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From: sto on 3 Jun 2010 01:06
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.

Thanks,
-sto
From: José Carlos Santos on 3 Jun 2010 04:30
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.

Best regards,

Jose Carlos Santos
From: David C. Ullrich on 3 Jun 2010 10:15
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...

>Best regards,
>
>Jose Carlos Santos

From: sto on 3 Jun 2010 10:39
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].

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

From: José Carlos Santos on 3 Jun 2010 11:07
On 03-06-2010 15:15, David C. Ullrich 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.

Of course! I always do that and I don't know why I didn't do it here. :-(

> 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 OP has already guessed the correct answer.

Best regards,

Jose Carlos Santos


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