Linear Spaces Normed Spaces Hilbert Spaces Part2.wmv

A Walk Through of | Linear Spaces | Vector Spaces | Metric Spaces | Normed Linear Spaces | Inner Product Spaces | Euclidean Spaces | Function Spaces | Functionals | Hilbert Spaces.

Category:

Education

Tags:

• Linear
• Spaces
• Vector
• Metric
• Normed
• Inner
• Product
• Hilbert
• Functionals

[Mathview]

Ah...great question. A convex functional p(x) is a function that maps elements of a Linear Space to real numbers and obeys three properties (1) p(x) >= 0 for all x in L. (2) p(a*x)=a*p(x) for scalars a. (3) p(x+y) <= p(x) + p(y). The third property is called sub-additive. So convex functionals are sub-additive functions that also obey (1) and (2).

[meme2342]

Great lectures. This and the preceding one have taught me more than the last week trying to understand this stuff via google! One question please: is a convex function the same as a subadditive function?

[Mathview]

@meme2342 In property (2) of linear functionals, the scalar a must be greater or equal , >= to zero.



[Mathview]

When we speak of Hilbert Space here, it is as defined in the video. One of many references gives this Definition: "Hilbert Space is a Euclidean space which is complete, separable, and infinite dimensional." We use this definition in our discussion. see Kolmogorov p. 155. If the term is used as defined it is consistent and OK.

[Mathview]

In some references, the term Hilbert Space is reserved for strictly infinite dimensional, complete, separable, Euclidean spaces. Other references extend the term to include finite dimensional linear spaces having inner product, norm, closure under limits of Cauchy sequences. I believe this is the source of confusion here. To avoid confusion, ALWAYS check the definition. Usage varies. Perhaps we can discuss fine points of these distinctions and their rational in future videos.

[kalebmckale]

Furthermore, Two Hilbert spaces H1 and H2 are isometrically isomorphic if and only if they have the same dimension.

[Mathview]

@kalebmckale This is not a problem if we define a Hilbert Space to be a complete, separable, infinite dimensional Euclidean Space. In this case, the infinite dimension is obtained by the limit n -> infinity of an n-dimensional Euclidean Space. I believe, the historical origin of the Hilbert space was as an infinite dimensional function space, the square integrable functions on an interval. Not sure why Hilbert would get the credit for finite dimensional spaces well known before Hilbert.

[Mathview]

@kalebmckale I believe the term Hilbert space is defined differently by some authors. Perhaps that's a source of confusion. We are using the definition of Kolmogorov and Fomin, which is standard in many references. The classic example is the L2 integrable function space of quantum mechanical wave functions, "overlap integral" inner product. An infinite dimensional basis is needed, such as the Fourier components. We use Euclidean Space to mean a Linear Space with Inner Product. Any dimension.

[Mathview]

@kalebmckale Not to belabor the point, but another reference is Dennery & Krzywicki Mathematics for Physicists Dover (1995) p. 197. Euclidean n-spaces are often defined as n dimensional spaces obtained from the n-fold Cartesian product of the real number field, with inner product defined in the usual way. So I don't see any problem as long as we are consistent. Does that answer your question?

[kalebmckale]

A Hilbert space is NOT a Euclidean space. Rather, a Euclidean space is an example of a Hilbert space. A Hilbert space is an inner product space which is complete with respect to the metric induced by the inner product. A Hilbert spaces does not have to be infinite dimensional. For example, Euclidean spaces are finite dimensional.