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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $K$.

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

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

  • 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).

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