# Hilbert Space: infinite or finite? - All real inner product spaces are Hilbert spaces?

I am confused about Hilbert spaces. I want to understand them better.

If I understand correctly a Hilbert space is an inner product space that has either finite or infinite dimension over real or complex numbers.

The background to this question is that I am trying to teach myself the mathematical linear algebra background to Fourier Series and Fourier Transforms.

I have read that a Hilbert Space is a vector space in $$\mathbb R^\infty$$ - that it is an infinite dimensional space, but I have also seen elsewhere references to 'finite Hilbert Space'. This makes me think that perhaps a Hilbert space is naturally assumed to be infinite unless qualified otherwise.

Looking more closely it seems to me that

• all real inner product spaces are Hilbert spaces
• all complex inner product spaces are Hilbert spaces provided that they follow the criteria listed, for example, on the wikipedia page under definition so that for example $$=^*$$
• If I understand correctly the term Hilbert space is useful collective term for 'well behaved' inner product spaces which include sets of sine and cosine functions, euclidean spaces etc.

My question is.. Are there any fundamentals errors in this understanding of Hilbert Spaces?

Sorry I realize that question is a bit vague.

I tried to check to see if parts this question had been asked before, but did not find it. For example, the question Dimension of a Hilbert space does not seem to address any of these particular points.

• There are no fundamental errors in your understanding. You will have no trouble recognizing the Hilbert spaces you need in your study. Just pay particular attention to the way each author defines things. – Ethan Bolker Oct 27 '18 at 13:01
• I don't know if this is relevant, but the space of real sequences with finitely many non-zero elements is a real inner product space as a subspace of the Hilbert space $\ell^2(\mathbb N)$, but is not Hilbert (it is not complete) – Calvin Khor Oct 27 '18 at 13:01
• Hilbert spaces are defined as complete inner product spaces. An inner product space is over $\mathbb{R}$ or $\mathbb{C}$. The complete means that the space "has all its limits", in the sense that every Cauchy sequence converges. This always happens for finite-dimensional inner product spaces, but it does rule out some badly-behaved infinite dimensional ones. – Joppy Oct 27 '18 at 13:39

Everything you said is correct: A Hilbert space $$H$$ is a real or complex vector space, equipped with a scalar product. But you forgot an essential point: This $$H$$ must be complete as a metric space. This extra condition is automatically fulfilled when $$H$$ is finite dimensional, because in this case $$H$$ is (in the real case) isometric with our standard euclidean $${\mathbb R}^n$$.
If, however, the given $$H$$ is infinite dimensional then the completeness has to be proved. Consider, e.g., the space $$X$$ of continuous functions $$f:\>[0,1]\to{\mathbb R}$$ with scalar product $$\langle f,g\rangle:=\int_0^1 f(t)\,g(t)\>dt$$. This particular $$X$$ is not complete.
• What about if $H = \mathbb{Q}$ with the usual Euclidean distance? Then $H$ is finite-dimensional yet not complete as a metric space. – Dave Oct 21 '20 at 22:21
• @Dave: Your $H$ is not a Hilbert space. See the first line of my answer. – Christian Blatter Oct 22 '20 at 8:10