Tag Archives: uniform continuous

Elementary property of uniform continuous functions

\mathbf{Problem:} Suppose f is an uniform continuous function on the whole \mathbb{R}^n, then there is a constant K such that

\displaystyle \sup\limits_{\mathbb{R}^n\times\mathbb{R}^n}\{f(x)-f(y)-K|x-y|\}<\infty

\mathbf{Proof:} By the assumption of f, we know \exists\,\delta>0 such that for any |x-y|\leq\delta, |f(x)-f(y)|<1 holds

Then if \displaystyle |x-y|=r, we can conclude that \displaystyle |f(x)-f(y)|<\frac r\delta+1.

Choosing \displaystyle K=\frac 1\delta,  we have \forall \,|x-y|=r,

\displaystyle f(x)-f(y)-K|x-y|<\frac r\delta+1-\frac 1\delta\, r<1

\text{Q.E.D}\hfill \square

\bf{Corollary:} Suppose f is an uniform continuous function on the whole \mathbb{R}^n, then there are constants K and C such that |f(x)|<C+K|x|.

Uniform continuity and uniqueness

In general it is not known that if H,u,f,g\in C(\mathbb{R}^{n}) and u is a viscosity solution of H(Du)-f=0 and H(Du)-g=0 in \mathbb{R}^n implies that f=g.
\bf{Proposition:} If H is uniformly continuous on \mathbb{R}^n, then f must equal g.

\bf{Proof:} Fix any x_0\in\mathbb{R}^n and real number r>0. Choose a function \theta\in C^\infty(\mathbb{R}^n) such that

\theta(x)=1 on |x|<\frac r2, \theta(x)=0 on |x|>r,

\displaystyle 0\leq \theta(x) \leq 1 and |D\theta|<\frac 2r,

Denote \displaystyle M=\sup\{|u(x)||x\in B_r(x_0)\}>0, otherwise f(x_0)=g(x_0). \theta_\epsilon(x)=\theta(\frac{x}{\epsilon}) and \delta=\sup\{|u(x)-u(y)||x\in B_r(x_0), |x-y|<\epsilon \}. Then \delta\to 0 if \epsilon\to 0.
Define \Phi on \mathbb{R}^n\times \mathbb{R}^n as

\Phi(x,y)=\theta(x-x_0)(u(x)-u(y))+3M\theta_\epsilon(x-y)+2\delta\theta(2(x-x_0)).

Claim: \Phi has a global maximum achieved only if  \displaystyle x\in B_{\frac{r}{2}}(x_0) and |x-y|\leq\epsilon.

  1.   if x\not\in B_r(x_0), then \Phi(x,y)\leq 3M
  2. if x\in \overline {B_r(x_0)}\backslash B_{\frac r2}(x_0), then
    \Phi(x,y)\leq u(x)-u(y)+3M\leq 3M+\delta
  3. if x\in B_{\frac r2}(x_0),
    \Phi(x,x)=3M+2\delta.
    If |x-y|>\epsilon, then \Phi(x,y)<2M+2\delta

claim holds.

Suppose \Phi achieves its maximum at (\bar{x}, \bar{y})
Then \Phi(\bar{x},\bar{y})=u(\bar{x})-u(\bar{y})+3M\theta_\epsilon(\bar{x}-\bar{y})+2\delta\theta(2(\bar{x}-x_0))
Define
\xi_1(x)=u(\bar{y})-3M\theta_\epsilon(x-\bar{y})-2\delta\theta(2(x-x_0))
Then
u(x)-\xi_1(x) has a maximum at \bar{x}. Since u is the viscosity subsolution of H(Du)=f

H(-3MD\theta_\epsilon(\bar{x}-\bar{y})-2\delta D\theta(2(\bar{x}-x_0)))\leq f(\bar{x})

Similarly we have \xi_2(y)=u(\bar{x})+3M\theta_\epsilon(\bar{x}-y)+2\delta\theta(2(\bar{x}-x_0)) and u(y)-\xi_2(y) has a minimum at \bar{y}

H(-3MD\theta_\epsilon(\bar{x},\bar{y}))\geq g(\bar{y}).

Because H is uniformly continuous

\displaystyle f(\bar{x})-g(\bar{y})\geq H(-3MD\theta_\epsilon(\bar{x}-\bar{y})-2\delta D\theta(2(\bar{x}-x_0)))-H(-3MD\theta_\epsilon(\bar{x},\bar{y}))\geq -\omega_H(4\delta|D\theta|)=-\omega_H(\delta\frac{8}{r})\quad (1) where \omega_H is the module of continuity of H.
Since going to subsequence if necessary, there exists z\in B_{\frac r2}(x_0), such that

\bar{x}=\bar{x}(\delta,\epsilon)\to z as \epsilon\to 0
\bar{y}=\bar{y}(\delta,\epsilon)\to z as \epsilon \to 0.

Letting \epsilon\to 0 in (1), we get f(z)-g(z)\geq 0 for some z\in B_r(x_0)
let r\to 0, we get f(x_0)\geq g(x_0).
By symmetry, f(x)=g(x).

 

\bf{Remark:}This proof is gleaned from M.G.Crandall, L.C.Evans and P.L.Lions, Properties of Viscosity solution of Hamilton-Jacobi equations. And L.C.Evans, Some man-min methods for the Hamilton-Jacobi equation. The original proof require u to be bounded. Here we remove the requirement.

This problem is a note to Micheal G. Crandall, Hitoshi Ishii, and Pierre Louis Lions, User’s guide to viscosity solutions of second order partial differential equations.