# Conservative, solenoidal vector fields and Hardy space

Theorem 1 Let ${E \in L^p ({\mathbb R}^n; {\mathbb R}^n), B \in L^q ({\mathbb R}^n ; {\mathbb R}^n)}$ with ${1 < p < \infty, \frac{1}{p} + \frac{1}{q} = 1}$, such that ${B = \nabla \phi}$, where ${\phi \in L^{\frac{nq}{n-q}}}$, if ${q < n}$ ; ${\phi \in L_{loc}^s}$ for all ${s < \infty}$, if ${q \ge n}$. If ${E}$ satisfies

$\displaystyle div E = 0 \quad \text{ in } D'({\mathbb R}^n),$

then ${E \cdot B \in \mathcal{H}^1}$.

# Two fixed point theorems in Banach space

Here is a famous fixed point theorem in finite dimension by Brouwer:

Theorem 1 (Brouwer fixed point theorem) Let ${M \subset {\mathbb R}^n}$ be a convex compact set, for any continuous function ${f : M \rightarrow M}$, there exists a point ${x_0 \in M}$ such that ${f (x_0) = x_0}$.

There are couple of ways to extend this theorem to Banach spaces. First recall that a mapping between two Banach spaces is called compact, if the mapping is continuous (not necessarily linear) and the images of bounded sets are pre-compact.

# 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.

# Fredholm Alternative and Riesz Schauder Theorem

Theorem (Analytic Fredholm Alternative) Let ${D}$ be a connected open subset of ${{\mathbb C}}$ and ${\mathcal{H}}$ be a separable Hilbert space. Suppose that ${f : D \longrightarrow \mathcal{L} (\mathcal{H})}$ is an analytic operator-valued function such that ${f(z)}$ is compact ${\forall z \in D}$. Then either

1. ${(I - f(z)) ^{-1}}$ does not exist ${\forall z \in D}$.
2. ${(I - f(z)) ^{-1}}$ exists ${\forall z \in D\setminus S}$, where ${S}$ is discrete.

# Quotient norm of self-adjoint element can be achieved

Theorem Let ${I}$ be a closed ideal of a C* algebra ${\mathcal{A}}$, then for any self-adjoint element ${a \in \mathcal{A}}$, there exists an ${i \in I}$ such that

$\displaystyle \|a-i\| = \inf \{ \|a -y\|: y \in I \}.$

Proof: Since ${I}$ is a closed ideal, ${\mathcal{A}/I}$ is also a C* algebra and one can define a natural projection homomorphism ${\pi : \mathcal{A} \rightarrow \mathcal{A}/I}$, with ${\| \pi(a)\| = \inf \{ \|a -y\|: y \in I \}}$. Fix an element ${a \in \mathcal{A}}$, define a continuous function ${f}$ as follow:

$\displaystyle f(x) = \begin{cases} \|\pi(a)\|, &x > \|\pi(a)\| \\ x, & -\|\pi(a)\| \le x \le \|\pi(a)\|\\ -\|\pi(a)\|, &x < -\|\pi(a)\| \end{cases}$

Let ${g(x) := x - f(x)}$, then ${g(x)}$ is identical ${0}$ on ${[-\|\pi(a)\|, \|\pi(a)\|]}$. By continuous functional calculus, ${g(a) \in \mathcal{A}}$. And since ${g}$ can be approximated by polynomials, we know ${\pi(g(a)) = g(\pi(a))}$. The spectral radius ${\nu(\pi(a)) \le \|\pi(a)\|}$, so we know ${g}$ is identical ${0}$ on the spectrum of ${\pi(a)}$, which implies ${\pi(g(a)) = g(\pi(a)) = 0}$. Therefore ${g(a)}$ is in the kernel of ${\pi}$, and hence in the ideal ${I}$. Since ${\|a - g(a)\| = \|f(a)\|}$ and ${\|f(a)\| \le \|\pi(a)\|}$ by the Gelfand Naimark isomorphism theorem, we know ${\|a - g(a)\| \le \| \pi(a)\|}$ and hence ${\|a - g(a)\| = \| \pi(a)\|}$. So ${g(a)}$ is the ${i}$ we are looking for. $\Box$

Note: this is a homework exercise given by Dr. Carlen.

# 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.