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Does continuous and increasing on $[0,1]$ and absolutely continuous on $[\varepsilon,1]$ imply absolutely continuous on $[0,1]$?

EDIT: I am trying to do this via the definition, but I am doing something wrong:

$\forall \epsilon >0, \exists \delta$ such that $\forall$ sequences $(a_n,b_n)$ if $\sum_{k=1}^{n}|b_n-a_n|<\delta =>\sum_{k=1}^{n}|f(b_n)-f(a_n)|<\epsilon$

Now, from continuity I have $|b_n-a_n|<\delta => |f(b_n)-f(a_n)|<\epsilon_1$ for some $\epsilon_1>0$.

and I can write the sum $\sum_{k=1}^{n}|f(b_n)-f(a_n)|<\epsilon_1n$

But this has to be wrong since continuity does not imply absolute continuity, I should be using monotonicity at some point in here. But I can't see where.

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  • $\begingroup$ "Absolutely continuous on $[\epsilon,1]$" for all $\epsilon>0$ one presumes. Why not just give it a try using the actual definition? (Yes, there are fancier ways, but if you can't get it with the definition then ... well you should!) $\endgroup$ Nov 23, 2015 at 2:10
  • $\begingroup$ @B.S.Thomson. Thanks, I edited the title. I'm working on doing this via the definition: $\forall \epsilon >0, \exists \delta$ such that $\forall$ sequences $(a_n,b_n)$ if $\sum_{k=1}^{n}|b_n-a_n|<\delta =>\sum_{k=1}^{n}|f(b_n)-f(a_n)|<\epsilon$ Now, from continuity I have $|b_n-a_n|<\delta => |f(b_n)-f(a_n)|<\epsilon_1$ for some $\epsilon_1>0$. and I can write the sum $\sum_{k=1}^{n}|f(b_n)-f(a_n)|<\epsilon_1n$ But this has to be wrong since continuity does not imply absolute continuity, I should be using monotonicity at some point in here. But I can't see where. $\endgroup$
    – Mike
    Nov 23, 2015 at 2:58
  • $\begingroup$ To use the definition for this function pick $\delta_1$ really small, small enough that $f(t)-f(0)$ is quite small for $0 \leq t < \delta_1$. Now on the interval $[δ_1/2,1]$ you have absolute continuity so you can use the definition there to pick a $δ_2$. Etc. {I know that sounds vague but if I am more specific it will do the proof for you. The idea is that close to the left endpoint exploit continuity; away from that endpoint exploit absolute continuity.] $\endgroup$ Nov 23, 2015 at 4:57
  • $\begingroup$ @B.S.Thomson I see what you mean! How does this look: Let for $(a_n,b_n)$ subintervals $\sum_{k=1}^{n}|b_n-a_n|<\delta$. Now, by assumption $f$ is AC on $[a_1,1]$. So $\forall \epsilon_c, \exists \delta_c$ s.t. $ \sum_{k=2}^{n}|f(b_n)-f(a_n)|<\epsilon_c$. By continuity, within the interval $[0,a1], \forall \epsilon_1, \exists \delta_1$ s.t. $f(b_1)-f(a_1)<\epsilon_1$. Now, choosing $\delta=\delta_1+\delta_c$ will force $ \sum_{k=2}^{n}|f(b_n)-f(a_n)|<\epsilon_1+\epsilon_c$. Donezo. Does this look alright? $\endgroup$
    – Mike
    Nov 23, 2015 at 6:15
  • $\begingroup$ @B.S.Thomson I only have one problem with what I wrote. Wouldn't I have to show that on the subinterval on the left endpoint, $[0,\delta_1/2]$, I have uniform continuity, not merely continuity, to satisfy absolute continuity everywhere? I'm thinking that somehow increasing and continuous on that interval would give me uniform continuity. Do they make the function monotonous just so it doesn't oscillate wildly? $\endgroup$
    – Mike
    Nov 23, 2015 at 10:08

1 Answer 1

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@Craig I'll answer here as the comments section is quite cramped.

"Do they make the function monotonous just so it doesn't oscillate wildly?" Exactly. Notice that if you are adding up $\sum_{i=1}^m |f(d_i)-f(c_i)|$ for nonoverlapping intervals in $[0,W]$ then $$\sum_{i=1}^m |f(d_i)-f(c_i)| =\sum_{i=1}^m [ f(d_i)-f(c_i)] \leq f(W)-f(0).$$ So all you need to control these for a monotonic increasing continuous function is to make sure that $f(W)-f(0)$ is small.

"Uniform continuity?" Well if $f$ is continuous on $[0,1]$ it is uniformly continuous there, if you feel you need it.


So now the idea of the proof. Start with (as always) "Let $\epsilon>0$.

First we set up the wall $W$. Choose $W>0$ so that $f(W)-f(0)< [small_1]$.

Now $f$ is absolutely continuous on $[W,1]$ (the right side of the wall) so we can select $\delta_r>0$ so that $$\sum_{i=1}^m |f(d_i)-f(c_i)| \leq [small_2] $$ whenever these intervals are nonoverlapping subintervals of $[W,1]$ with total length less than $\delta_r$.

Now $f$ is continuous at $W$ (the wall) so there is a $\delta_W$ with $$f(W+\delta_W) - f(W-\delta_W) < [small_3] .$$

Now consider any finite sequence of intervals $\{[c_i,d_i]\}$ of $[0,1]$ with total length less than $\delta_?$ [still to be determined as this isn't the actual proof just our heuristics to find a proof].

That sum has three parts: the pieces in $[0,W]$ on the left side of the wall (ii) the pieces in $[W,1]$ on the right side of the wall and (iii) maybe one solitary piece straddling the wall. Give an $\epsilon/3$ to each of these possibilities and we are done and can write up a tidy $\epsilon$, $\delta$ proof.

P.S. Don't leave the problem without a counterexample showing that the hypotheses that you used (and that were stated in the problem) cannot be dropped). This is proper mathematical etiquette: think of it as saying "please" and "thankyou" in social situations. Add "counterexample" to your solutions whether the problem asks it or not. Never be impolite and just answer the problem, rushing away for more important things. Obviously continuity cannot be dropped. Find an example with the oscillatory behaviour that you mentioned above happening at the left endpoint. The example will have to have unbounded variation on $[0,1]$ even though it is absolutely continuous (and hence BV) on $[W,1]$ for all $W>0$.

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