Is function $f(x, y, z)= \frac{x^2(1-y)}{1+y}$ uniformly continuous?

Question

Can you please help me with this problem. I have function: $$f(x, y, z)= \frac{x^2(1-y)}{1+y}$$ (I didn't forget $z$, it is not in the expression)

I am having trouble figuring out whether the function $f$ is uniformly continuous or not. I know that a function is uniformly continuous if it is continuous on its domain and if the domain is a closed and bounded set (this is the only method I have in my book). But in this case domain is: $$D:=\{x, y, z \in R^3 \ | \ y\neq-1\}$$ So I'm stuck. Can you maybe help me explain how I can solve this problem using the definition? Can I use some other simpler methods (if there are any)?

Answer 1

No, it is not uniformly continuous. Let $g(y)=\frac{1-y}{1+y}$ and note that $g(y)=f(1,y,0)$. Therefore, if $f$ was uniformly continuous, then $g$ would be uniformly continuous too. But it isn't. Take $\delta>0$. Since $\lim_{y\to-1^+}g(y)=+\infty$, it is easy to find two numbers $x,y\in(-1,-1+\delta)$ such that $\bigl|g(x)-g(y)\bigr|\geqslant1$. Therefore, if you take $\varepsilon=1$,$$(\forall\delta>0)(\exists x,y\in\mathbb{R}\setminus\{-1\}):|x-y|<\delta\wedge\bigl|g(x)-g(y)\bigr|\geqslant\varepsilon.$$This is the negation of the uniform continuity of the function $g$.

Answer 2

If $y\to -1$ then (for $x\ne 0$) the distances $|f(x,y_1)-f(x,y_2)|$ could be arbitrarily large.

Answer 3

In the bounded set $\{1\}\times (-1,0)\times \{0\},$ $f$ is unbounded (the limit as $y\to -1^+$ is $\infty$). Such a function is never uniformly continuous.