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.

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.