# Two Continuous Extensions of an Operator

Let $$T: V \to U$$ be a linear operator, where $$V \subset W$$, and suppose there exists two norms on $$W$$, denoted $$\| \cdot \|_1$$ and $$\| \cdot \|_2$$, such that $$V$$ is a dense subspace of $$W$$ with respect to either norm. If $$T$$ is continuous with respect to both operators, are the unique extensions of $$T$$ to a continuous map on $$W$$ with either norm the same?

• Do you have any more information on the norms? If the norms are equivalent it follows from the construction of the extension itself (assuming U is complete). – Joe Thomas Oct 4 '19 at 20:25
• The norms are not necessarily equivalent. – Jacob Denson Oct 4 '19 at 20:41

Let $$W = \ell^1(\mathbb{Z})$$ and define the additional norms on $$W$$ given by $$\|x\|_+ = \sum_{n < 0} \frac{|x_n|}{|n|} + \sum_{n \geq 0} |x_n| \qquad \text{and} \qquad \|x\|_- = \sum_{n \leq 0} |x_n| + \sum_{n > 0} \frac{|x_n|}{|n|}$$ Define $$S : W \to \mathbb{R}$$ by $$S(x) = \sum_n x_n$$, and let $$V$$ be the subspace of $$x \in W$$ with $$S(x) = 0$$. To show that $$V$$ is dense under either norm, for $$n \in \mathbb{Z}$$ let $$e_n \in W$$ be the vector with $$n$$-th entry $$1$$ and all other entries $$0$$, i.e. $$(e_n)_i = \delta_{ni}$$. Then for any $$x \in W$$, setting $$y_n = x - S(x)e_n \in V$$, we have $$\|x - y_n\|_+ = |S(x)|\|e_n\|_+ \to 0$$ as $$n \to -\infty$$ and similarly, $$\|x - y_n\|_- = |S(x)| \|e_n\|_- \to 0$$ as $$n \to \infty$$. It follows that $$V$$ is dense under both $$\|\cdot\|_+$$ and $$\|\cdot\|_-$$.
Now let $$U = \mathbb{R}$$, and define $$T_+ : W \to U$$ by $$x \mapsto \sum_{n \geq 0} x_n$$ and $$T_- : W \to U$$ by $$x \mapsto -\sum_{n < 0} x_n$$. It's clear that $$T_+$$ is continuous under $$\|\cdot\|_+$$ and $$T_-$$ is continuous under $$\|\cdot\|_-$$, since for $$x \in W$$ we have $$|T_+(x)| \leq \|x\|_+$$ and $$|T_-(x)| \leq \|x\|_-$$. Note also that $$T_+$$ and $$T_-$$ agree on $$V$$, since $$T_+ - T_- = S = 0$$ on $$V$$, i.e. they both extend a map $$T : V \to U$$ which is continuous under both norms -- thus $$T_+$$ is the unique extension of $$T$$ under $$\|\cdot\|_+$$, while $$T_-$$ is the unique extension of $$T$$ under $$\|\cdot\|_-$$. Despite this, the two maps disagree: we have $$T_+(e_0) = 1$$, while $$T_-(e_0) = 0$$.