0
$\begingroup$

I am seeking for a continuous of bounded variation function in $[0,1]$ that is absolutely continuous in $(a,1]$ for all $a\in(0,1)$, but is not in $[0,1]$.

The function $x\sin\left(\frac{1}{x}\right)$, for instance, is absolutely continuous and of bounded variation in $(a,1]$, but is not bounded in $[0,1]$.

$\endgroup$
1
  • $\begingroup$ The latter function is continuous so it must be bounded, did you mean something else? $\endgroup$
    – copper.hat
    Feb 19, 2022 at 20:27

1 Answer 1

4
$\begingroup$

There does not exist such a function. We recall that a function $f:[a,b] \rightarrow \mathbb{R}$ is absolutely continuous if and only if it is continuous, of bounded variation and has the property that $f(A)$ has Lebesgue measure $0$ whenever $A\subset [0,1]$ has Lebesgue measure $0$ (this last property is known as the Luzin-N-property). Now, if $f:[0,1] \rightarrow \mathbb{R}$ is as in your hypothesis and $A\subset [0,1]$ has Lebesgue measure $0$ then $$f(A)=f(A\cap \{0\})\cup \bigcup\limits_{n=1}^\infty f(A\cap [1/n, 1]).$$ The assumption that $f$ is a.c. when restricted to any subinterval $(a, 1]$, $a>0$, gives that all the sets in the union on the RHS have measure $0$, thus the LHS has measure $0$. Thus $f$ satisfies the Luzin-N-property and is a.c.

Edit: One can also prove it directly. Let $\epsilon>0$. We claim that there is a $\delta>0$ such that $V_0^\delta (f)\leq\epsilon$. Indeed, if this isn't the case we have $V_0^\delta (f)>\epsilon$ for all $\delta>0$. Using the continuity of $f$, pick $\delta'>0$ such that $|f(x)-f(0)|<\epsilon/2$ for $x\in [0, \delta')$. Then there is a partition $$0=x_{1, 0}<x_{1,1}<\ldots <x_{1,n_1}=\delta'$$ with $\sum\limits_{i=0}^{n_1-1}|f(x_{1,i+1})-f(x_{1,i})|>\epsilon$. Then $$\sum\limits_{i=1}^{n_1-1}|f(x_{1,i+1})-f(x_{1,i})|>\epsilon/2.$$

Now use that $V_0^{x_{1,1}} (f)>\epsilon$ to obtain a partition $$0=x_{2, 0}<x_{2,1}<\ldots <x_{2,n_2}=x_{1,1}$$ with $\sum\limits_{i=0}^{n_2-1}|f(x_{2,i+1})-f(x_{2,i})|>\epsilon$. Then again $$\sum\limits_{i=1}^{n_2-1}|f(x_{2,i+1})-f(x_{2,i})|>\epsilon/2.$$ Continuing like this we obtain infinitely many non-overlapping partitions with total variation at least $\epsilon/2$. This contradicts that $f$ has bounded variation on $[0,1]$. The conclusion is that there is a $\delta>0$ such that $V_0^\delta (f)\leq\epsilon$. Since $f$ is a.c. on $[\delta, 1]$ there is a $\delta''>0$ witnessing the $\epsilon$ condition in absolutely continuity on $[\delta, 1]$ . Then $\min(\delta, \delta'')$ will witness the $2\epsilon$ condition in absolutely continuity on $[0,1]$.

$\endgroup$
2
  • 2
    $\begingroup$ +1 Very nice. ${}$ $\endgroup$
    – copper.hat
    Feb 19, 2022 at 20:39
  • 1
    $\begingroup$ Terrific, thank you so much $\endgroup$ Feb 19, 2022 at 22:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .