# Can we have an isometric embedding of this metric space into an Hilbert space?

A metric space (from this Q&A), is defined below. I'd like to know if its possible to have an isometric embedding of this metric space into an hilbert space? As per Schoenberg theorem $-d^2(x,y)$ need to be conditionally positive definite. So I want to verify if $d(x,y)$ of this metric space satisfy this condition.

Consider $L^2(\mathbb{R}) \setminus \{0\}$ and impose the following equivalence class on this.

Equivalence class $\sim$

Let $f$ be a function and we define its equivalence class as $f\sim g$ iff $g(x) = kf_{\theta}(x+\tau)$, for some $\theta \in [-\pi,\pi], k \in \mathbb{R}\setminus \{0\}$,$\tau \in \mathbb{R}$ and $$f_{\theta}(x) = f(x) \cos(\theta) + f_h(x) \sin(\theta)$$ where $f_h(x)$ is the hilbert transform of $f(x)$.

Lets us denote this set together with this equivalence class as $R = (L^2(\mathbb{R}) \setminus \{0\})/ \sim$.

Now we define metric..

Given $f,g \in R$, let $$h(x) = \frac{\int \limits_{-\infty}^{\infty}f(t)g(t+x)\mathrm dt}{\sqrt{\int \limits_{-\infty}^{\infty}f(t)^2\mathrm dt \int \limits_{-\infty}^{\infty}g(t)^2\mathrm dt}}.$$

Now define $$H(x) = \sqrt{h(x)^2 + h_h(x)^2}$$ where $h_h(x)$ is the hilbert transform of $h(x)$.

and our similarity measure is given as $$s(f,g) = \sup_x H(x).$$

Define metric as $$d(f,g) = \arccos(s(f,g)).$$

(proof that $(R,d)$ is a metric space)...

Note we can move from this space to a projective Hilbert space with translations identified by making the below transition

Replace $f$ by $F(x)=f(x)+if_h(x)$. Thats it!

If we denote Willie's cosine metric on $M$ as $w$, then given any $f,g \in R$, we create objects $F,G \in M$ given as $F(x) = f(x) + if_h(x), G(x) = g(x) + ig_h(x)$.

Now it can be seen that $$d(f,g) = w(F,G)$$ Hence $(R,d)$ is a metric space.

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we assume sufficient regularity on $\mathcal{L}^2(\mathbb{R})$ for Hilbert transform to exist. – Rajesh Dachiraju Jan 20 '14 at 5:12