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Riesz lemma for Lebesgue differentiation theorem

  1. Mar 19, 2006 #1
    Please anyone out there, I need your assistance. I am trying to follow the Lebesgue differentiation theorem in Riesz & Nagy book "Lessons of Functional Analysis". It is the first theorem of the book and they use this lemma:

    "Lemma.- Let g(x) be a continuous function defined in the closed interval (a, b), and let E be the set of points x interior to this interval and such that there exists a W lying to the right of x with g(W) > g(x). Then the set E is either empty of an open set, i.e., it decomposes into a finite number or a denumerable infinity of open and disjoint intervals (a_k, b_k), and g(a_k) < (or equal) g(b_k) for all these intervals."

    In a previous thread I asked why the "openness" of a set implied that it decomposes into a finite number or a denumerable infinity of open and disjoint intervals. I have found an answer in Apostol 3.11 theorem.

    My problem is other now. I will copy exactly Riesz's proof word by word.

    "Proof.- To prove this lemma we first observe that the set E is open, since if W > x_0 and g(W) > g(x_0), then, in view of the continuity of the function g on x_0, the relations W > x, g(W) > g(x) remain valid when x varies in the neighborhood of the point x_0 (OK, no problem understanding this).

    This being true, let (a_k, b_k) be any one of the open intervals into which E decomposes; the point b_k will not belong to this set E (OK also).

    Let x be a point between a_k and b_k; we shall prove that g(x) < (or equal to) g(b_k); the result of the lemma will follow by letting x tend to a_k (OK).

    To prove that g(x) < (or equal to) g(b_k), let x_1 be the largest number between x and b_k for which g(x_1) > (or equal to) g(x); we have to show that x_1 coincides with b_k. If this were not true, the point W_1 which correspond to x_1 by the hypothesis of the lemma would lie beyond b_k and, since b_k does not belong to the set E, we would have g(x_1) < g(W_1) < (or equal to) g(b_k) < g(x_1), which yields a contradiction.

    Well I think I can understand anything of this proof save one thing:

    They say: if x_1 is by definition "the largest number between x and b_k for which g(x_1) > (or equal to) g(x)", then x_1 is naturally less or equal to b_k. Our work is to find a contradiction when we assume that x_1 is < b_k. So it is b_k.

    But my question is: why they can assume that there is a largest number between x and b_k for which g(x_1) > (or equal to) g(x) ? How do they know there is not an infinity of numbers which fullfill that condition? In such case there would not be a "largest" number. How do they supress that posibility (the infinity of numbers fullfilinf that condition)???. Does the continuity of the function g backs their assumption?

    May be this kind of question is too bizantine or maybe I am asking things out of fashion (I said this because my last questions have got not answer) but I suppose that the item must be in the scope of many of you, therefore I request your assistance.

    My PC is spoiled. I am in a cyber café and I will go out in some minutes. So, if you answer me, don't think I am uninterested if I don't reply promptly. Tomorrow at 8.00 am I will have a PC again.

    Sincerely, thanks.

    Castilla.
     
  2. jcsd
  3. Mar 19, 2006 #2

    Hurkyl

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    That's not true! After all, the interval [0, 1] is an infinite set, and it has a largest number.

    I don't want to spoil this as a good exercise, so I'll just try to restate what you need to show in a clearer manner:

    Lemma:
    (1) Let f be a continuous function on the interval [p, q].
    (2) Let [itex]S = \{ s \in [p, q] \, | \, f(s) \geq f(p) \}[/itex]
    Then S has a maximum element.

    (This could actually be stated with much weaker hypotheses)
     
    Last edited: Mar 19, 2006
  4. Mar 19, 2006 #3

    mathwonk

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    it seems rather obvious that if x < y and f(x) < f(y) and f is continuous then a<y and f(a) < f(y) for all a near x. doesn't that do it?
     
  5. Mar 20, 2006 #4
    Mathwonk:

    I think that your statement is mentioned (and understood) in the first paragraph of the proof, but I fail to see how that implies the existence of "the largest numer between x and b_k for which g(x_1) > (or equal to) g(x)".

    Hurkyl:

    With a sense of shame, I request some aditional hint.

    Thanks to both of you.

    Castilla
     
  6. Mar 20, 2006 #5

    mathwonk

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    i am just saying my statement immediately implies the statement claimed in the lemma, what else do you want?
     
  7. Mar 20, 2006 #6
    Hm, are you saying that Riesz' proof could and should finish in its first paragraph?
     
  8. Mar 20, 2006 #7

    Hurkyl

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    Well, if you can't prove something... then start with something easier! Instead of trying to prove that the S in my lemma has a maximum, can you prove it has a supremum?

    (don't read the next sentence unless you need another hint!)

    What ways might you show that the supremum of a set is actually a maximum?
     
  9. Mar 20, 2006 #8

    mathwonk

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    i am too impatient to read riesz's proof, but there are 3 statements to be proved.
    1) the set is opoen. i am saying only that this is clear, and i noticed that riesz also says so in the first sentence of his proof.

    2) any open set is a disjoint union of a countable collection of intervals. this i am assumed known, as riesz also seems to assume.

    3) if (c,d) is one of those intervals, the g(c) is not greater than g(d).

    the proof seems to be occupied with showing this, which appeareed obvious to me, but probably isn't, if riesz proves it.

    i will think about it for a few minutes.

    i have not read hurkyls hint but he may be trying to explain how to prove the asumed fact about the decomposition of an open set into a collection of disjoint intervals.
     
  10. Mar 20, 2006 #9
    S has a supremum because it is bounded by "q".

    Let s_1 be the supremum of S. So:
    1) s_1 > (or equal to) any s.
    2) For all delta > 0 there exists an s (of S) / s_1 - delta < s.

    3. Let's supose that s_1 does not belong to S; then f(s_1) < f(p).

    4. But the continuity of function f in s_1 implies that if I take s sufficiently near to s_1 (say s_1 - delta < s < s_1 + delta, then f(s) < f(a).

    2 and 4 are incompatible, because they imply some "s" that belong and do not belong to S. Then the hypothesis fail ad the supremum does belong to S. Then it is its maximum.

    Am I ok, Hurkyl and Mathwonk? Dont leave me with the doubt.
     
  11. Mar 20, 2006 #10

    mathwonk

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    3) seems "obvious" too after a 1 minute drunken reverie with eyes closed.

    let (c,d) be one of the open components of E. assume g(c) > g(d).

    Since d is not a point of our set, g(d) is at least as great as g(x) for all x to the right of d.

    hence if g(c) > g(d), then g(c) > g(x) for all x to the right of d also.

    g continuous, so g has a max on [c,d] but not at d, so at some z between c and d, maybe at z = c.

    but then g(z) is at least as great as g(c), and since c is not in E, g(z) is at least as great as g(x) for all x > c.


    Since there is no point to the right of z where g has a greater value than at z, z does not belong to the set E. Since z is in [c,d], hence z = c.


    now choose a small nbhd of c, say (c,c+e), on which g never varies by as much as (1/2)(g(c)-g(d)).

    then we have a problem. i.e. g has a max say at w, on [c+e,d]. but g(c+e) > g(d) so the max is not at d. but then g(w)>g(d), and so g(w) is as great as g(x) for all x > w. hence w is not in E, contradiction.

    ARRRGGHHH! i hate these picky analysis proofs. you have to be too smart to do this all day.


    now where is my error?
     
  12. Mar 20, 2006 #11

    Hurkyl

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    Upon first glance this looks right. Very good. But there is a shorter way!

    I think the essence of your proof is that any limit point of S is, in fact, an element of S. In other words, you're proving that S is closed! Can you find a much easier way to prove S is a closed set?
     
  13. Mar 20, 2006 #12

    Hurkyl

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    P.S. I don't mean to ignore mathwonk -- I'm just tired, and I want to respond to the part I already follow. :smile:
     
  14. Mar 21, 2006 #13
    Let s* be a number / f(s*) < f(p). As the function f is continuous in s*, in a neighborhood of s* there is a s** such that f(s**) < f(p). This is generalizable.
    Therefore the set {s belongs to closed (a, b) / f(s) < f(p)} is open,

    Therefore the set {s belongs to closed (a, b) / f(s) > (or equal to) f(p)} is closed,

    Therefore the supremum of this last set belongs to it.

    It is fine?
    (Excuse I dont use latex, some problem in machines here).
     
  15. Mar 21, 2006 #14

    Hurkyl

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    That works. The fastest way I know is this:

    [tex]S = f^{-1}( \,[f(p), +\infty)\, )[/tex]

    And since S is the inverse image of a closed set...
     
  16. Mar 21, 2006 #15
    I dont remember to have studied that the inverse image of a closed set is also a closed set, but anyway I have filled the gap in my understanding of Riesz Lemma.

    Thanks Hurkyl and Mathwonk.
     
  17. Mar 21, 2006 #16

    Hurkyl

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    It follows directly from the fact the inverse image of an open set is an open set!
     
  18. Mar 21, 2006 #17
    A good info, Hurkyl. Thanks again.
     
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