{-1}y\|_{\infty}/\|y\|_{\infty} = N$ can be made as large as we like by taking $N$ larger, so no fixed constant bounds it, confirming $T^{-1}$ is unbounded.

Let's summarize:

• Diagonal operator with multipliers $1/n$: bounded (multipliers bounded by 1) and one-to-one (no multiplier is zero).
• Its inverse has multipliers $n$, an unbounded sequence, so the inverse exists on the range of $T$ but is not bounded.

So $T$ is bounded, one-to-one, and its inverse on the range exists but is unbounded, which is option (C).

\[\boxed{\text{Option (C): bounded, injective, inverse exists but unbounded}}\]