Discussion
Let
\[ \begin{aligned} c_{00} = \{&(x_{1},x_{2},x_{3},\ldots) : x_{i} \in \mathbb{R},\ i \in \mathbb{N}, \\ &x_{i} \ne 0 \text{ only for finitely many indices } i\}. \end{aligned} \]
For \((x_{1},x_{2},x_{3},\ldots) \in c_{00}\), let \(\|(x_{1},x_{2},x_{3},\ldots)\|_{\infty} = \sup\{|x_{i}| : i \in \mathbb{N}\}\).
Define \(F, G : (c_{00}, \|\cdot\|_{\infty}) \to (c_{00}, \|\cdot\|_{\infty})\) by
\[ \begin{aligned} F((x_{1},\ldots,x_{n},\ldots)) = (&(1+1)x_{1}, (2+\tfrac{1}{2})x_{2}, \\ &\ldots, (n+\tfrac{1}{n})x_{n}, \ldots), \end{aligned} \]
\[ \begin{aligned} G((x_{1},\ldots,x_{n},\ldots)) = (&\tfrac{x_{1}}{1+1}, \tfrac{x_{2}}{2+\frac{1}{2}}, \\ &\ldots, \tfrac{x_{n}}{n+\frac{1}{n}}, \ldots), \end{aligned} \]
for all \((x_{1},x_{2},\ldots,x_{n},\ldots) \in c_{00}\).
Then
Answer: B
F is NOT continuous but G is continuous.
A linear map between normed spaces is continuous exactly when it is bounded.
F multiplies the n-th coordinate by n + 1/n. For the unit vector \(e_{n}\), \(\|F(e_{n})\|_{\infty} = n + 1/n\), which is unbounded while \(\|e_{n}\|_{\infty} = 1\). So F is unbounded and not continuous.
G divides the n-th coordinate by n + 1/n, which is at least 2, so \(\|G(x)\|_{\infty} \le \tfrac{1}{2}\|x\|_{\infty}\). G is bounded and therefore continuous. (G is the inverse of F on \(c_{00}\), a bounded map whose inverse is unbounded.)
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