# Harnack inequality implies Holder continuity

We will show a basic argument why Harnack inequality implies Holder continuity. Assume ${u \in H^1(B_2)}$ is a weak solution to a uniform elliptic equation

$\displaystyle -\partial_i(a^{ij} \partial_j u) = 0,$

in ${B_2}$. Assume we have Harnack inequality (Indeed we do have due to Moser), i.e. for any positive weak solution ${u}$ we will have

$\displaystyle \sup_{B_1} u \le C \inf_{B_1} u,$

where ${C}$ is a constant independent of ${u}$.

First one will know ${u \in L^\infty(B_1)}$, see this note.

If ${u}$ is a weak solution, so are ${\sup_{B_2} u - u}$ and ${u - \inf_{B_2} u}$. And they are strictly positive in ${B_1}$ due to the strong maximum principle, otherwise ${u}$ will be a constant. By Harnack inequality, we will have

$\displaystyle \sup_{B_2} u - \inf_{B_1} u \le C(\sup_{B_2} u - \sup_{B_1} u),$

$\displaystyle \sup_{B_1} u - \inf_{B_2} u \le C(\inf_{B_1} u - \inf_{B_2} u).$

Adding these two inequalities up we will have

$\displaystyle \text{osc}_{B_2} u + osc_{B_1} u \le C(osc_{B_2} u - osc_{B_1}),$

hence

$\displaystyle osc_{B_1} u \le \frac{C-1}{C+1} osc_{B_2} u.$

Denote ${\theta = \frac{C-1}{C+1} < 1}$, and choose ${\alpha > 0}$ such that ${\theta = 2^{- \alpha}}$, by interation we know

$\displaystyle osc_{B_{2^{-k}}} u \le \theta^k osc_{B_1} u \le 2 \theta^k \|u\|_{L^\infty(B_1)}.$

Then for any ${0 < r < 1}$, choose ${k \in {\mathbb N}}$ such that ${2^{-k-1} < r \le 2^{-k}}$, we will have

$\displaystyle osc_{B_r} u \le osc_{B_{2^{-k}}} u \le 2^{-\alpha k + 1} \|u\|_{L^\infty(B_1)} = 2^{1+\alpha} \cdot 2^{- \alpha ( k +1)} \|u\|_{L^\infty(B_1)}< 2^{1+\alpha} \|u\|_{L^\infty(B_1)} r^\alpha.$

So Holder continuity follows.

# Newtonian Potential

— 1. Definition of Newtonian Potential —

Definition 1 Let ${\Omega}$ be a bounded open set, we define the Newtonian potential of ${f}$ is the function ${N_f}$ on ${{\mathbb R}^n}$ by

$\displaystyle N_f(x) = \int_\Omega \Gamma(x-y) f(y)\, dy,$

where ${\Gamma}$ is the fundamental solution of Laplace’s equation.

# 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.

# Regularity of scalar elliptic equation by Moser iteration

Theorem Suppose ${v \in W^{1,2}(B_R)}$ is a subsolution of ${-\partial_i (a^{ij}(x) \partial_j u) =0}$, ${\lambda I \le (a^{ij}(x)) \le \Lambda I}$. Then for all ${p>0}$, ${0<\theta <1}$,

$\displaystyle \sup_{B_{\theta R}} v \le C(n, \lambda, \Lambda, p) (1- \theta)^{- \frac{n}{p}} \left( \frac{1}{|B_R|} \int_{B_R} (v^+)^p \right)^{\frac{1}{p}}.$

# 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.