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Let $f:[0,1]\rightarrow \mathbb{R}$ be continuous. Suppose that $f'(x)$ exists and satisfies $|f'(x)|\leqslant \frac{1}{\sqrt{x}}$ for each $x$ in $(0,1]$.

I have to show the following:
1. for each $\varepsilon \gt 0$, $f$ is absolutely continuous on $[\varepsilon, 1]$.
2. $|f(1)-f(0)|\leqslant 2.$

My Attempt.

  1. $|f'(x)|=\lim_{y\rightarrow x}|\frac{f(y)-f(x)}{y-x}|\leqslant 1/\sqrt{x}$. So $$|f(y)-f(x)|\leqslant \frac{|y-x|}{\sqrt{x}}\lt \delta /\sqrt{x}$$
    Let $\varepsilon \gt 0$. Let $\delta = \varepsilon\cdot \sqrt{x}$. Let $\{[x_i-y_i]\}$ be a collection of nonoverlapping intervals with $\sum |x_i-y_i|\lt \delta$. Then we have $$\sum |f(x_i)-f(y_i)|\lt \varepsilon.$$ So $f$ is absolutely continuous.

  2. Since $f$ is absolutely continuous, it is a definite integral and $$f(t)=f(a)+\int_a^t f'(x)~dx.$$ Then $$ \begin{align*} |f(1)-f(0)| & = |\int_a^1 f'(x)~dx-\int_a^0 f'(x)~dx|\\ & = |\int_0^1 f'(x)~ dx|\\ & \leqslant \int_0^1 |f'(x)|~dx\\ & \leqslant \int_0^1 1/\sqrt{x}\\ & = 2. \end{align*} $$
    Is what I've done okay? Thanks.

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    Do you assume $y\le x$ in ht first part?2011-11-28
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    @MartinSleziak: do I necessarily have to, since I'm told that the derivative exist.2011-11-28
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    @Bass You have to assume one bigger than the other, as the upper limit of the derivative has to be assumed to be as big as possible for the inequalities to work out. Just be careful of which is which, and that little detail should work out.2011-11-28
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    @Arthur: Okay. Is everything else fine? thanks.2011-11-28
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    Part 1 is not quite right. Your $\delta$ depends on $x$ (which $x$ is not specified), but $\delta$ is supposed to depend on $\epsilon$ and not the partition.2011-11-28

1 Answers 1

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Hint for part 1: For $x\in[\epsilon,1]$ we know that $|f'(x)|\le\frac{1}{\sqrt{\epsilon}}$. Thus, For any intervals $[a_i,b_i]\subset[\epsilon,1]$, we have $$ |f(b_i)-f(a_i)|\le\frac{|b_i-a_i|}{\sqrt{\epsilon}} $$ Part 2 looks fine.

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    Thanks for your answer. So choosing $\delta=\epsilon^{3/2}$ does the trick?2011-11-28
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    Perhaps, you are getting the $\epsilon$ from $[\epsilon,1]$ and the $\epsilon$ from $\delta,\epsilon$ confused. Try the interval $[\alpha,1]$ instead; or perhaps do a $\delta,\alpha$ proof for absolute continuity.2011-11-28
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    Okay, now I'm really confused.2011-11-28
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    For $x\in[\alpha,1]$, we know that $|f'(x)|\le\frac{1}{\sqrt{\alpha}}$. Thus, for any intervals $[a_i,b_i]\subset[\alpha,1]$, we have $$|f(b_i)-f(a_i)|\le\frac{|b_i-a_i|}{\sqrt{\alpha}}$$ Now what can you say about a partition where $$\sum_i|b_i-a_i|<\delta$$2011-11-28
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    I realize you are reluctant to provide a complete solution because the question was tagged as homework. Well, the homework was due 3 hours ago and I already turned it in. However, if you don't mind, I'll appreciate it if you could provide the solution as I still want to know how to solve it. Thanks.2011-11-29
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    For absolute continuity, you need to show that for any $\epsilon>0$ you can find a $\delta>0$ so that for any set of disjoint intervals $\{[a_i,b_i]\}$ for which $$\sum_i|b_i-a_i|<\delta$$ we also have $$\sum_i|f(b_i)-f(a_i)|<\epsilon$$ Since $$|f(b_i)-f(a_i)|\le\frac{|b_i-a_i|}{\sqrt{\alpha}}$$ if we take $\delta=\epsilon\sqrt{\alpha}$, we get $$\begin{align}\sum_i|f(b_i)-f(a_i)|&\le\frac{1}{\sqrt{\alpha}}\sum_i|b_i-a_i|\\&<\frac{1}{\sqrt{\alpha}}\delta\\&=\epsilon\end{align}$$2011-11-29
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    Thanks very much. I really appreciate it.2011-11-29
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    @robhjon: Dear robjohn, Please how did you get $$|f(b_i)-f(a_i)|\le\frac{|b_i-a_i|}{\sqrt{\alpha}} $$? Thanks.2012-02-24
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    @Nana: The [Mean Value Theorem](http://en.wikipedia.org/wiki/Mean_value_theorem) says that if $f$ is differentiable on $(a,b)$, then $$\frac{f(b)-f(a)}{b-a}=f'(c)$$ for some $c\in(a,b)$. Note that since $a, we have that $f'(c)\le\frac{1}{\sqrt{a}}$.2012-02-24