Hausdorff Measure and Capacity

Definition 1 Let {E \subset {\mathbb R}^n, 0 \le d < \infty, 0 < \delta \le \infty}. Define

\displaystyle \mathcal{H}^d_\delta(E) : = \inf \bigg\lbrace (1/2)^d \alpha(d) \sum_{i=1}^\infty diam(A_i)^d | E\subset \bigcup_{i=1}^\infty A_i, diam(A_i)\le \delta \bigg\rbrace,

where {\alpha(d) := \frac{\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2} + 1)}}.

\displaystyle \mathcal{H}^d (E) := \lim_{\delta \rightarrow 0} \mathcal{H}^d_\delta(E) = \sup_{\delta >0} \mathcal{H}^d_\delta(E)

Then {\mathcal{H}^d} is called d-dimensional Hausdorff measure in {{\mathbb R}^n}.

Remark 1 Hausdorff measure is a Borel outer measure, it is not Radon for {0 \le d < n} because {{\mathbb R}^n} is not {\sigma-}finite with respect to {\mathcal{H}^d}. And it is a measure if restricted on Lebesgue measurable sets (by Caratheodory condition: We say a set {E} satisfies Caratheodory condition, if for any set {A \subset {\mathbb R}^n}, {\mathcal{L}^n(A) = \mathcal{L}^n(A \cap E) + \mathcal{L}^n(A \setminus E)}). Moreover, note that {\alpha(d)} gives the volume of unit ball in {d} dimension if {d} is an integer, we naturally have (not trivially) {\mathcal{H}^n = \mathcal{L}^n }, where {\mathcal{L}^n} is the n-dimensional Lebesgue measure.

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High order BV function has continuous representative

As known, bounded variation (BV) functions are not continuous, but evidently higher order BV functions on corresponding dimension have continuous representative. As a consequence, {W^{n,1}({\mathbb R}^n) \hookrightarrow C^0({\mathbb R}^n)}.

Definition A function {f \in BV_n({\mathbb R}^n)}, if {f \in W^{n-1,1}({\mathbb R}^n)}, and the nth order distributional derivative {D^n f} is a finite Radon measure.

Theorem If {f \in BV_n({\mathbb R}^n)}, then {f} has a continuous representative.

Proof: Note that {C_c^\infty({\mathbb R}^n)} is dense in {BV_n({\mathbb R}^n)}, i.e. for any {f \in BV_n({\mathbb R}^n)}, there exists a sequence {f_k \in C_c^\infty({\mathbb R}^n)} such that {\lim_{k \rightarrow \infty} f_k = f} in {W^{n-1,1}}, and {\lim_{k\rightarrow \infty} \| \nabla^n f_k\|_{L^1({\mathbb R}^n)} = \|D^n f \| ({\mathbb R}^n)}. (See for example, book by Evans and Gariepy). So it suffice to show for all {f \in C_c^\infty({\mathbb R}^n)},

\displaystyle \|f\|_{L^\infty({\mathbb R}^n)} \le \| \nabla^n f\|_{L^1({\mathbb R}^n)}

Indeed,

\displaystyle f(x_1, \cdots , x_n) = \int_{-\infty}^{x_1} \partial_1 f(s_1,x_2, \cdots , x_n)\,ds_1 = \int_{-\infty}^{x_1} \cdots \int_{-\infty}^{x_n} \partial_1 \cdots \partial_n f

This gives us the desire inequality. Then by density result we know that {BV_n({\mathbb R}^n)} admits continuous representative for every element. \Box

Corollary {W^{n,1}({\mathbb R}^n) \hookrightarrow C^0({\mathbb R}^n)}.

Remark If the dimension is greater than the order of BV function, then the above theorem will fail. For example, let {f(x) = |x|^{-\frac{1}{2}}} be defined in a neighborhood {\Omega} around 0, with smooth boundary in {{\mathbb R}^3}. {|\nabla f| \sim |x|^{-\frac{3}{2}}} and {|\nabla^2 f| \sim |x|^{-\frac{5}{2}}}. So {f \in W^{2,1}(\Omega)}. Then we can extend {f} on {{\mathbb R}^3}, since {\partial \Omega} is smooth (See for example, PDE book by Evans). But then {f \rightarrow \infty} as {|x| \rightarrow 0}, which doesn’t admit a continuous representative.

Equivalence of compactness in metric space

Lemma In a metric space {(X,d)}, if {\{C_k\}_{k=1}^\infty \subset X} is a sequence of compact sets such that {C_{k+1} \varsubsetneq C_k} for {k \in {\mathbb N}} and {\lim_{k \rightarrow \infty}} diam{C_k = 0}. Then {\cap_{k=1}^\infty C_k = \{c\}}, where {c} is a point in {X}.

Proof: First note that {\cap_{k=1}^\infty C_k} cannot contain more than 1 points. If {a,b \in \cap_{k=1}^\infty C_k}, then {a,b \in C_k \quad \forall k}. Since {\lim} diam{C_k}=0, it forces {a=b}.

It remains to show {\cap_{k=1}^\infty C_k} is nonempty. Assume it is empty, then {\cup_{k=1}^\infty (X \setminus C_k ) = X \setminus (\cap_{k=1}^\infty C_k) = X}. Since metric space is Hausdorff, {C_k} is closed for all {k} and {\{ X \setminus C_k \}} is an open cover for {C_1}. Since {C_1} is compact, there exists finite subcover such that {C_1 \subset \cup_{i=1}^m (X\setminus C_{k_i}) = X \setminus C_{k_m}}, which contradicts {C_{k_m} \varsubsetneq C_1}. \Box

Theorem In a metric space {(X,d)} (no matter it has countable basis or not), let {E} be a subset in {X}. The following statement are equivalent by assuming axiom of choice:

  1. {E} is compact;
  2. {E} is sequentially compact;
  3. {E} is complete and totally bounded.

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Kolmogorov-Riesz theorem

I will show an {L^1} version of Arzela-Ascoli theorem, which is called Kolmogorov-Riesz theorem, by using a fact in topology as following.

Definition In a metric space, a set is totally bounded, if for any fixed {\varepsilon>0}, the set can be covered by finitely many open balls of radius {\varepsilon}.

Lemma In a metric space, a set {E} is compact if and only if {E} is complete and totally bounded.

Proof of the lemma can be find here.

Theorem 1 (Kolmogorov Riesz Theorem (finite measure domain)) Let {\mathcal{F}} be a bounded set of {L^1({\mathbb R}^n)} functions, Assume that {\forall \varepsilon >0}, there exists a {\delta >0} such that

\displaystyle \| f(\cdot + y) - f \|_{L^1({\mathbb R}^n)} \le \varepsilon

whenever {|y| < \delta}, for all {f \in \mathcal{F}}. Then {\mathcal{F}_{| \Omega}} has a compact closure in {L^1(\Omega)} for any finite measure set {\Omega \subset {\mathbb R}^n}.

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Composition of sobolev functions

The following materials are exercise 8.10 and 8.11 from the functional analysis book by Prof. Brezis. The questions arise from whether we can well define a composition of two W^{1,p} functions or not. From Corollary 8.11 we know if G \in C^1, u \in W^{1,p}, G(u) \in W^{1,p} is well defined. However bad things might happen if G \in W^{1,p}. For instance, G(x) = |x|, then G(u) might be bad on the set \{ u = 0\}. I thought we could still define things in the sense of distribution, but Prof. Brezis said in this special case we could have something more than that, because of the following propositions. The proof is not hard by following the hints, but the idea is really clever (IMO)!

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