Uniform continuity of a function: multidimensional distribution function

Question

If a distribution function $F$ defined on $\mathbb{R},$ is continuous then it's uniformly continuous, this is easy to prove, since more generally if $f$ is continuous on $\mathbb{R}$ and has finite limits on $+\infty$ and $-\infty,$ then it's uniformly continuous on $\mathbb{R}.$

Is it true that continuous multidimensional distribution functions on $\mathbb{R^d},$ $F(x_1,...,x_d)=P(X_1\leq x_1,...,X_d \leq x_d)$ for a vector random variable $(X_1,...,X_d)$?

In other word, if we have a continuous function $f:\mathbb{R}^d \rightarrow \mathbb{R},$ having finite limits at $\lim_{x \rightarrow +\infty}f(x)=l_1,\lim_{x \rightarrow -\infty} f(x)=l_2$ (meaning that each term is tending to $\infty$), then $f$ is uniformly continuous.

Is there are any references or a proof for this statement?

Answer 1

It is true that for $ F(x_1,\cdots, x_d) =\mathsf{P}(X_1\le x_1,\cdots,X_d\le x_d)$, a multidimensional distribution function of random vector $\mathbf{X}= (X_1,\cdots,X_d) $, if $ F(\mathbf{x})=F(x_1,\cdots, x_d) $ is continuous on $\mathbb{R}^d $, the $ F $ is uniformly continuous on $\mathbb{R}^d $. To prove it, letting that \begin{align*} \mathbf{I}_M(\mathbf{x})&=(I_M(x_1),\cdots,I_M(x_d))\\ &=(x_1\wedge M\vee (-M),\cdots,x_d\wedge M\vee (-M)) \end{align*} then $ \mathbf{I}_M(\mathbf{x}) $ is an $ \mathbb{R}^d \mapsto [-M,M]^d $ continuous map, \begin{align*} &\mathbf{I}_M(\mathbf{x})=\mathbf{x},\qquad \text{if } \mathbf{x}\in [-M,M]^d,\\ &\|\mathbf{I}_M(\mathbf{x}')-\mathbf{I}_M(\mathbf{x}'')\| =\max_{1\le i\le d}|I_M(x'_i)-I_M(x''_i)|\\ &\quad\le \|\mathbf{x}'-\mathbf{x}'' \|= \max_{1\le i\le d}|x'_i-x''_i|. \tag{1} \end{align*} Also let $F_M(\mathbf{x})=F(\mathbf{I}_M(\mathbf{x})) $, then $ F_M=F\circ \mathbf{I}_M $ (as a composite function of continuous functions $ F $ and $ \mathbf{I}_M(\mathbf{x})$) is continuous in $\mathbb{R}^d $ and \begin{align*} &F_M(\mathbf{x})=F(\mathbf{x}),\qquad \text{if } \mathbf{x}\in [-M,M]^d,\tag{2}\\ & F \text{ is continous in }\mathbb{R}^d \\ &\implies F \text{ is continous in }[-M,M]^d\\ &\implies F \text{ is uniformly continous in }[-M,M]^d\\ &\implies F_M \text{ is uniformly continous in }[-M,M]^d \quad (\text{ by (2)})\\ &\implies F_M \text{ is uniformly continous in }\mathbb{R}^d. \quad(\text{by (1))}\tag{3} \end{align*} The (3) is derived from following relation \begin{align*} &\sup_{\|\mathbf{x}'- \mathbf{x}''\|<\delta}|F_M(\mathbf{x}')-F_M(\mathbf{x}'')|\\ &\quad=\sup_{\|\mathbf{x}'- \mathbf{x}''\|<\delta}|F(\mathbf{I}_M(\mathbf{x}')) -F(\mathbf{I}_M(\mathbf{x}''))|\\ &\quad =\sup_{\|\mathbf{y}'- \mathbf{y}''\|<\delta, \mathbf{y}',\mathbf{y}''\in [-M,M]^d} |F(\mathbf{y}') -F(\mathbf{y}'')| \end{align*} Now we prove the following \begin{gather*} |F_M(\mathbf{x})-F(\mathbf{x})|\le \mathsf{P}(\|\mathbf{X}\|\ge M) \tag{4} \end{gather*} It is clear that \begin{align*} \{X_i\le x_i\}&\subset \{X_i\le I_M(x_i)\}\cup \{|X_i|\ge M\}\\ &\subset\{X_i\le I_M(x_i)\}\cup \{\|\mathbf{X}\|\ge M\}\\ \{X_i\le x_i,1\le i\le d\}&\subset \{X_i\le I_M(x_i),1\le i\le d\}\cup\{\|\mathbf{X}\|\ge M\}\\ F(\mathbf{x})&\le F_M(\mathbf{x}) + \mathsf{P}(\|\mathbf{X}\|\ge M).\tag{5} \end{align*} Similarly, \begin{align*} F_M(\mathbf{x})&= \mathsf{P}(X_i\le I_M(x_i),1\le i\le d)\\ & \le \mathsf{P}(X_i\le x_i, \|\mathbf{X}\|\le M)+\mathsf{P}(\|\mathbf{X}\|\ge M)\\ F_M(\mathbf{x})&\le F(\mathbf{x})+\mathsf{P}(\|\mathbf{X}\|\ge M) \tag{6} \end{align*} From (5),(6) we get (4) and \begin{equation*} |F(\mathbf{x}'')-F(\mathbf{x}')|\le |F_M(\mathbf{x}'')-F_M(\mathbf{x}')|+2 \mathsf{P}(\|\mathbf{x}\|\ge M) \end{equation*}

Now for fixed $ M $ letting $ \delta\to0 $ in above expression, using (3) and the uniform continuity of $ F_M $ we have $$ \varlimsup_{\delta\to0}\sup_{\|\mathbf{x}'-\mathbf{x}''\|\le\delta}|F(\mathbf{x}') -F(\mathbf{x}'')| \le 2\mathsf{P}(\|X\|\ge M). $$ At last, letting $M\to+\infty $, we get $$ \varlimsup_{\delta\to0}\sup_{\|\mathbf{x}'-\mathbf{x}''\|\le\delta}|F(\mathbf{x}') -F(\mathbf{x}'')| = 0, $$ and the $ F $ is uniformly continuous.

Answer 2

There is a following counterexample for the statement. Put $d=2$ and $A=\{(x_1,x_2)\in\Bbb R^2:x_1x_2\ge 0\}\cup \{(n,-n):n\in\Bbb Z\}$. Define a function $f:A\to\Bbb R$ putting $f(x_1,x_2)=0$, if $x_1 x_2\ge 0$, and $f(n,-n)=n^2$ for each $n\in\Bbb Z$. Since $A$ is a closed subset of a metrizable (and, hence, normal) space $\Bbb R^2$, by Tietze extension theorem the function $f$ can be extended to a continuous function on $\Bbb R^2$. Then $$\lim_{x_1,x_2\to +\infty} f(x_1,x_2)= \lim_{x_1,x_2\to -\infty} f(x_1,x_2)=0,$$ but $f$ is not uniformly continuous.

Answer 3

Counterexample: On $\mathbb R^2$ define

$$f(x,y)=\frac{x^2}{1+(y+x)^4}.$$

Let $\epsilon>0.$ Let $M=\dfrac{1}{\sqrt \epsilon}.$ Then for $x,y>M,$

$$\frac{x^2}{1+(y+x)^4} < \frac{x^2}{x^4} =\frac{1}{x^2} <\frac{1}{M^2} =\epsilon.$$

Thus $\lim_{x,y\to \infty} f(x,y)=0.$ Since $f(-x,-y)=f(x,y)$ we have the same result for $x,y\to-\infty.$

However, $f(x,-x) = x^2,$ so $f(x,y)$ is not uniformly continuous.