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

Mathematics Asked on January 3, 2022

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.

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 $$frac{d}{dx}sqrt{f(x)} = frac{f'(x)}{2sqrt{f(x)}}$$

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 $$g(x) = frac{f'(x)}{2sqrt{f(x)}}$$ 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 $$frac{f'(x)}{2sqrt{f(x)}}$$ 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.

Answered by Rivers McForge on January 3, 2022

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.

Answered by Greg Martin on January 3, 2022

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