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From: David C. Ullrich on 4 Jun 2010 08:11
On Thu, 03 Jun 2010 10:39:43 -0400, sto <sto(a)address.invalid> wrote:

>David C. Ullrich wrote:
>> On Thu, 03 Jun 2010 09:30:40 +0100, Jos� Carlos Santos
>> <jcsantos(a)fc.up.pt> wrote:
>>
>>> On 03-06-2010 6:06, sto wrote:
>>>
>>>> Can anyone give a rigorous proof that the restriction of any absolutely
>>>> continuous (real valued) function to a bounded interval is of bounded
>>>> variation? I can't. The two books I have looked in state this is
>>>> "obvious", but give no proofs. I am trying to construct a proof from
>>>> basically just the definition of absolute continuity (and function
>>>> variations), without having to utilize measures or Radon-Nikodym.
>>> Let _f_ be an absolutely continuous function from [a,b] into R.
>>>
>>> Given e > 0, there is a d > 0 such that
>>>
>>> sum_{1 <= k <= n}|f(b_k) - f(a_k)| < e
>>>
>>> whenever {(a_k,b_k) | 1 <= k <= n} is a set of pairwise disjoint
>>> intervals of [a,b] such that the sum of their lengths is smaller
>>> than _d_. So, the variation of _f_ on any interval of length smaller
>>> than or equal to _d_ is smaller than _e_. Now, express [a,b] as the
>>> union of N such intervals. Then the variation of f (on [a,b]) will be
>>> smaller than N*e.
>>
>> Precisely. I think it would be a better idea to say "take e = 1"
>> instead of "Given e > 0", just to clarify what matters and what
>> doesn't.
>>
>> And, if we're starting from nothing but the definitions,
>> there's a slightly subtle, "obvious", and easily proved
>> fact that you're using here without mentioning it.
>> It should be mentioned and proved - instead of saying
>> what the slight gap is I'll let you consider the question...
>
>The gap is that the variation function V of f is an additive set
>function: V[a,b] = V[a,x] + V[x,b].

Precisely. This is quite easy to prove, but I don't think it
counts as _literally_ trivial, since it does require an idea -
a person _could_ know the definitions but not see how
to prove it.

(Hint for anyone in that situation: The inequality
V[a,b] >= V[a,x] + V[x,b] is literally trivial from
the definitions. For the other inequality, given
a partition of [a,b], _add_ the point x as another
endpont to get a partition of [a,x] and a partition
of [x,b]...)

>Thanks,
>-sto
>>
>>> Best regards,
>>>
>>> Jose Carlos Santos
>>

From: sto on 4 Jun 2010 10:25
David C. Ullrich wrote:
> On Thu, 03 Jun 2010 10:39:43 -0400, sto <sto(a)address.invalid> wrote:
>
>> David C. Ullrich wrote:
>>> On Thu, 03 Jun 2010 09:30:40 +0100, José Carlos Santos
>>> <jcsantos(a)fc.up.pt> wrote:
>>>
>>>> On 03-06-2010 6:06, sto wrote:
>>>>
>>>>> Can anyone give a rigorous proof that the restriction of any absolutely
>>>>> continuous (real valued) function to a bounded interval is of bounded
>>>>> variation? I can't. The two books I have looked in state this is
>>>>> "obvious", but give no proofs. I am trying to construct a proof from
>>>>> basically just the definition of absolute continuity (and function
>>>>> variations), without having to utilize measures or Radon-Nikodym.
>>>> Let _f_ be an absolutely continuous function from [a,b] into R.
>>>>
>>>> Given e > 0, there is a d > 0 such that
>>>>
>>>> sum_{1 <= k <= n}|f(b_k) - f(a_k)| < e
>>>>
>>>> whenever {(a_k,b_k) | 1 <= k <= n} is a set of pairwise disjoint
>>>> intervals of [a,b] such that the sum of their lengths is smaller
>>>> than _d_. So, the variation of _f_ on any interval of length smaller
>>>> than or equal to _d_ is smaller than _e_. Now, express [a,b] as the
>>>> union of N such intervals. Then the variation of f (on [a,b]) will be
>>>> smaller than N*e.
>>> Precisely. I think it would be a better idea to say "take e = 1"
>>> instead of "Given e > 0", just to clarify what matters and what
>>> doesn't.
>>>
>>> And, if we're starting from nothing but the definitions,
>>> there's a slightly subtle, "obvious", and easily proved
>>> fact that you're using here without mentioning it.
>>> It should be mentioned and proved - instead of saying
>>> what the slight gap is I'll let you consider the question...
>> The gap is that the variation function V of f is an additive set
>> function: V[a,b] = V[a,x] + V[x,b].
>
> Precisely. This is quite easy to prove, but I don't think it
> counts as _literally_ trivial, since it does require an idea -
> a person _could_ know the definitions but not see how
> to prove it.
>
> (Hint for anyone in that situation: The inequality
> V[a,b] >= V[a,x] + V[x,b] is literally trivial from
> the definitions. For the other inequality, given
> a partition of [a,b], _add_ the point x as another
> endpont to get a partition of [a,x] and a partition
> of [x,b]...)

I did see this proof in a book before attempting to prove my theorem,
and I thought it might provide a mechanism for demonstrating bounded
variation but for some strange reason I just could not get all the
epsilons and deltas to work out. The key part of the proof "the
variation of f (on [a,b]) will be smaller than N*e", just didn't occur
to me. For some reason I often understand the complicated theorems but
get stuck on the minutiae.


>
>> Thanks,
>> -sto
>>>> Best regards,
>>>>
>>>> Jose Carlos Santos
>

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