Surjection of norms

Question

Let $V$ be an infinite dimensional $\mathbb{C}$ (or $\mathbb{R}$) vector space. Suppose there exists two norms on $V$ such that \begin{equation*} \| \cdot\|_1 \leq \| \cdot \|_2. \end{equation*}

Is it true that there always exists a surjection between the two different completions of $V$? That is, does there always exist a continuous $\mathbb{C}$-linear surjection ($\mathbb{R}$-linear surjection) \begin{equation*} \overline{V}^{\|\cdot \|_2} \twoheadrightarrow \overline{V}^{\|\cdot \|_1} \, ? \end{equation*}

The particular example I have in mind is when $G$ is an infinite discrete countable group and $V = \mathbb{C}[G]$, the complex valued functions on $G$ with finite support. Then there is a surjection from the maxmial group $C^*$-algebra and the reduced $C^*$-algebra.

Answer 1

Yes. Because you consider the identity map $(V,\|\cdot\|_2)\longmapsto(V,\|\cdot\|_1)$. The inequality between the norms shows that it is bounded.

It is not immediately obvious to me that it is a surjection in the generality of your question. But it is in the concrete case you mention, because a C$^*$-homomorphism has closed image (when the domain is a C$^*$-algebra).

Answer 2

In general there is no such surjection (the C*-case is quite special in that respect).

Indeed, let $V=c_{00}$ be the vector space of finitely supported vectors. For $(\xi_n)$ in $V$ define $\|(\xi_n)\|_1 = \sup_n |\xi_n|$ and $\|(\xi_n)\|_2^p = \sum_{k=1}^\infty |\xi_k|^p$ for some fixed $p\in (1,\infty)$. Conspicuously, the respective completions are $c_0$ and $\ell_p$. Of course, $c_0$ is not a quotient of $\ell_p$ as the latter space is reflexive.