uniformly continuous of multivariable function

Art of Problem Solving

AoPS Online

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

uniformly continuous of multivariable function

G H J

Reveal topic tags

real analysis unsolved

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

9 posts

#1 h Nov 2, 2012, 4:34 AM • 2 Y

Y by Adventure10, Mango247

How can I determine which of the following functions are uniformly continuous on the given domain A?

$f \left( x, y \right) = \frac{x^3 + y^2}{x^2 + y}$ , $A = \left\{ \left( x, y \right) \in \mathbb m{R}^2 \left| \right. \left| y \right| \leq \frac{x^2}{2} %Error. "nocomma" is a bad command. , x^2 + y^2 < 1 \right\}$

$g \left( x, y \right) = \frac{y^2 + 4 x^2}{y^2 - 4 x^2 - 1}$ , $A = \left\{ \left( x, y \right) \in \mathbb m{R}^2 \left| 0 \leq x^2 - y^2 \leqslant 1 \right\} \right.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

895 posts

#2 h May 14, 2025, 1:42 PM

Y by

For $f(x,y)$ , the domain $A$ is comprised of a region that is bounded between the parabolas $-\frac{x^2}{2} \le y \le \frac{x^2}{2}$ and the circle $x^2+y^2<1$ . Furthermore, $f$ is defined for all $(x,y) \in \mathbb{R}^{2} \backslash \{y=-x^2\}$ . However, $y=-x^2$ intersects both $y = \pm\frac{x^2}{2}$ at $(0,0) \in A$ , and $\lim_{(x,y) \rightarrow (0,0)} f(x,y) = DNE$ (does not exist). Thus, $f$ is not uniformly continuous over $A$ due to $(x,y)=(0,0)$ .

For $g(x,y)$ , the domain $A$ is comprised of a region that is bounded between the lines $-x \le y \le x$ and the right branch of the hyperbola $x^2-y^2 \le 1$ (which coincidentally has asymptotes $y = \pm x$ ). Furthermore, $g$ is defined for all $(x,y) \in \mathbb{R}^{2} \backslash \{y^2-4x^2=1\}$ , which the intersection of this latter hyperbola and $A = \varnothing$ . Hence, $\lim_{(x,y) \rightarrow (p,q)} g(x,y) = L$ exists for any $(x,y) = (p,q) \in A$ , and $g$ is uniformly continuous over $A$ .

This post has been edited 12 times. Last edited by Mathzeus1024, Yesterday at 4:23 PM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a