- #1
Castilla
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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 anyone 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.
thanks.
Castilla.
"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 anyone 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.
thanks.
Castilla.