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From my Professor's Lectures, a Hilbert Space is defined as an inner product space $\bigoplus$ complete. In the same lecture, 'complete' here means, a system $\{\varphi_j\}_{j=1}^\infty \subset V$ is complete if $\nexists g \in V\setminus\{0\} $ such that $g$ is perpendicular to $\varphi_j, \forall j $.

It is also explicitly stated that 'complete' here doesn't mean 'complete space' where all Cauchy sequences are convergent. However, in the proof for a lemma that states:

  • An orthonormal system $\{\varphi_j\}_{j=1}^\infty \subset V$ that is complete is a basis in $V$,

We came to a point where we proved that $S_n(f) = \sum_{k=1}^{\infty} C_k\varphi_k$ is a Cauchy sequence, and then my Professor said that because we're in a Hilbert Space (complete), thus $S_n(f)$ is convergent. Is there something I'm missing here? Are both definition of 'complete' true in Hilbert Spaces?

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3 Answers 3

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There's a notion of completeness of a metric space and a notion of completeness of a basis, and they're not the same thing. Hilbert spaces are defined to be, in particular, complete metric spaces. Completeness of a basis means something different. It means what you said, and another way of stating it is that the span of the given system is dense in the Hilbert space. So, one is an intrinsic property of the space itself and the other is a property of particular subspaces.

For the record, I think a more clear way to state completeness of a system than you did is that if $\langle g,\varphi_j\rangle=0$ for all $j$, then $g=0$.

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  • $\begingroup$ Does one of the two imply the other? I mean, 1) A metric space is complete, then there is an orthonormal system of it; 2) If a metric space has an orthonormal system, then it is complete. Is any one statement above true? $\endgroup$
    – narip
    Nov 15, 2022 at 14:15
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    $\begingroup$ @narip orthonormality requires an inner product. If an inner product space is complete, then it does have an orthonormal basis. The other implication is false, see e.g. math.stackexchange.com/questions/201119/…, math.stackexchange.com/questions/2864274/… $\endgroup$
    – cmk
    Nov 16, 2022 at 0:27
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In the context of Hilbert spaces, the term 'complete' is used in both ways.

A metric space is complete if every Cauchy sequence in this space is convergent.

An orthonormal system $M:=\{\varphi_{\alpha}\}_{\alpha}$ in a Hilbert space $\mathcal{H}$ is complete if $M^{\perp}=\{0\}$, that is, if $\langle\varphi, \varphi_{\alpha}\rangle = 0$ for every $\alpha$ implies $\varphi = 0$

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The following may or may not be the exact definitions your professor is using. However, even when the terminology is slightly different, the versions in use have more or less the same nuances.

  • A Hilbert space is an inner product space such that the distance $\sqrt{\langle x-y,x-y\rangle}$ makes it a complete metric space, i.e. a metric space where sequences are convergent if and only if they are Cauchy. In that sense it is said to be complete.

  • A sequence $\{x_i\}_{i\in I}$ in an inner product space $(E,\langle\cdot,\cdot\rangle)$ is said to be orthogonal if $\langle x_i,x_j\rangle=0$ for all $i\ne j$.

  • A sequence $\{x_i\}_{i\in I}$ is said to be orthonormal if it is orthogonal and $\langle x_i,x_i\rangle=1$ for all $i\in I$.

  • A sequence in a Hilbert space $H$ is said to be a complete orthonormal system (or a Hilbert basis) if it is orthonormal and $\{x\in H\,:\, \forall i\in I, \langle x_i, x\rangle=0\}=\{0\}$. Equivalently, a Hilbert basis is an orthonormal sequence with dense linear span.

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