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.

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    $\begingroup$ It is a 1-dimensional real Hilbert space,. $\endgroup$
    – user491874
    Dec 3, 2017 at 16:52
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    $\begingroup$ $\mathbb{R}^n$ 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. $\endgroup$
    – user123
    Dec 3, 2017 at 16:53

1 Answer 1


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.


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