Example of absolutely continuous function $f$ with $\sqrt{f}$ not absolutely continuous

Question

I'm looking for an example of a function $f$ that is absolutely continuous, but $\sqrt{f}$ is not absolutely continuous.

I've been playing around with the Cantor-Lebesgue function, but I feel like there should be something simpler.

Answer 1

I believe $f(x) = x^2 (\cos \frac1x)^4$ is an example on the interval $(0,1)$. While a proof is certainly needed, the key observation is that sum of the infinitely many local maxima of $f$ converges (indeed $f'$ is uniformly bounded), but the sum of the infinitely many local maxima of $\sqrt f$ does not converge.

Answer 2

No Cantor-Lebesgue type of example will work, at least not on a closed interval $[a, b]$. This is because if $f(x)$ is absolutely continuous on $[a, b]$, then so is $(f(x))^p$ for any $p > 0$. The argument goes as follows: $f(x)$ is absolutely continuous on $[a, b]$ iff $f'(x)$ exists and is Lebesgue integrable almost everywhere on $[a, b]$, and $$f(x) = f(a) + \int_a^x f'(t) dt$$ for all $x \in [a, b]$.

Suppose $f(x)$ is a nonnegative, absolutely continuous function on $[a, b]$. $f(x)$ is absolutely continuous only if it is absolutely continuous on every finite subcollection of closed intervals of $[a, b]$. Since $f(x) \equiv 0$ on any subinterval of $[a, b]$ implies $\sqrt{f(x)}$ is also identically zero on that subinterval, hence automatically absolutely continuous there, WLOG we can assume the support of $f(x)$ is $[a, b]$. This implies that the set $Z_f := \{ x \in [a, b]: f(x) = 0 \}$ has measure zero in $[a, b]$.

Since $f(x)$ is differentiable almost everywhere on $[a, b]$, $\sqrt{f(x)}$ also has a Lebesgue integrable derivative almost everywhere on $[a, b]$, as $$\frac{d}{dx}\sqrt{f(x)} = \frac{f'(x)}{2\sqrt{f(x)}}$$

wherever $f'(x)$ exists and $f(x) \neq 0$, and $Z_f$ has measure zero. So to show $\sqrt{f(x)}$ is absolutely continuous, we just need to show that, if $g(x) = \frac{f'(x)}{2\sqrt{f(x)}}$ wherever this expression is defined and $g(x) = 0$ elsewhere, then $$\sqrt{f(x)} = \sqrt{f(a)} + \int_a^x g(t) dt$$ for any $x \in [a, b]$, since the set where $\frac{f'(x)}{2\sqrt{f(x)}}$ is undefined has measure zero. Since $[a, b] \setminus Z_f$ is open and has full measure, it follows that we may write $(a, b) \setminus Z_f$ as a finite or countable union of open intervals $\{ I_j \}_{j \in S} = \{ (a_j, b_j) \}_{j \in S}$ such that:

• $\sum_{j \in S} (b_j - a_j) = 1$
• $f(a_j) = f(b_j) = 0$ for all $j \in S$, except possibly if $a_j = a$ or $b_j = b$.

Here $S = \{1, ..., n \}$ for some $n$, or $S = \Bbb{N}$.

Clearly $g$ is Lebesgue integrable on each of the $I_j$, and $\sqrt{f(x)} - \sqrt{f(y)} = \int_y^x g(t) dt$ whenever $a_j < y < x < b_j$ for some $j \in S$. If we let $y \rightarrow a_j$, by continuity of $\sqrt{f}$, we can evaluate the improper integral $\int_{a_j}^x g(t) dt$ as

$$\int_{a_j}^x g(t) dt = \lim_{y \rightarrow a_j} \int_y^x g(t) dt = \lim_{y \rightarrow a_j} \sqrt{f(x)} - \sqrt{f(y)} = \sqrt{f(x)} - \sqrt{f(a_j)}.$$

A similar limiting operation letting $x \rightarrow b_j$ shows that $\int_{I_j} g(t) dt = \int_{a_j}^{b_j} g(t) dt = 0$ whenever $f(a_j) = f(b_j) = 0$. If $U$ is the $I_j$ with left endpoint $a$, we also find that $\int_U g(t) dt = -\sqrt{f(a)}$. So either $x \in U$, in which case $\int_a^x g(t) dt = \sqrt{f(x)} - \sqrt{f(a)}$, or else $x \in \operatorname{cl}(I_j)$ for some $I_j = (a_j, b_j) \neq U$, in which case

\begin{align*} \int_a^x g(t) dt &= \int_a^{a_j} g(t) dt + \int_{a_j}^x g(t) dt \\ &= \int_U g(t) dt + \sum_{I_k \neq U: I_k \subseteq [a, a_j]} \int_{I_k} g(t) dt + \sqrt{f(x)} - \sqrt{f(a_j)} \\ &= -\sqrt{f(a)} + 0 + \sqrt{f(x)} \\ &= \sqrt{f(x)} - \sqrt{f(a)}, \ \end{align*}

which shows that $\sqrt{f}$ is absolutely continuous. A similar proof shows that if $f(x)$ is absolutely continuous on $[a, b]$, then so is $(f(x))^p$ for any $p > 0$. That also implies that no positive power of the Cantor function can be absolutely continuous on an interval, since otherwise the Cantor function itself would be.