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.
|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.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.