3
$\begingroup$

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.

  • 0
    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

3

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.

  • 0
    @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