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?
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?
Show 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.
Updated On: Jul 21, 2026
- \(T\) is bounded but not one-to-one.
- \(T\) is one-to-one but not bounded.
- \(T\) is bounded and the inverse (from the range of \(T\)) exists but not bounded.
- \(T\) is bounded and the inverse (from the range of \(T\)) is bounded.
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Check that T is bounded.
For \(x = (x_n) \in \ell^{\infty}\), the \(n\)-th entry of \(Tx\) is \(x_n/n\). Since \(n \geq 1\), \(|x_n/n| \leq |x_n|\) for every \(n\), so
\[ \|Tx\|_{\infty} = \sup_n \left| \frac{x_n}{n} \right| \leq \sup_n |x_n| = \|x\|_{\infty}. \]
So \(T\) is a bounded linear operator with \(\|T\| \leq 1\). Taking \(x = (1, 0, 0, \ldots)\) gives \(Tx = (1,0,0,\ldots)\) with equal norms, so in fact \(\|T\| = 1\).
Step 2: Check that T is one-to-one.
Suppose \(Tx = 0\). Then \(x_n/n = 0\) for every \(n\), and since \(n \neq 0\), this forces \(x_n = 0\) for every \(n\). So only \(x=0\) maps to \(0\), meaning \(T\) is injective (one-to-one). This already rules out options (A) and (B): (A) says \(T\) is not one-to-one and (B) says \(T\) is not bounded, both false.
Step 3: Look at the inverse on the range of T.
Since \(T\) is injective, \(T^{-1}\) is well defined from the range of \(T\) back to \(\ell^{\infty}\), by \((T^{-1}y)_n = n y_n\). We need to check if this inverse is bounded, meaning whether some constant \(C\) satisfies \(\|T^{-1}y\|_{\infty} \leq C\|y\|_{\infty}\) for every \(y\) in the range of \(T\).
Step 4: Build a sequence that breaks boundedness of the inverse.
For each \(N\), let \(y^{(N)}_n = 1/\sqrt{n}\) for \(n = 1, \ldots, N\) and \(y^{(N)}_n = 0\) for \(n > N\). Then \(\|y^{(N)}\|_{\infty} = 1\) for every \(N\) (the largest value is at \(n=1\)). This \(y^{(N)}\) is in the range of \(T\), since its preimage \(x^{(N)}_n = n y^{(N)}_n = \sqrt{n}\) for \(n \leq N\) (and \(0\) after) is a genuinely bounded sequence, bounded by \(\sqrt{N}\). But
\[ \|T^{-1} y^{(N)}\|_{\infty} = \|x^{(N)}\|_{\infty} = \sqrt{N}, \]
which grows without bound as \(N \to \infty\), while \(\|y^{(N)}\|_{\infty}\) stays fixed at \(1\). So no single constant works for every \(y\) in the range, and \(T^{-1}\) is not bounded.
Final Answer:
\(T\) is bounded and one-to-one, but its inverse on the range of \(T\) exists and is unbounded, matching option (C). \[ \boxed{\text{Option (C)}} \]
For \(x = (x_n) \in \ell^{\infty}\), the \(n\)-th entry of \(Tx\) is \(x_n/n\). Since \(n \geq 1\), \(|x_n/n| \leq |x_n|\) for every \(n\), so
\[ \|Tx\|_{\infty} = \sup_n \left| \frac{x_n}{n} \right| \leq \sup_n |x_n| = \|x\|_{\infty}. \]
So \(T\) is a bounded linear operator with \(\|T\| \leq 1\). Taking \(x = (1, 0, 0, \ldots)\) gives \(Tx = (1,0,0,\ldots)\) with equal norms, so in fact \(\|T\| = 1\).
Step 2: Check that T is one-to-one.
Suppose \(Tx = 0\). Then \(x_n/n = 0\) for every \(n\), and since \(n \neq 0\), this forces \(x_n = 0\) for every \(n\). So only \(x=0\) maps to \(0\), meaning \(T\) is injective (one-to-one). This already rules out options (A) and (B): (A) says \(T\) is not one-to-one and (B) says \(T\) is not bounded, both false.
Step 3: Look at the inverse on the range of T.
Since \(T\) is injective, \(T^{-1}\) is well defined from the range of \(T\) back to \(\ell^{\infty}\), by \((T^{-1}y)_n = n y_n\). We need to check if this inverse is bounded, meaning whether some constant \(C\) satisfies \(\|T^{-1}y\|_{\infty} \leq C\|y\|_{\infty}\) for every \(y\) in the range of \(T\).
Step 4: Build a sequence that breaks boundedness of the inverse.
For each \(N\), let \(y^{(N)}_n = 1/\sqrt{n}\) for \(n = 1, \ldots, N\) and \(y^{(N)}_n = 0\) for \(n > N\). Then \(\|y^{(N)}\|_{\infty} = 1\) for every \(N\) (the largest value is at \(n=1\)). This \(y^{(N)}\) is in the range of \(T\), since its preimage \(x^{(N)}_n = n y^{(N)}_n = \sqrt{n}\) for \(n \leq N\) (and \(0\) after) is a genuinely bounded sequence, bounded by \(\sqrt{N}\). But
\[ \|T^{-1} y^{(N)}\|_{\infty} = \|x^{(N)}\|_{\infty} = \sqrt{N}, \]
which grows without bound as \(N \to \infty\), while \(\|y^{(N)}\|_{\infty}\) stays fixed at \(1\). So no single constant works for every \(y\) in the range, and \(T^{-1}\) is not bounded.
Final Answer:
\(T\) is bounded and one-to-one, but its inverse on the range of \(T\) exists and is unbounded, matching option (C). \[ \boxed{\text{Option (C)}} \]
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