Prove that the function $f(x,y)=\frac{1}{x^2+y^2}$ is not uniformly continuous over the domain:

$D$={$(x,y):x^2+(y-2)^2 <2^2$}

Using the definition:

$\forall \delta>0 , \exists \epsilon(\delta) > 0 / \forall (x_1,y_1) , (x_2,y_2) \in D : 0< ||(x_1,y_1),(x_2,y_2)||<\delta \implies |f(x_1,y_1)-f(x_2,y_2)|>\epsilon$

Let :



Conditions on $\delta$:

$x_1^2+(y_1-2)^2<2^2 \implies 0<\delta<8$

$x_2^2+(y_2-2)^2<2^2 \implies 0<\delta<8$


$\forall \delta>0 , \exists \epsilon(\delta) \in ]0,\frac{12}{\delta^2}[ / \forall (x_1,y_1) , (x_2,y_2) \in D : 0< ||(x_1,y_1),(x_2,y_2)||<\delta \implies |f(x_1,y_1)-f(x_2,y_2)|>\epsilon$

However I'm not sure if this is correct nor I am sure if It's possible to put conditions on $\delta$ while proving uniform continuity.

I would be grateful to whoever can point out my mistakes or whoever has a much cleaner solution to this question.

Thanks in advance.


Your mistakes

  1. The order of $\epsilon$ and $\delta$ is inverted in your proof.
  2. You've made a typo in the coordinates of $(x_2,y_2)$. It should be $(0,\dfrac{\delta}{4})$.

Your proof continued

The statement to be proved is

$$\exists\, \epsilon_0 > 0, \forall\, \delta > 0, \exists\, (x_1,y_1), (x_2,y_2) \in D : 0< ||(x_1,y_1)- (x_2,y_2)||<\delta \implies |f(x_1,y_1)-f(x_2,y_2)|>\epsilon_0.$$

Fix $\epsilon_0 > 0$ at the beginning.

From your choice of $(x_1,y_1), (x_2,y_2)$, you have $|f(x_1,y_1)-f(x_2,y_2)|=\dfrac{12}{\delta^2}$.

$$\frac{12}{\delta^2} > \epsilon_0 \iff 0 < |\delta| < \sqrt{\frac{12}{\epsilon_0}} \tag1 \label1$$

When $\delta$ satisfies \eqref{1}, we're done. Otherwise, for "large values" of $\delta$, since the quantifier of $(x_1,y_1), (x_2,y_2)$ is "there exists", you can choose points $(x_1,y_1), (x_2,y_2)$ closer to the origin with a smaller $\delta' < \delta$ so that $$0<||(x_1,y_1)- (x_2,y_2)||<\delta'<\delta.$$ This proves the statement for all $\delta > 0$.

A shorter answer

Take a sequence in $D$: $(x_n, y_n) = (0, \dfrac1n), n \in \Bbb{N}$. Then

$$\lim_{n\to\infty} ||(x_n,y_n) - (x_{2n},y_{2n})|| = \lim_{n\to\infty} \frac{1}{2n} = 0,$$


$$\lim_{n\to\infty} ||f(x_n,y_n) - f(x_{2n},y_{2n})|| = \lim_{n\to\infty} |n^2 - (2n)^2| = \lim_{n\to\infty} 3n^2 = \infty,$$

so $f$ is not uniformly continuous on $D$.


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