Let $(F,\|\cdot\|)$ be a semi-normed space, such that the kernel $E$ of $\|\cdot\|$ is finitely dimensional. Then $F/ E$ is a normed space and $E$ has a unique Hausdorff locally convex topology, since it is finitely-dimensional.

I want to generate a topology on $F$ from this data by identifying it with $F/E\oplus E$.

Is there any canonical choice of this identification?

Alternatively one can take an algebraic complement $H$ of $E$. Then $(H,\|\cdot\|)$ is a normed space, and taking an arbitrary norm on $E$ we get a norm on $F=E+H$. Off-course this norm depends on the choice of $H$, but perhaps all such norms are equivalent.

Does the topology on $F$ constructed as above depends on the choice of $H$?

  • $\begingroup$ Your claim that $E$ has a unique topology is not true: As a finite-dimensional space is has unique Hausdorff vector space topology. The relative topology of the semi-norm is the coarsest topology. $\endgroup$ – Jochen Aug 3 '17 at 9:05
  • $\begingroup$ @Jochen you are right, I have been sloppy. Now corrected $\endgroup$ – erz Aug 3 '17 at 17:59

I'm afraid there is no canonical choice.

Let $F$ be the set of $C^1$-smooth functions $f: \mathbb{R}\to\mathbb{R}$ for which the seminorm $$\|f\|_F = \sup_{x\in\mathbb{R}} |x^2f'(x)|$$ is finite. Then $E = \{f\in F:\|f\|=0\}$ is one-dimensional, the space of constant functions.

However, different choices of complements of $E$ yield inequivalent norms on $F$. For example, we could take $H=\{f\in F: f(1)=0 \}$, leading to the norm $$\|f\|_1 = \|F\|_F + |f(1)|$$ Or take $H=\{f\in F: f(-1)=0 \}$, leading to the norm $$\|f\|_{2} = \|F\|_F + |f(-1)|$$ These two are not comparable, and do not yield the same topology on $F$. Indeed, the sequence of functions $$ f_n(x) = \begin{cases} 0,\quad &x\le 0, \\ n^2x^2,\quad &0\le x\le 1/n \\ 2-n^2(x-2/n)^2,\quad & 1/n \le x \le 2/n \\ 2 \quad & x\ge 2/n \end{cases} $$ converges to $0$ in the second norm but not the first. To check this, notice that $$ f_n'(x) = \begin{cases} 0,\quad &x\le 0, \\ 2n^2x,\quad &0\le x\le 1/n \\ -2n^2(x-2/n),\quad & 1/n \le x \le 2/n \\ 0 \quad & x\ge 2/n \end{cases} $$ hence $\|f_n\|_F\to 0$. Since $f_n(-1)=0$, we get $\|f_n\|_2 \to 0$. But $f_n(1) = 2$ for large $n$, so $\|f_n\|_1\to 2$.

The underlying problem is that to form a norm (or locally convex topology) one essentially needs some linear functionals that are not constant on $E$, i.e., not continuous with respect to $F$-seminorm. And there is no canonical choice of discontinuous linear functionals on a normed space... as Tolstoy wrote, every discontinuous linear functional is discontinuous in its own way.


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