# $l(\alpha)=\int_a^b\|\alpha$ '$(t)\| \;\mbox{dt}$ , $\alpha$ a differentiable function whose derivative is integrable.

Let $\alpha:[a,b]\longrightarrow \mathbb{R}^n$ a differentiable function whose derivative is integrable.

We can say that $\displaystyle l(\alpha)=\int_a^b\|\alpha$'$(t)\| \;\mbox{d}t\;$? , $\alpha$' not necessarily continuous

If $\alpha$' is integrable $\Longrightarrow$ $\alpha$ rectifiable ?

If $\alpha$' is integrable and $\alpha$ rectifiable $\Longrightarrow$ $\displaystyle l(\alpha)=\int_a^b\|\alpha$'$(t)\| \;\mbox{d}t\;$?

$l(\alpha)=\text{Sup}\;l(\alpha,P)$ , length of $\alpha$

$P:$ partition of $[a,b]$

Any hints would be appreciated.

• What is the question? – gerw Apr 6 '13 at 17:56

HINT: start taking a partition of $[a,b]$, $P=\{a=t_0, t_1,\ldots,t_m=b\}$ and try to calculates $$\sum_{i=1}^m\Vert \alpha(t_i)-\alpha(t_{i-1})\Vert$$
You know $\alpha$ is $C^1$, in that case you can write $$\Vert\alpha(t_i)-\alpha(t_{i-1})\Vert=\Vert\alpha'(t_{i-1})\cdot(t_i-t_{i-1})+r_i(t_i-t_{i-1})\Vert\leq$$ $$\leq\Vert\alpha'(t_{i-1})\cdot(t_i-t_{i-1})\Vert+\Vert r_i(t_i-t_{i-1})\Vert$$
such that for every $\varepsilon$, you can choose $P$ to get $\Vert r_i(t_i-t_{i-1})\Vert/(t_i-t_{i-1})<\varepsilon$.