Separable Hilbert spaces - L03 - Frederic Schuller

[@rifathkhan9995]

I think in his point 2 of definition of shauder basis, he must require a “unique” sequence {\psi^i} in complex numbers. This would resolve dominic’s objection. Let me know if anyone disagrees with me.

[@sardanapale2302]

You are right : the unicity requirement resolves it. The problem with his definition is that he wants to allow for uncountable Schauder(?) basis (because later we will need those to study unbounded operators) but... Schauder basis, as far as I have checked, are assumed countable. Its the Hamel basis that ALWAYS are uncountable... Professor Schuller is not very confortable with the subject hehe... and this not a critique, this kind of stuff can be very technical at times.

[@jimnewton4534]

Dr Schuller mentions several times that these lectures accompany another set of lectures from the physics point of view. Are those lectures available?

[@millerfour2071]

9:27 (need squares of the norms in the parallelogram/polarization identity), 34:47, 42:55, 1:04:11

[@rifathkhan9995]

I agree with the first mistake. What are the other 3 mistakes you are pointing out?

[@millerfour2071]

@@rifathkhan9995 They are not mistakes, just timestamps for review :)

[@alexlittle8115]

Unitary maps are automatically bounded because || U psi ||^2 = || psi ||^2 and hence the norm is 1.

[@kaaraasindhi1751]

He is love ❤️

[@jimnewton4534]

Another question about Schauder basis. Schuller stated condition ii that any element of the Hilbert space can be expressed as an infinite sum of elements of the basis, each scaled by a complex number. However, it is not clear whether the converse is true (as it was in the finite case). I.e., given any sequence of complex numbers such that this sequence converges (say to x) is x necessarily in H? This question is important in control theory. I.e., I can take a sequence of continuous functions, scale them by carefully chosen coefficients, and the sequence converges to a discontinuous function. A possibility which is often/sometimes glossed over when some presentations of control theory.

[@ccreutzig]

The partial sums form a sequence of vectors in the space, being finite sums. For those choices of coefficients where these sequences are Cauchy, the series converges to an element in the vector space.

[@asaflevif]

I want to be someday like Dominic!

[@TheNonHiddenSingularity]

at 52:35 when he is explaining the orthonormality, he dies = 1 if e1 = e2 but they are 2 different elements of S and in S, e1 and e2 are requested to be linerarly independant, right ?

[@ANSIcode]

The notion of "structure preserving map" at 1:07 is used in a way that differs from common practice. A structure preserving map is a morphism in the respective category. This means: For sets it's just general functions, for groups it's group homomorphisms, for vector spaces (without additional structure) it's linear maps, etc. Here the term refers to the respective isomorphisms of the categories, rather than the morphisms.

[@tofu-munchingCoalition.ofChaos]

You mean 1:07:00

[@jelmar35]

At 54:23 Schuller talks about his relativist training. Does anyone know what he means by that?

[@alexleviyev]

Physicists primarily learn about how to manipulate tensors when they are taught general relativity, and prefer manipulating tensors by working with their components in some basis instead of the tensor itself. The "upstairs" vs "downstairs" position of an index thus has very significant meaning in terms of the object to which the component corresponds. He's just making a reference that his previous training in relativity makes him acutely aware which objects are which depending on its index placement.

[@MrPpppp55555]

Is there any place where we can find the Problem sets ?

[@teretx566]

at 10:00 the parallelogram law is wrong, as already noticed. But the polarizazion identity is wrong as well... No idea of what is written on the problem sheets ... Also, Obi comments on the Shauder basis are correct: Shauder bases are ordered bases.

[@thephysicistcuber175]

48:00 I think the word you're looking for is "minimal".

[@thomassinha5301]

Is the norm equation written at 5:33 using the complex square root? Or is the inner product of an element of the dual space w itself guaranteed to be real?

[@barneylovesyou8878]

Inner product is a map to ℝ

[@jimnewton4534]

Perhaps this question is answered later, but I'll ask it anyway. at about minute 30:00 he starts talking about the Schauder basis, being a set of factors and a sequence of complex numbers such that the sum of the corresponding scaled vectors converges to a chosen element. However, if the sequence is not absolutely summable, then you might arrive at different sums depending on the order. Is this an issue? can I sum in different ways and get different results? if not, why not? where is the restriction? I don't see it in the definition anywhere.

[@USAboysp]

The assumption is that the series converges to an element of the Hilbert space this in fact implies that the series is absolutely convergent, this easily follows from the completeness of the space and use of the triangle inequality of the norm.

[@obi6822]

Jim Newton: You are correct. That is why Schauder bases are always considered ordered. The different summation result that originates from changing the order of summation has the correspondingly changed coordinates. So, yes, you can sum in different ways to get a different result, but still any point in the space H has uniquely defined coordinates with respect to the basis. Basis with the property you mention are called "unconditional Schauder basis". Look for Schauder basis in wikipedia

[@obi6822]

Sean Prudhoe: No. What you state is false. The converse of what you state is true though and follows as you say.

[@aeroscience9834]

It seems like in your definition of Schauder basis, you could have countable and uncountable basis for the same vector space

[@aeroscience9834]

44:10 I see you figured that

[@OhadAsor]

wrt axiom 3 of schauder basis, simply take the intersection of all such S, then you got the minimal one

[@FaridSalazar]

This will not suffice; in general, the intersection would be empty. You may have two disjoint Schauder basis.

[@LaureanoLuna]

'The least such set' would be right if by 'least' we refer to the set-inclusion ordering. In other words: no proper subset of a Schauder basis is a Schauder basis.

[@bonsairobo]

I've looked around for other definitions of Schauder basis, and all of them require S to be a countable sequence such that each vector can be written uniquely in that sequence. You might be able formulate another definition where S is uncountable, but it seems unnecessary.

[@numspacsym]

I guess a Hilbert space H can have many Schauder bases but each will be minimal in the power set of H partially ordered by set inclusion.

[@BruhaspathiRoark]

awesome lecture

[@tulliusagrippa5752]

If a norm can be constructed from an inner product and conversely, how is a Banach space different from an Hilbert space?

[@Andre-fm9zx]

not every norm satisfies the parallelogram identity

[@tulliusagrippa5752]

@@Andre-fm9zx So are there norms that cannot be constructed from an inner product? Did the lecturer misspeak? Examples or references would be useful.

[@Andre-fm9zx]

@@tulliusagrippa5752 It is the case that every inner product induces a norm, but not viceversa. Not every norm can induce an inner product. Let's take a look into the l1 norm over a finite vector space (this is just the sum of absolute values of the components of the vector). It's easy to see that this is a norm. It will be 0 if and only if every component is zero and this corresponds to the zero vector. It also satisfies the triangle inequality (this is a property of the absolute value). Now, this norm does not satisfy the parallelogram identity (check yourself). Which means that you cannot construct an inner product from it and viceversa

[@tulliusagrippa5752]

@@Andre-fm9zx. Thank you for taking the trouble to respond. It is greatly appreciated. I must have misunderstood Dr Schuller.

[@brydust]

If we start with a Hilbert space then the argument in this video says that a Schuader basis will in general be smaller than a Hamel basis for the same space because we can use infinite sums. What if we start the other way around? If we fix a basis's size then in general it appears the Schuader induced Hilbert space will be larger because we can use infinite sums (i.e. SchuaderSpan(Basis) > HamelSpan(Basis) ). But I suspect that this last line of reasoning is fallacious because then we cannot, in general, even generate a Hilbert space from HamelSpan(Basis) precisely because we do not have the infinite sums... and therefore, we need to then increase the Hamel basis to even begin to make the space Hilbert

[@FaridSalazar]

In general every vector space (regardless if it is Hilbert or not) has a Hamel basis. This can be proven using Zorn's lemma. However, this basis can be humongous. For a Hilbert space, the Schauder basis is a "relatively small" basis that incorporates the topological data to "span" all the space with fewer elements.

[@FrivolousMatter]

So there is only one Separable Hilbert space? I'm confused because the title says "spaces".

[@JacobGaiter]

There are many different separable Hilbert spaces. They are all isometrically isomorphic to l^2, but not in any meaningful way.

[@FrivolousMatter]

@@JacobGaiter Thanks.

[@nicolasbanks7871]

@@JacobGaiter Not sure what you mean by "not in any meaningful way". They are isometrically isomorphic to each other, so they have the same structure as Hilbert spaces. However, it is true that in analysis, one is less often concerned with identifying objects up to equivalence (in this case, isometric isomorphism), but instead would like to work with the particular elements of the Hilbert space of interest, so this identification is not often as useful. This is in contrasts to algebraists, who are usually quite happy to treat isomorphic objects interchangeably.

[@JacobGaiter]

@@nicolasbanks7871 I did not make it clear that I was taking an analytical view of them. Though, from a physics view point, finding the eigenvalues/eigenfunctioms of a Hamiltonian does give one a meaningful isomorphism from the position space representation of a quantum mechanical system into the energy eigenbasis representation which is essentially l^2. I was somewhat ignorant to this view point until relatively recently.

[@jimnewton4534]

at 46m00 Schuller out of the blue claims something about uniqueness which wasn't clear to me. Is it true that each element of the Hilbert space is uniquely representable as an "infinite linear combination" of a subset of the basis. That can't be true. Certainly there are many different sequences converging to any given element of H. right? In a finite dimensional space, every element is uniquely representable in terms of the basis, but that seems NOT TRUE for the infinite dimensional space, especially if the span of the basis is dense in H.

[@jimnewton4534]

I see in the lecture notes (drive.google.com/file/d/1I7rIH7Rtm0cCKVuLNeWfFMdKurX123x5/view) Schuller has clarified this point by claiming that an element must be expressible as a unique linear combination of the basis elements. And when this is not possible, then a Schauder basis simply does not exist for the space.

[@rockingrekhta361]

Sir at 7:49... Is parallelogram equality is true?

[@sreejithanair3026]

The identity he wrote down is wrong. There should be squares on all the norms. The same goes for the polarisation relation.

[@antonsupalov3855]

which semester is that?

[@aeroscience9834]

1:12:15 unitary maps are automatically bounded

[@hassaannaeem4374]

series are serious haha.

[@jimnewton4534]

So every Hilbert is unitarily isomorphic to the set of square summable sequences. How does this relate to the question of which functions are equal to their Fourier series?

[@ULRecordings]

Instead of looking at l^2(N), check out the space of periodic L^2 functions. This lecture only covered inf. dim. “seperable” Hilbert spaces. You want to look into square integrable functions. Every L^2-function can be expanded as a Fourier series. Every Hilbert space admits an orthonormal basis, and each vector in the Hilbert space can be expanded as a series in terms of this orthonormal basis.

[@jimnewton4534]

@@ULRecordings yes but isn't L-2 only a subset of functions which are equal to their Fourier series?

[@ULRecordings]

​@@jimnewton4534 Yes true. If you look up the condition for a function to write in terms of Fourier series, the function must be absolute integrable over some compact interval [-m, m], so it's classes are in L^1([-m,m]). It is a Banach space though, because it's a complete normed vector space.

[@jimnewton4534]

@@ULRecordings I think even that is not necessary, but only sufficient. The obvious example, the cos function on the real line, which is not a compact interval.

[@michielsnoeken5596]

Why is the norm of an element of a seperable Hilbert Space finite?

[@JacobGaiter]

By definition, a Hilbert space is a complete normed inner product space with a compatible norm induced by its inner product. This assumes that any element of the space has a well defined norm, i.e. has a finite norm.

[@michielsnoeken5596]

@@JacobGaiter Thankyou lots!

[@jimnewton4534]

I'd like an example of a Hilbert space which is not separable.

[@rockingrekhta361]

Sir please give an example of separability in real life.... How can we see in our life....

[@Miyelsh]

Look into the Fourier Series. Every bounded function over some interval can be constructed as a countable series of sines and cosines.

[@sardanapale2302]

real numbers (rational are a dense subsequence). I think this is the simplest example.

[@sayantansaha9047]

smart women..... im triggered.. :3