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A Hilbert space is, by definition, a complete inner product space. If $(V,|.|)$ is finite dimensional inner product space of dimension $n$ then it is (topologically) isomorphic to $\mathbb{R}^n$ which is of course complete. My instinct here is to say "and therefore, $V$ is also a Hilbert space". I'm not sure about this step however since completeness depends on the norm and the norm depends on the selected inner product (assuming the induced norm $||x|| = (x | x)^{1/2}$ is used). So, must we place some condition on the inner product or is the topological isomorphism between $\mathbb{R}^n$ enough to guarantee that V is Hilbert?

user1234
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ItsNotObvious
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    $(V,\langle\cdot,\cdot\rangle)$ is a Hilbert space if it's complete for the norm $\lVert v\rVert :=\sqrt{\langle x,x\rangle}$. If $V$ is finite dimensional, then we can find $a,b>0$ such that $a\lVert \cdot\rVert_2\leq \lVert \cdot\rVert\leq b\lVert \cdot\rVert_2$ (where $\lVert \cdot\rVert_2$ is the euclidian norm). It shows that $(V,\lVert \cdot\rVert)$ is complete, without any condition on the inner product. – Davide Giraudo Jun 01 '11 at 21:57

4 Answers4

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An $n$-dimensional real inner product space is not just topologically isomorphic to $\mathbb{R}^n$ (this would indeed not imply completeness, since it is not a topological property). It is isometric to it. Indeed, let $V$ be such a space and let $v_1,\dots,v_n$ be an orthonormal basis in $V$. Let $e_1,\dots,e_n$ be the standard basis in $\mathbb{R}^n$. Define a linear transformation $T:V \to \mathbb{R}^n$ by declaring $T(v_i)=e_i$ for $i=1,\dots,n$ and extending linearly. Then $T$ is clearly a bijection and $\left\langle T\left(v_{i}\right),T\left(v_{j}\right)\right\rangle =\left\langle e_{i},e_{j}\right\rangle =\delta_{ij}=\left(v_{i},v_{j}\right)$, so $T$ respects the inner product structures. That is, $T$ is an isomorphism of inner product spaces and therefore any property of $V$ (as an inner product space) is implied by the corresponding property in $\mathbb{R}^n$. In particular it is Hilbert.

Mark
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    To nitpick a bit: Every topological vector space has a uniform structure (since it has an underlying group structure). The Hausdorff topological structure on $\mathbb{R}^n$ is unique and so is its uniform structure. With respect to that structure $\mathbb{R}^n$ is complete. This has little to do with isometry. But, as pointed out in other answers, with the existence of a compact neighborhood of $0$. – t.b. Jun 02 '11 at 05:29
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The step is incomplete, but the gap isn't hard to patch. Let $V$ be a finite-dimensional real vector space equipped with a Hausdorff topology such that addition and scalar multiplication are continuous, and let $e_1, ... e_n$ a basis of it. Consider the function

$$f : \mathbb{R}^n \ni (x_1, ... x_n) \mapsto x_1 e_1 + ... + x_n e_n \in V.$$

By assumption, $f$ is continuous. The image of the closed ball $B_r$ of radius $r$ about the origin is Hausdorff by assumption, so $f$ restricted to $B_r$ is a continuous bijection from a compact space to a Hausdorff space, hence a homeomorphism. Hence $f$ is a homeomorphism.

If $V$ is a finite-dimensional inner product space, then run the above argument with $e_1, ... e_n$ an orthonormal basis. Then $f$ is an isometry.

More generally, it's true that any two norms on a finite-dimensional real or complex vector space are equivalent (so if one is complete, they all are).

Qiaochu Yuan
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  • How are you defining 'equivalent'? Also why is the image of the closed ball $B_{r}$ about the origin Hausdorff? – Lucio D Jun 10 '14 at 21:46
  • @Lucio: the usual meaning (http://en.wikipedia.org/wiki/Norm_(mathematics)#Definition). Any subspace of a Hausdorff space is Hausdorff. – Qiaochu Yuan Jun 10 '14 at 21:57
  • Of course yes. If you have a chance could you check my MSE post 'Convergence of characteristic functions on hypercube' and let me know what you think. – Lucio D Jun 11 '14 at 14:35
  • @QuiaochuYuan I just want to confirm something regarding the continuity of the function $f$ that you define. Since we assume that vector addition $V \times V \rightarrow V$ and scalar multiplication $\mathbb{R} \times V \rightarrow V$ are continuous is $f$ continuous since by composing vector addition and scalar multiplication we get that $(\mathbb{R} \times V) \times (\mathbb{R} \times V) \times ... \times (\mathbb{R} \times V) \rightarrow V$ is continuous. If we then fix the basis as being the members taken from $V \times ... \times V$ then we reduce to the continuous function $f$? – Lucio D Jun 11 '14 at 19:54
  • @Lucio: yes, that's right. – Qiaochu Yuan Jun 12 '14 at 00:30
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    @QiaochuYuan How does the conclusion $f$ is a homeomorphism follow from your argument that $f$ restricted to $B_{r}$ is a continuous bijection froma compact space to a Hausdorff space is a homeomorphism? Also I just want to confirm that I have the right idea. Any finite-dimensional normed space, say $V$, is isomorphic(has a linear homeomorphism) with $\mathbb{R}^{n}$ for some $n$. But it's only when we endow $V$ with an inner product that we can show that inner product space $V$ is isometrically isomorphic with $\mathbb{R}^{n}$. Is that reasoning fine? Thanks for your time. –  Jun 15 '14 at 16:22
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    @John: hmm. It looks like I do need more details here. The point is that a subset of $\mathbb{R}^n$ is open iff its intersection with all $B_r$ is open, so $f$ restricted to $B_r$ being a homeomorphism for all $r$ implies that $f$ is open, and a continuous bijection is a homeomorphism iff it is open. The second thing is fine. – Qiaochu Yuan Jun 15 '14 at 17:01
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    @QiaochuYuan Thanks for your response. To show that $f$ is open can I do the following based on what you said: Take any open $U \subset \mathbb{R}^{n}$ then we have that $U = \cup_{r > 0}(B_{r} \cap U)$. It follows that $f(U) = f(\cup_{r > 0}(B_{r} \cap U)) = \cup_{r > 0}f(B_{r} \cap U)$. The result follows since $f$ is a open on $B_{r}$. Is this fine? –  Jun 15 '14 at 18:05
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    @John: yes, that's right. – Qiaochu Yuan Jun 15 '14 at 18:35
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    @QiaochuYuan Hi Qiaochu could I ask what argument exactly you are referring to when you say "then run the above argument with $e_{1},...,e_{n}$ an orthonormal basis"? I can't quite see how to show that $f$ is then an isometry. –  Jun 19 '14 at 20:10
  • @John: I mean consider the map $\mathbb{R}^n \ni (x_1, ..., x_n) \mapsto x_1 e_1 + ... + x_n e_n \in V$ where $V$ is an inner product space, $e_1, ..., e_n$ is an orthonormal basis of $V$, and $\mathbb{R}^n$ is given its standard inner product. Then this map is an isometry; this is a straightforward exercise. – Qiaochu Yuan Jun 19 '14 at 20:15
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As a matter of fact, the compactness of the unit sphere in a finite dimensional space can tell us that all norms on that space are equivalent. Therefore, in any norm (induced by inner product or otherwise), finite dimensional normed linear spaces are always complete.

ncmathsadist
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Well, how does proving that $V$ and $\mathbb{R}^n$ are topologically equivalent work? You conctruct an orthonormal basis and map it to $e_1,e_2,...,e_n$ right? Then you extend linearly and prove that this map is a continuous linear bijection. But it is even more: It's an Isometry, and thus preserves completeness, as is easily checked.

amueller
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