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.

  • $\begingroup$ Do you assume $y\le x$ in ht first part? $\endgroup$ – Martin Sleziak Nov 28 '11 at 15:31
  • $\begingroup$ @MartinSleziak: do I necessarily have to, since I'm told that the derivative exist. $\endgroup$ – Bass Nov 28 '11 at 15:36
  • $\begingroup$ @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. $\endgroup$ – Arthur Nov 28 '11 at 15:54
  • $\begingroup$ @Arthur: Okay. Is everything else fine? thanks. $\endgroup$ – Bass Nov 28 '11 at 16:18
  • $\begingroup$ 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. $\endgroup$ – robjohn Nov 28 '11 at 17:04

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.

  • $\begingroup$ Thanks for your answer. So choosing $\delta=\epsilon^{3/2}$ does the trick? $\endgroup$ – Bass Nov 28 '11 at 17:22
  • $\begingroup$ 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. $\endgroup$ – robjohn Nov 28 '11 at 17:35
  • $\begingroup$ Okay, now I'm really confused. $\endgroup$ – Bass Nov 28 '11 at 18:18
  • $\begingroup$ 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 $$\sum_i|b_i-a_i|<\delta$$ $\endgroup$ – robjohn Nov 28 '11 at 18:50
  • 1
    $\begingroup$ 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. $\endgroup$ – Bass Nov 29 '11 at 0:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.