Let $f:[0,1] \to \mathbb{R}, G = graph(f)$.

 

If $\sum_{j = 1}^m |f(x_i)-f(x_{i-1})|^s \le c$ for all partitions $0 = x_0< \ldots < x_m = 1 $ then $H^s(G) < \infty$

What technique can I use to prove this result?

Can it be reduced to the theorem stating that a rectifiable curve $\Gamma$ has $H^1(\Gamma) < \infty$?

Let $f:[0,1] \to \mathbb{R}, G = graph(f)$.

If $\sum_{j = 1}^m |f(x_i)-f(x_{i-1})|^s \le c$ for all partitions $0 = x_0< \ldots < x_m = 1 $ then $H^s(G) < \infty$

What technique can I use to prove this result?

Can it be reduced to the theorem stating that a rectifiable curve $\Gamma$ has $H^1(\Gamma) < \infty$?

Let $f:[0,1] \to \mathbb{R}, G = graph(f)$.

If $\sum_{j = 1}^m |f(x_i)-f(x_{i-1})|^s \le c$ for all partitions $0 = x_0< \ldots < x_m = 1 $ then $H^s(G) < \infty$

What technique can I use to prove this result?

Can it be reduced to the thorem stating that a rectifiable curve $\Gamma$ has $H^1(\Gamma) < \infty$?

Let $f:[0,1] \to \mathbb{R}, G = graph(f)$.

If $\sum_{j = 1}^m |f(x_i)-f(x_{i-1})|^s \le c$ for all partitions $0 = x_0< \ldots < x_m = 1 $ then $H^s(G) < \infty$

What technique can I use to prove this result?

Can it be reduced to the theorem stating that a rectifiable curve $\Gamma$ has $H^1(\Gamma) < \infty$?