Functional Analysis 25 | Hahn-Banach Theorem

[@brightsideofmaths]

Reupload: I corrected one mistake :)

[@moroccanevil]

Thank you very much

[@lkapo7047]

Sehr hilfreich. Vielen Dank!

[@brightsideofmaths]

Gerne :)

[@cc.quantr]

Functional analysis is also one of my favourites!

[@tahernom9207]

but I find difficult to study it

[@tobiemmanuel8306]

Difficult for me

[@wesleyrm]

​​@@tobiemmanuel8306 Same, difficult for me. Couldn't yet get my degree because I took this course since I was told it was interesting lol. This time I am studying for real. The real problem was that I did not master the main concepts in Topology since I studied mostly Physics, Statistics and Informatics in university. Now I at least know enough of the basics of Topology to be able to advance in the subject.

[@vpambs1pt]

At 3:50 why is lambda assumed to be 1?

[@chriswunder5420]

Great Videos! I'm alwas enjoying the videos after working through theroems in literatur, and it really helps to remember them better as you explain them more playfully. I'd love you see more aboute sobolev spaces and/or integral transformations. Anyway great content!

[@alvarojaramillo9285]

Thank you very much, excellent material.

[@roozbehr93]

It would be beneficial if you make another video of Hahn-Banach Theorem, possibly with more examples maybe and a little explanation (getting into the proof). also how to find all dual norms of a sub space in R^3 maybe. like (x,y,0) with f(x) = a.x and a=(b,c,0). ( just an example ). besides, i think Dual spaces worth to be explained more in my opinion. Thanks in advance. i enjoy watching your Series ;)

[@wesleyrm]

3:00 -Also, you could have wrote the map u' from U to F as u'(u) = ||u||_X directly.- -That is, the norm is a linear functional itself.- Of course, it is always positive, unlike your function that maps u=λx to λ||x||_X, which can be negative and seems more useful. From then Hahn-Banach directly gives us norm preservation in the dual space, so we have x' with ||x'||_X' = ||u'||_U' = sup({|u'(u)|=||u||, for u in U with ||u||=1}) = 1. The rest of the proof is identical: since x lies in U, x'(x)=u'(x)=||x||_X.

[@brightsideofmaths]

The norm is not a linear functional :)

[@wesleyrm]

​​​ Yes, obviously haha. Sorry for wasting your time. And thanks for the response!

[@vanrltv]

I am a bit confused by all these different norms. I have a few questions if you would be so kind to answer. 1. In part (a), why is ||u'||_u' = 1 ? Where does this 1 come from? The same question for part (c) with sup||x'||_x' . 2. In (c), i am bit confused by notation here. What exactly is a difference between ||x'|| and ||x'||_x'. I feel like i should know this stuff from previous videos, but i cant seem to connect all the dots.

[@wesleyrm]

1. It comes from the fact that for u= λx, u'(u) = λ||x||, thus |u'(u)| = |λ| ||x|| = ||u||. Finally, ||u'||_U' = sup({|u'(u)|, ||u||=1}). But |u'(u)| = ||u||. Given that ||u|| has to be 1, ||u'||_U' = 1. You could also have noted that for u to have ||u||=1, λ can only be + or - 1/||x|| and went from there.

[@wesleyrm]

And I don't think you need results from previous videos. Your 2 questions relate to the same thing, the operator norm. Which is defined as ||f||_X' = sup({|f(x)|, ||x||=1}).

[@trondsaue7860]

In your final example you do not explain why you need a _closed_ subspace U. I believe it is needed to show that the quotient space X/U is a normed vector space.

[@mathiasbarreto9633]

I am a little bit confused in part b) of the applications. What exactly is x'(x_1) and x'(x_2) ? Using the property of the map x' in part a), aren't they simply their respective norms? If true won't this mean that different but equidistant vectors (with the same norm) give x'(x_2) - x'(x_1) = 0 although x_1 and x_2 are not the same vectors?

[@brightsideofmaths]

Part (b) tells you something about the dual space. If you have two different points x1, x2, you always find a functional with different values at both points. Regarding your question: We don't know what x'(x_1) or x'(x_2) is exactly. The property in (a) just works for the one vector x.

[@trondsaue7860]

A small question: In part d) you define the quotient space X/U as the equivalence class [x] for x in X. Does the same restriction as two lines above still hold, that is, x is not element of the closed subspace U ? I would say yes, since otherwise the definition becomes strange.

[@brightsideofmaths]

No, you don't need this restriction there. If x is an element of U, then [x] = U.

[@trondsaue7860]

@@brightsideofmaths I believe that at 9:08 you say that if you put in a point u into the equivalence class, then the quotient space is the zero vector...

[@brightsideofmaths]

@@trondsaue7860 Yes, U is the zero vector in the quotient space :)

[@trondsaue7860]

@@brightsideofmaths Thanks ! Also for your reactivity !

[@trondsaue7860]

@@brightsideofmaths ..since [x] + [u] = [x] for u in U, right ?

[@kimanthony1667]

In part (a), $x`(x) := ||x||_X$ seems non-linear function. This seems to go against the definition of dual space. If anyone knows, could you please explain? Thanks.

[@brightsideofmaths]

Yes, you missed the order of the choice of the variables. For each x we choose a linear x' with the given property.

[@wesleyrm]

@@brightsideofmaths Ah, good! You are right, x'(x) = ||x|| only for one point, x.

[@nguyenvietdung7588]

brilliant

[@jaewooknam8765]

Why do we need a *closed* subspace for (d)? Can you help me to figure it out?

[@brightsideofmaths]

In the proof: X/U needs to be a normed space with the given norm. Therefore, we need that U is closed.

[@chriswunder5420]

I think in the context of the Video in d) when y' is defined, we can conclude from a) that y'([x]) is ||[x]|| by choice of x, which is 0 only when x is in U, because the norm is defined as the distance of x to U and U is closed. So thats why y'(x) is not 0. But if course in order to define the normed factor space, U needs to be closed, otherwise ||[•]|| is only a seminorm. The infimum can still be zero even if x is not in U and thus [x] not the null-vector in X/U.

[@chriswunder5420]

Now that I think about it, doesn't it matter i we define the norm as infimum over u ||x+u|| or ||x-u||? U is a vector space so it probably doesn't Matter. But with -, I think of it more as a distance.

[@brightsideofmaths]

@@chriswunder5420 Good point. In order to understand the concept, x-u would be the better choice.

[@StratosFair]

Very nice, if you have the time I think some more applications and/or the proof of this theorem would be very helpful