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)}


\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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