# Two problems on Lipschitz continuity in Hausdorff measure

Let $$H^p(A)=\lim_{\delta\to 0}H^p_{\delta}(A)$$ be the $$p$$-dimensional Hausdorff measure, where $$H^p_{\delta}(A)=\inf\{\sum_{i}diam(E_i)^p:A\subset\bigcup_{i}E_i,diam(E_i)<\delta\}$$

(1)

Suppose $$f:A\to\mathbb{R}^k$$ satisfies Lipschitz condition: $$\|f(x)-f(y)\|\leq M\|x-y\|$$ for constant $$M$$. Show that for p-dimensional Hausdorff measure, $$H^p((f(A))\leq M^p \cdot H^{p}(A)$$.

Here is my proof:

Let $$A\subset\bigcup_{j}E_j$$, and $$diam(E_j)<\delta$$, satisfying $$H^p(A)+\epsilon\ge\sum_{j}diam^p(E_j)$$.

By Lipschitz condition, $$diam((f(E_j))\leq M\cdot diam(E_j)$$ and $$H^p_{\delta}(A)+\epsilon\ge\frac{1}{M^p}\sum_{j}diam^{p}(f(E_j))$$.

Since $$f(A)\subset\bigcup_{j}f(E_j)$$, and $$diam(f(E_j))\leq M\cdot\delta$$,

$$M^p(H^p_{\delta}(A)+\epsilon)\ge H_{M\delta}^{p}(f(A))$$. Let $$\epsilon, \delta\to 0$$, $$H^p((f(A))\leq M^p\cdot H^p(A)$$.

Can someone help me check whether the proof is correct?

(2)

Suppose $$f:[a,b]\to\mathbb{R}$$ has continuous derivatives. Let $$A=\{(x,f(x):x\in[a,b]\}\subset\mathbb{R}^2$$. Show that $$H^1(A)=\int_{a}^{b}\big((1+f')^2\big)^{1/2}dx$$.

By the first question, $$b-a=H^{1}([a,b])\leq H^{1}(A)\leq M\cdot H^{1}([a,b])=M(b-a)$$. How to use this to show the equation?

By the way, I am not so sure about how to solve question 2. Please leave your solutions if possible.

Thanks!

Your proof for (1) is alright except you wrote $$M$$ instead of $$M^p$$ at the first character of the last line.

For (2), first note that $$A$$ is compact and since we can suppose the $$E_i$$ to be open sets, given a cover of $$A$$ by $$E_i$$'s, we can find a finite subcover.
But then, we can also find a uniform neighbourhood of $$f$$ such that the graphs of all the functions near $$f$$ are contained in the finite subcover we found above.

Consider a partition $$P=\{x_0, x_1, ..., x_n \}$$ of $$[a, b]$$.

Let $$m_i = \inf_{x_{i - 1} \le x \le x_i} f(x)$$ and $$M_i = \sup_{x_{i - 1} \le x \le x_i} f(x)$$.
Analogously, let $$m'_i = \inf_{x_{i - 1} \le x \le x_i} f'(x)$$ and $$M'_i = \sup_{x_{i - 1} \le x \le x_i} f'(x)$$.

As a consequence of the initial remarks, and the fact that $$f$$ is $$C^1$$, we can take $$E_i = [x_{i-1}, x_i] \times [m_i, M_i]$$.
Observe that $$diam E_i = ((x_{i-1} - x_i)^2 + (M_i - m_i)^2)^\frac{1}{2}$$, since it's a rectangle.

From mean value theorem, we have:

$$((x_{i-1} - x_i)^2 + (m'_i)^2(x_{i-1} - x_i)^2)^\frac{1}{2} \le diam E_i \le ((x_{i-1} - x_i)^2 + (M'_i)^2(x_{i-1} - x_i)^2)^\frac{1}{2}$$.

Summing over $$i$$ and letting $$|P| \to 0$$, we get to $$H_{\delta}^1(A)=\int_a^b(1 + f'(x)^2)^\frac{1}{2}dx$$(again, we can do that because $$f'$$ is continuous on a compact set, hence integrable). It follows immediately that $$H^1(A)=\int_a^b(1 + f'(x)^2)^\frac{1}{2}dx$$.

• Thanks! But how do you show that $A$ is compact?
– user758472
Commented May 9, 2020 at 4:10
• $A$ is the image of the compact set $[a, b]$ by the continuous function $x \mapsto (x, f(x))$. Commented May 9, 2020 at 4:12
• Should it be $m_i^2$ and $M_i^2$ in the mean value theorem step?
– user758472
Commented May 10, 2020 at 10:54
• Yep. I'll edit it. Commented May 10, 2020 at 17:02