Completion of a seminormed space

Question

Let $E$ be a normed space. A completion for $E$ is a pair $(\bar{E},\varphi)$ consisting of a Banach space $\bar{E}$ and a continuous linear map $\varphi:E\to \bar{E}$ that is injective, preserves norms, and such that $\varphi(E)$ is dense in $\bar{E}$. In his book Lang proves the existence of such a completion.

Now suppose $E$ is only a seminormed space. Lang defines the completion of $E$ as the completion of the quotient space $E/E_0$, where $E_0$ is the subspace of all $x\in E$ with $|x|=0$. The quotient space $E/E_0$ is a normed space with norm defined by $|x+E_0|=|x|$.

But when applying the completion proof to the seminormed space $E$ it looks like the pair $(\bar{E},\varphi)$ we obtain consists of a complete normed (not seminormed) space $\bar{E}$ and a continuous linear map $\varphi:E\to \bar{E}$ that preserves norms, and such that $\varphi(E)$ is dense in $\bar{E}$. However $\varphi$ fails to be injective.

Can I produce a continuous linear bijection between these two completions that preserves norms? (The two completions being the completion of the quotient space $E/E_0$ and the completion of the seminormed space $E$ described in the last paragraph)

Answer 1

It depends on what you mean by the completion of a seminormed space, but I guess that you want to take the space $\tilde E$ of all Cauchy-sequence modulo the subspace of al null sequences, and as the canonical map $\varphi:E\to \tilde E$ the equivalence class of the constant sequence. But this map is injective if and only if $E$ is properly normed, the kernel of $\varphi$ is in fact the kernel of the norm.

And yes, then $\tilde E$ will be isomorphic to the completion of $E/E_0$.

It may help to think about the role (or the universal property) of the completion in the theory (or category) of seminormed spaces: It should be a complete space $\tilde E$ together with a continuous linear $\varphi:E\to\tilde E$ such that, for every complete $F$ and every $T:E\to F$ there is a unique $\tilde T:\tilde E\to F$ with $T=\tilde T\circ\varphi$. For normed spaces, the required uniqueness implies injectivity of $\varphi$, but for semi-normed spaces it does not.

Answer 2

Define the completion of a seminormed space as a pair $(\bar{E},\varphi)$ consisting of a Banach space $\bar{E}$ and a continuous linear map $\varphi:E\to \bar{E}$ that preserves norms and such that $\varphi(E)$ is dense in $\bar{E}$. (the only difference with the completion of a normed space is that we don't require injectivity.)

The existence of such a completion for $E$ can be proved in the same way as for normed spaces by considering Cauchy sequences modulo null sequences and the canonical map $\varphi:E\to\bar{E}$ returning the equivalence class of the constant sequence.

Just like normed spaces, there is also a uniqueness property for the completion of a seminormed space :

Let $(F,\psi)$ be another completion for the seminormed space $E$. Then there exist a unique continuous linear bijection $\lambda:\bar{E}\to F$ that preserves norms and such that $\psi=\lambda\circ \varphi$.

The proof of this fact is the same as for normed spaces. Now we can prove what we want:

Let $\overline{E/E_0}$ denote the completion of the normed quotient space $E/E_0$. We need to show that $\bar{E}$ and $\overline{E/E_0}$ are isomorphic. Let $f:E\to E/E_0$ be defined by $f(x)=x+E_0$. We check that $f$ is continuous, linear, surjective and preserves norms. ($f$ might not be injective since $E$ is only seminormed).

Then $\psi\circ f:E\to \overline{E/E_0}$ is continuous, linear, preserves norms, and $(\psi\circ f) (E)=\psi(E/E_0)$ is dense in $\overline{E/E_0}$. It follows that the pair $(\overline{E/E_0},\psi\circ f)$ is a completion for $E$. Hence by the uniqueness property there exist a unique continuous linear bijection $\lambda:\bar{E}\to \overline{E/E_0}$ that preserves norms and such that $\psi\circ f=\lambda\circ \varphi$. This $\lambda$ is the required isomorphism.