# Behavior of harmonic functions comparing with a Green’s function

Theorem Let ${E \subset \subset B_1}$, ${u \in C^0 (\bar{B_1} \setminus E) \cap C^2 (B_1 \setminus E)}$ satisfies

$\displaystyle - \Delta u =0, \quad \text{in } B_1 \setminus E, \text{ with } u \ge 0 \text{ on } \partial B_1,$

If there exists ${G \in C^2(\bar{B_1} \setminus E)}$ such that

$\displaystyle - \Delta G =0, \quad \text{in } B_1 \setminus E \text{ and } \lim_{dist(x,E) \rightarrow 0} G(x) = +\infty.$

If ${\frac{u^-}{G(x)} \rightarrow 0}$ as ${dist(x,E) \rightarrow 0}$, then ${u \ge 0}$ in ${(B_1 \setminus E)}$.

Proof: Without loss of generality assume ${G \ge 0}$ on ${\partial B_1}$, or we can add a large constant on ${G}$ to make it positive on boundary. ${\forall \varepsilon >0}$, there exists ${0 < \delta(\epsilon) < \epsilon}$ such that

$\displaystyle \frac{u^-}{G(x)} \le \varepsilon \text{ whenever } 0 < dist(x,E) \le \delta.$

Since ${\frac{-u}{G} \le \frac{\max(0, -u)}{G} = \frac{u^-}{G} \le \varepsilon}$, we have

$\displaystyle u(x) \ge - \varepsilon G(x) \text{ for } 0 < dist(x,E) \le \delta.$

Also ${u \ge - \varepsilon G(x)}$ on ${\partial B_1}$ since ${G}$ is non-negative on boundary. Then maximum principle tells ${ u \ge - \varepsilon G}$ in ${B_1 \setminus E_\delta}$, where ${E_\delta = \{x\in B_1: dist(x,E) \le \delta \}}$. Then we can first take ${\delta \rightarrow 0^+}$ and then take ${\varepsilon \rightarrow 0^+}$ to get ${u \ge 0}$ in ${B_1 \setminus E}$. $\Box$

Remark The condition ${\frac{u^-}{G(x)} \rightarrow 0}$ is essential, otherwise one can take ${-G}$ plus a large constant as a counterexample.

# 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.

# 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 ${\Omega \subset {\mathbb R}^n}$.

# Basic spectral properties of banach algebra

Definition Let ${B}$ be a banach space, ${\mathcal{L}(B)}$ be the space of bounded linear operator from ${B \rightarrow B}$. If ${T \in \mathcal{L}(B)}$, then say a complex number ${\lambda}$ is in the resolvent set ${\rho(T)}$ of ${T}$ if ${\lambda I - T}$ has a bounded inverse. Then we define the resolvent of ${T}$ at ${\lambda}$ to be ${R_\lambda(T) = (\lambda I -T)^{-1}}$. If ${\lambda \not\in \rho(T)}$, then ${\lambda}$ is said to be in the spectrum ${\sigma(T)}$ of T.

We will learn some properties of the resolvent and the spectrum.

Definition ${X,Y}$ are Banach spaces, we say ${X}$ is embedded in ${Y}$, denoted by ${X \hookrightarrow Y}$, if ${X \subset Y}$ and ${\| \cdot \|_Y \le C \| \cdot \|_X}$.
Notation: ${\langle \xi \rangle := \sqrt{1+\xi^2}}$.