Question

Let \(c_{00} = \{(x_1, x_2, x_3, \dots) : x_i \in \mathbb{R}, i \in \mathbb{N}, x_i \neq 0 \text{ only for finitely many indices } i\}\).
For \((x_1, x_2, x_3, \dots) \in c_{00}\), let \(||(x_1, x_2, x_3, \dots)||_\infty = \sup\{|x_i| : i \in \mathbb{N}\}\).
Define \(F, G: (c_{00}, ||.||_\infty) \to (c_{00}, ||.||_\infty)\) by
\[ F((x_1, x_2, \dots, x_n, \dots)) = ((1+1)x_1, (2+\frac{1}{2})x_2, \dots, (n+\frac{1}{n})x_n, \dots), \]
\[ G((x_1, x_2, \dots, x_n, \dots)) = \left(\frac{x_1}{1+1}, \frac{x_2}{2+\frac{1}{2}}, \dots, \frac{x_n}{n+\frac{1}{n}}, \dots\right), \]
for all \((x_1, x_2, \dots, x_n, \dots) \in c_{00}\).
Then

• A. F is continuous but G is NOT continuous
• B. F is NOT continuous but G is continuous
• C. both F and G are continuous
• D. NEITHER F NOR G is continuous

Hint:

For a "diagonal" linear operator on a sequence space (one that just multiplies each component by a scalar), the operator is bounded if and only if the sequence of scalars is bounded. The operator norm is the supremum of the absolute values of these scalars. For F, the scalars are \(n+1/n\), which is an unbounded sequence. For G, the scalars are \(1/(n+1/n)\), which is a bounded sequence.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a linear operator \(A\) between normed spaces, continuity is equivalent to boundedness. An operator A is bounded if there exists a constant \(C \ge 0\) such that \(||A(x)|| \le C||x||\) for all \(x\) in the domain. The operator norm is the smallest such C, given by \(||A|| = \sup_{||x||=1} ||A(x)||\). If the operator norm is finite, the operator is bounded and thus continuous. If it is infinite, the operator is unbounded and not continuous.
Step 3: Detailed Explanation:
Let \(x = (x_1, x_2, ......)\). Analysis of Operator F: \(F(x) = y = (y_1, y_2, ......)\) where \(y_n = (n + 1/n)x_n\). To check if F is bounded, we look for its operator norm: \[ ||F|| = \sup_{||x||_\infty = 1} ||F(x)||_\infty = \sup_{||x||_\infty = 1} \sup_n |(n+\frac{1}{n})x_n| \] Let's test this with a sequence of vectors \(e_k \in c_{00}\), where \(e_k\) has a 1 in the kth position and zeros elsewhere. Note that \(||e_k||_\infty = 1\). The image is \(F(e_k)\). The kth component of \(F(e_k)\) is \((k+1/k) . 1 = k+1/k\). All other components are zero. So, \(||F(e_k)||_\infty = \sup_n |(F(e_k))_n| = k + 1/k\). The norm of the operator is the supremum of these values over all unit vectors. \[ ||F|| = \sup_{k \ge 1} ||F(e_k)||_\infty = \sup_{k \ge 1} (k + 1/k) \] As \(k \to \infty\), the term \(k+1/k \to \infty\). The supremum is infinite. Since the operator norm is infinite, F is unbounded. An unbounded linear operator is not continuous. Thus, F is NOT continuous.
Analysis of Operator G: \(G(x) = y = (y_1, y_2, ......)\) where \(y_n = \frac{x_n}{n + 1/n}\). Let's find the operator norm of G. \[ ||G(x)||_\infty = \sup_n |y_n| = \sup_n \left|\frac{x_n}{n+1/n}\right| = \sup_n \left(\frac{1}{n+1/n} |x_n|\right) \] Since \(|x_n| \le \sup_k |x_k| = ||x||_\infty\), we have: \[ ||G(x)||_\infty \le \sup_n \left(\frac{1}{n+1/n} ||x||_\infty\right) = \left(\sup_n \frac{1}{n+1/n}\right) ||x||_\infty \] The term \(\frac{1}{n+1/n}\) is maximized when the denominator \(n+1/n\) is minimized. For \(n \ge 1\), the minimum of \(n+1/n\) occurs at \(n=1\), where the value is \(1+1/1 = 2\). So, \(\sup_n \frac{1}{n+1/n} = \frac{1}{1+1/1} = \frac{1}{2}\). Therefore, \(||G(x)||_\infty \le \frac{1}{2}||x||_\infty\). This shows that G is a bounded linear operator with norm \(||G|| \le 1/2\). Since G is bounded, it is continuous. Thus, G is continuous.
Step 4: Final Answer:
F is NOT continuous but G is continuous. Step 5: Why This is Correct:
The continuity of the linear operators F and G is equivalent to their boundedness. The norm of F was shown to be infinite by considering the sequence of standard basis vectors, proving it is not continuous. The norm of G was shown to be finite (\(\le 1/2\)), proving it is bounded and therefore continuous.