## 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)|.

### 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.