Is it possible extend a continuous seminorm on a dense subspace of a Banach space to complete space?

Question

Let $V$ be a infinite dimensional separable Banach space over $\mathbb{C}$

Let $U$ a subspace dense in $V$

Let $p : U \to \mathbb{R}$ be a continuous seminorm on $U$

I would like to know if is it possible extend $p$ to a continuos seminorm on $V$

Thanks.

Answer 1

Uniqueness is clear since any continuous function is determined by its action on a dense subspace. It remains to prove existence.

This can be done by mimicking the usual proof that a continuous linear map defined on dense subspace have a unique continuous extension. As this is an adaptation of a standard proof, I'll sketch the details here and leave details for you to fill in.

Indeed, there is only really one sensible definition of our extension. We define $\tilde{p}$ by $$\tilde{p}(x) = \lim_{n \to \infty} p(x_n)$$ where $x_n \to x$ and $x_n \in U$ for every $n$. To see that this is well defined we need to check that

1. whenever $x_n$ is a sequence in $U$ converging to $x$ in $V$, $\lim p(x_n)$ exists.
2. If $x_n, y_n$ are sequences in $U$ converging to $x$ in $V$, then $\lim p(x_n) = \lim p(y_n)$.

To show the first of these, you should attempt to show that if $x_n$ is a Cauchy sequence in $U$ then $p(x_n)$ is a Cauchy sequence in $\mathbb{R}$. For the second, one approach is to consider the sequence $z_n$ defined by $z_{2n} = x_n$ and $z_{2n+1} = y_n$. This converges to $x$ in $V$ and hence, by the first part, $p(z_n)$ converges to some $\alpha \in \mathbb{R}$. In particular, the subsequences $p(z_{2n}) = p(x_n)$ and $p(z_{2n+1}) = p(y_n)$ both converge to the same limit, $\alpha$.

Once you've done this, we have a well-defined extension $\tilde{p}$ of $p$ to $V$ (to see that $\tilde{p}$ extends $p$ consider the constant sequences in $U$). It remains to check that $\tilde{p}$ is a continuous seminorm. It is clear that $\tilde{p}$ is a seminorm, by passing to the limit in the relevant properties for $p$.

To prove continuity, suppose that $x^{(n)} \to x^{(\infty)}$ in $V$ and take $x_k^{(n)} \to x^{(n)}$ as $k \to \infty$, where $x_k^{(n)} \in U$ for $1 \leq n \leq \infty$. First show that for every $\varepsilon > 0$, $\| x_k^{(n)} - x_k^{(\infty)}\| < \varepsilon$ for large enough $n$ and $k$ (easy application of triangle inequality). Then \begin{align} |\tilde{p}(x^{(n)}) - \tilde{p}(x^{(\infty)})| \leq& |\tilde{p}(x^{(n)} - x^{(\infty)})| \\ \leq &|\tilde{p}(x^{(n)} - x_k^{(n)})| + |p(x_k^{(n)} - x_k^{(\infty)})| + |\tilde{p}(x_k^{(\infty)} - x^{(\infty)})|. \end{align} You can finish the proof by using continuity of $p$ and convergence of the various sequences to estimate the right hand side in the above inequality.