Let, $f,g:(0,1)\times (0,1)\to \mathbb R$ be two continuous functions defined by $f(x,y)=\dfrac{1}{1+x(1-y)}$ and $g(x,y)=\dfrac{1}{1+x(y-1)}$. Then which is correct?

  1. $f$ and $g$ both are uniformly continuous.
  2. $f$ is uniformly continuous but $g$ is not.
  3. $g$ is uniformly continuous but $f$ is not.
  4. neither $f$ nor $g$ is uniformly continuous.

From definition, I can't show the uniform continuous. So how we can show the uniform continuity?

We know that for a function $F\colon (a,b)\to \mathbb R$ is uniformly continuous if $F$ is continuous in $(a,b)$ and the limit of $F$ exists at $a$ as well as at $b$.

Is it applicable for this problem?

That means , I want to say that both the functions are continuous on $(0,1)\times (0,1)$. If we can show that the limit exists at $(0,0)$ and at $(1,1)$ then can we say that the functions are uniformly continuous?

Please suggest me about the problem.

  • $\begingroup$ Can you check whether they are continuous atleast? $\endgroup$
    – user87543
    Jan 4 '15 at 16:45
  • $\begingroup$ Yes.they are continuous. $\endgroup$
    – Empty
    Jan 4 '15 at 16:47
  • 3
    $\begingroup$ how did you prove that? "$F:(a,b):→ℝ$ is uniformly continuous if F is continuous in $(a,b)$ & the limit of F exists at a as well as at b" does not work here becasue then you should check existence of limit on whole boundary and not just on $(0,0)$ and $(1,1)$ $\endgroup$
    – user87543
    Jan 4 '15 at 16:49
  • $\begingroup$ Seeing the functions we can say that there are no singularity in whole of the domain. $\endgroup$
    – Empty
    Jan 4 '15 at 16:52

Note that $x(1-y)\geq0$ on the compact set $K:=[0,1]^2$. It follows that $$f(x,y):={1\over 1+x(1-y)}$$ is continuous on $K$, whence uniformly continuous on $K$. A fortiori $f$ is uniformly continuous on the interior of $K$.

Things are different with $g$: Consider the sequence ${\bf z}_n:=\left(1-{1\over n},{1\over n}\right)$ $\>(n\geq1)$. This sequence is obviously convergent with limit $(1,0)$. One computes $$g({\bf z}_n)=g\left(1-{1\over n},{1\over n}\right)={1\over 1+\bigl(1-{1\over n}\bigr)\bigl({1\over n}-1\bigr)}={n\over2}{1\over 1-{1\over 2n}}\ .$$ This shows that $g({\bf z}_{n+1})-g({\bf z}_n)\geq1$ for all large enough $n$, whereas at the same time ${\bf z}_{n+1}-{\bf z}_n\to 0$. It follows that $g$ cannot be uniformly continuous on the interior of $K$.

  • $\begingroup$ As you may have noticed in my answer below $f$ is even a Lipschitz function. $\endgroup$
    – aly
    Mar 30 '15 at 13:16

Let $(x_1,y_1),(x_2,y_2)\in(0,1)\times(0,1)$. Then \begin{align*} |f(x_1,y_1)-f(x_2,y_2)|&=\frac{|x_1(y_1-y_2)+(y_2-1)(x_1-x_2)|}{(1+x_1(1-y_1))(1+x_2(1-y_2))}\\ &\le\frac{x_1}{1+x_1(1-y_1)}|y_1-y_2|+\frac{1-y_2}{1+x_2(1-y_2)}|x_1-x_2|\\ &\le\frac{1}{1+x_1(1-y_1)}|y_1-y_2|+\frac{1}{1+x_2(1-y_2)}|x_1-x_2|\\ &\le|y_1-y_2|+|x_1-x_2|\\ &\le\sqrt{2}\|(x_1,y_1)-(x_2,y_2)\|. \end{align*} These inequalities show that $f$ is a Lipschitz function. Hence it is also uniformly continuous.

By the other hand $g$ is not uniformly continuous. We can use the well known characterization of uniform continuity via sequences. More precisely we will find sequences $(u_n)$ and $(v_n)$ auch that $u_n-v_n\to 0$ and $g(u_n)-g(v_n)\not\to 0$. Let $u_n=(1-\frac{1}{n},\frac{\sqrt{n}-1}{n-1})$ and $v_n=(1-\frac{1}{n},\frac{\sqrt{n}-2}{2(n-1)})$ ($n\ge 5$). Then \begin{equation*} \|u_n-v_n\|=\frac{\sqrt{n}}{2(n-1)}\to 0\text{ and }g(v_n)-g(u_n)=\sqrt{n}\to\infty. \end{equation*}

The correct answer is (b).

  • 1
    $\begingroup$ I'm not so sure about $g$. It becomes unbounded near (1, 0). $\endgroup$
    – D Poole
    Mar 30 '15 at 4:20


  • A uniformly continuous function on a bounded domain must be bounded (can you show why?)
  • See if you can bound one of your functions by something of the form $|h(x,y_0) - h(z,y_0)| < K_1|x-z|$ and $|h(x_0,y) - h(x_0,w)| < K_2|y-w|$. Could you combine these inequalities to show your function is uniformly continuous?

Let's justify my first hint. Suppose $h$ is a uniformly continuous function on a bounded domain $D \subset \mathbb{R}^n$ which maps to $\mathbb{R}$. Then, there must exists a $\delta$ such that when $|x-y| < \delta$, $|h(x) - h(y)| < 1$. Consider such a $\delta$. Then, $\overline{D}$ is compact, so the cover $\{B_\delta(x)| x \in D\}$ consisting of balls of radius $\delta$ about every point in $D$ has a finite subcover, call it $U = \{B_\delta(x_1), \cdots, B_\delta(x_n)\}$. But then, by construction, $$h(D) \subseteq \bigcup_{k=1}^n B_1(h(x_k)),$$ and since the left hand side is bounded, the image of $h$ is bounded.

Notice that $g$ is unbounded at $(x,y) \to (1,0)$. From our previous argument, $g$ cannot be uniformly continuous.

For $f$, observe that we can extend $f$ to the function $\tilde{f}:[0,1]\times[0,1] \to \mathbb{R}$ (since $1+x(1-y)$ does not vanish on the boundary of the domain). Then, $\tilde{f}$ is continuous, and since it is defined on a compact set, it is uniformly continuous. Thus, $f$, which is the restriction of $\tilde{f}$ to $(0,1)\times(0,1)$, is also uniformly continuous.

(I probably thought you could use my second hint to prove $f$ was uniformly continuous when I posted it several months ago. I still think you could, but I find the above argument much simpler).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.