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?

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  • $\begingroup$ 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). $\endgroup$ – Joe Thomas Oct 4 '19 at 20:25
  • $\begingroup$ The norms are not necessarily equivalent. $\endgroup$ – Jacob Denson Oct 4 '19 at 20:41

Update: I've added a much cleaner construction of the counterexample.

No, the two extensions may be different.

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 $T_-(e_0) = 0$.

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