# Are the real numbers an example of a Hilbert space or is it rather the real number plane?

Reason I ask, I am studying lecture on Hilbert spaces and the real numbers are the first example of a Hilbert space. The the lecture talks about completeness and demonstrates the concept using a real number line.

Naturally I wonder is the real number line itself a Hilbert space. ? I would say not since I can't make vectors from a horizontal line. Unless we make the vertical component the 0 vector. If that is the case shouldn't the lecture say the real plane is a Hilbert space and NOT the real numbers are a Hilbert space? Cab someone explain ? Thank you.

• It is a 1-dimensional real Hilbert space,.
– user491874
Dec 3, 2017 at 16:52
• $\mathbb{R}^n$ can be seen as a Hilbert space for any $n \geq 1$. As long as you're in a vector space and you have some kind of inner product, and you have completeness, then you should be good. Dec 3, 2017 at 16:53

A Hilbert space is an inner product space that is also complete as a metric space (with respect to the norm induced by the inner product). The real numbers $$\mathbb{R}$$ are a one-dimensional vector space over themselves, and we can define an inner product on $$\mathbb{R}$$ by $$\langle x, y \rangle = xy.$$ This inner product induces the usual norm on $$\mathbb{R}$$, i.e. the absolute value, which we know is complete. Thus $$\mathbb{R}$$ is a Hilbert space.

Indeed, for any $$n\in\mathbb{N}$$, we can define an inner product on $$\mathbb{R}^n$$ (or $$\mathbb{C}^n$$) by $$\langle x, y \rangle = \sum_{j=1}^{n} x_j \overline{y}_j,$$ where $$x_j$$ and $$y_j$$ are the $$j$$-th components of $$x$$ and $$y$$, respectively (conjugation is required if the base field is complex, and is trivial if the base field is real). This inner product induces the usual Euclidean norm on $$\mathbb{R}^n$$ (or $$\mathbb{C}^n$$), which is again complete. Hence all of these spaces are (finite dimensional) Hilbert spaces.