Let $\Omega\subseteq\mathbb{R}^2$, $(x_0,y_0)\in\Omega$, $f:\Omega\to\mathbb{R}$. Then $f$ is $\epsilon-\delta$ continuous at $(x_0,y_0)\in\Omega$ if, for any $\epsilon>0$, there exists a $\delta>0$ [$\delta$ depends on $\epsilon$ and $x_0$] such that \begin{align*}\tag{1} \forall x\in\Omega, |(x,y)-(x_0,y_0)|<\delta\Longrightarrow |f(x,y)-f(x_0,y_0)|<\epsilon \end{align*} Moreover, if (1) holds for all $(x_0,y_0)$ in $\Omega$, the $f$ is said to be continuous on $\Omega$.

Furthermore, if $\delta$ independent to $x_0$, the $f$ is said to be uniformly continuous on $\Omega$.

Consider the following example.

Let $\Omega\subseteq\mathbb{R}^2$ and $f: \Omega\to\mathbb{R}$ given by \begin{align*} \Omega :=\{(x,y)\in [0,1]\quad\text{and}\quad(x,y)\ne (0,0)\}\quad f(x,y)=\frac{1}{x+y} \end{align*} This is an example I saw in a book and it was claimed that $f$ is continuous but not uniform continuous.

I tried to show $f$ is continuous from definition and I got stuck. What I have so far is the following.

\begin{align*} \left\lvert\frac{1}{x+y}-\frac{1}{x_0+y_0}\right\rvert<\epsilon\quad\text{whenever}\quad 0<|x-x_0|<\delta,0<|y-y_0|<\delta \end{align*} Step 1 is to express $|f(x,y)-f(x_0,y_0)|$ in term of $|x-x_0|+|y-y_0|$. So I have \begin{align*} \left\lvert\frac{1}{x+y}-\frac{1}{x_0+y_0}\right\rvert&=\left\lvert\frac{x_0+y_0-(x+y)}{(x+y)(x_0+y_0)}\right\rvert\\ &=\left\lvert\frac{x_0-x+y_0-y}{(x+y)(x_0+y_0)}\right\rvert=\cdots \end{align*} I was stuck at this point. Hope anyone could help finishing the proof.


$$\left|\frac{1}{x+y}-\frac{1}{x_0+y_0}\right|=\left\lvert\frac{x_0+y_0-(x+y)}{(x+y)(x_0+y_0)}\right\rvert\\ =\left\lvert\frac{x_0-x+y_0-y}{(x+y)(x_0+y_0)}\right\rvert\leq \frac{|x-x_0|+|y-y_0|}{|x+y||x_0+y_0|}\leq \frac{2\delta}{(|x_0+y_0|-2\delta)|x_0+y_0|}\to 0$$ when $\delta\to 0$

Because, $$-\delta<x-x_0<\delta\\ -\delta<y-y_0<\delta$$ $$\Rightarrow -2\delta<x+y-(x_0+y_0)<2\delta$$ $$|(x+y)-(x_0+y_0)|<2\delta\Rightarrow |x_0+y_0|-|x+y|\leq |(x+y)-(x_0+y_0)|<2\delta$$

Note that $|x_0+y_0|-2\delta >0$ for $(x_0,y_0)\neq (0,0)$ and $\delta$ small enough. It is not uniformly continuous, because $f$ is unbounded on the bounded set $[0,1]\times [0,1]\setminus \{(0,0)\}$ -see Does there exist an unbounded function that is uniformly continuous?

If you assume that $f(x,y)=\frac{1}{x+y}$ is uniformly continuous in $[0,1]\times[0,1]\setminus \{(0,0)\}$, then for any $\epsilon>0 \,\exists\delta >0: \|(x_1,y_1)-(x_2,y_2)\|_{l^2}<\delta$ (implying also) $|x_1-x_2|<\delta$ and $|y_1-y_2|<\delta\Rightarrow |f(x_1,y_1)-f(x_2,y_2)|<\epsilon$.

In $\|(x_1,y_1)-(x_2,y_2)\|<\delta$ you can take whatever norm $\|.\|$ in $\mathbb R^2$ you want, because all norms on finite dimensional spaces are equivalent and you will still have the implication $|x_1-x_2|<C\delta$ and $|y_1-y_2|<C\delta$, where $C$ is a constant coming from the equivalence between the norm you chose and the $\|.\|_\infty$ norm.

Now, take $\epsilon=1$ and find the corresponding $\delta$. Take for example $y_1=y_2=0$ ($y_1=y_2=1$ or any other $y_1=y_2\in[0,1]$ does the job, too). Then you have $|f(x_1,0)-f(x_2,0)|<1$ for $|x_1-x_2|<\delta$ (note that $|y_1-y_2|=0<\delta$). Now, take $0<x_2<\delta$ -fixed and an arbitrary $0<x_1<\delta$ so that $0<x_1,x_2<\delta$. Clearly you have $|x_1-x_2|<\delta$ which implies $|f(x_1,0)-f(x_2,0)|<1\,\,\forall x_1\in (0,\delta)$. But this is not true since letting $x_1\to 0$ you get $|f(x_1,0)-f(x_2,0)|\to \infty$.

Finally, it is seen that the point $(0,0)$ prevents the function of being uniformly continuous. If you remove this point together with a whole $\rho>0$ neighbourhood (for example a small ball $B_0(\rho)$ around the origin), then $f$ is uniformly continuous on $[0,1]\times[0,1]\setminus B_0(\rho)$

  • $\begingroup$ Could you expand your answer that $f(x,y)$ is Not uniform continuous? $\endgroup$
    – math101
    Oct 2 '15 at 11:05
  • $\begingroup$ I added the uniform continuity to my answer. $\endgroup$
    – Svetoslav
    Oct 2 '15 at 11:36
  • $\begingroup$ okay, i need to go home and digest your answer, will be back $\endgroup$
    – math101
    Oct 2 '15 at 11:52

For the second claim that $f(x,y)$ fails to be uniform continuous, as an alternative, I would like to prove using sequences. Consider the sequences $((x_n,y_n))$ and $((u_n,v_n))$ in $D$ given by $$(x_n,y_n):=(\frac{1}{n+1},0),\quad\text{and}\quad (u_n,v_n):=(1/n,0)\quad\forall\,\,n\in\mathbb{N}$$ We have $$|(x_n,y_n)-(u_n,v_n)|=\frac{1}{(n+1)n}\to 0,$$, but $$|f(x_n,y_n)-f(u_n,v_n)|=|n+1-n|=1\not\to 0$$


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