Functional Analysis 25 | Hahn-Banach Theorem

Поділитися
Вставка
  • Опубліковано 17 лис 2024

КОМЕНТАРІ • 36

  • @brightsideofmaths
    @brightsideofmaths  4 роки тому +17

    Reupload: I corrected one mistake :)

  • @lkapo7047
    @lkapo7047 3 роки тому +11

    Sehr hilfreich. Vielen Dank!

  • @alvarojaramillo9285
    @alvarojaramillo9285 6 днів тому

    Thank you very much, excellent material.

  • @cc.quantr
    @cc.quantr 3 роки тому +5

    Functional analysis is also one of my favourites!

    • @tahernom9207
      @tahernom9207 2 роки тому +4

      but I find difficult to study it

    • @tobiemmanuel8306
      @tobiemmanuel8306 3 місяці тому

      Difficult for me

    • @wesleyrm
      @wesleyrm 2 місяці тому

      ​​@@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.

  • @chriswunder5420
    @chriswunder5420 3 роки тому +3

    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!

  • @roozbehr93
    @roozbehr93 3 роки тому +5

    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
    @wesleyrm 2 місяці тому

    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
      @brightsideofmaths  2 місяці тому

      The norm is not a linear functional :)

    • @wesleyrm
      @wesleyrm 2 місяці тому

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

  • @trondsaue7860
    @trondsaue7860 2 роки тому

    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.

  • @vanrltv
    @vanrltv 2 роки тому +3

    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
      @wesleyrm 2 місяці тому

      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
      @wesleyrm 2 місяці тому

      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}).

  • @nguyenvietdung7588
    @nguyenvietdung7588 3 роки тому +2

    brilliant

  • @mathiasbarreto9633
    @mathiasbarreto9633 3 роки тому +4

    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
      @brightsideofmaths  3 роки тому +4

      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
    @trondsaue7860 2 роки тому

    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
      @brightsideofmaths  2 роки тому +1

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

    • @trondsaue7860
      @trondsaue7860 2 роки тому

      @@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
      @brightsideofmaths  2 роки тому

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

    • @trondsaue7860
      @trondsaue7860 2 роки тому

      @@brightsideofmaths Thanks ! Also for your reactivity !

    • @trondsaue7860
      @trondsaue7860 2 роки тому +1

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

  • @kimanthony1667
    @kimanthony1667 Рік тому

    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
      @brightsideofmaths  Рік тому +2

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

    • @wesleyrm
      @wesleyrm Рік тому

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

  • @StratosFair
    @StratosFair 2 роки тому

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

  • @jaewooknam8765
    @jaewooknam8765 3 роки тому

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

    • @brightsideofmaths
      @brightsideofmaths  3 роки тому

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

    • @chriswunder5420
      @chriswunder5420 3 роки тому

      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
      @chriswunder5420 3 роки тому

      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
      @brightsideofmaths  3 роки тому +1

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