Proving $1\over x^2$ is not uniformly continuous

Question

I need to show that $1 \over x^2$ is not uniformly continuous on the interval $(0,2]$ using the definition of uniform continuity.

Definition of Uniform Continuity on a set A:

Let $A \subseteq \Bbb R$, let $f:A \rightarrow \Bbb R$.

$f$ is uniformly continuous if $ \forall \epsilon > 0$ $\exists \delta > 0$ such that if for any $x \in A$, $x$ satisfies the inequality:

$|x - u| < \delta \implies |f(x) - f(u)| < \epsilon$

To prove this function is not uniformly continuous by the given definition, I negated the definition such that:

$f$ is not uniformly continuous on the interval $(0,2]$ whenever

$\exists \epsilon_0 > 0$ $ \forall \delta > 0$ $\exists$ $x_\delta \in A$ satisfies $|x_\delta-u_\delta| < \delta $ but $|f(x_\delta) - f(u_\delta)| > \epsilon_0$

I chose to let $\epsilon_0 = 2$

For any $ 0< \delta < 2$ choose $x_\delta = \delta$ and $u_\delta$ = $\delta \over 2$

Then $|x_\delta - u_\delta|$ = $|\delta -$ $\delta \over 2|$ = $\delta \over 2$

But $|f(x_\delta) - f(u_\delta)|$ = |$1 \over \delta^2$ - $\frac 1 1 \over \delta^2$ | = |$(\frac 1\delta)^2 - \frac 2\delta^2)|$ = $3 \over \delta^2$ $> 2 = \epsilon_0$

But on the interval $(0,2]$ this does not always hold true.

Can anyone help me with this? I am thinking that my choice of $\delta$ is wrong but I'm not sure.

Answer 1

Here is another way:

If $f$ was uniformly continuous, then for some $\delta>0$ we would have $f(x)-f(x+\delta) < 1$ for all $x$ such that $x,x+\delta \in (0,2]$.

Let $f(x) = {1 \over x^2}$, and let $\phi(x) = f(x)-f(x+\delta)$ for $x \in (0,2-\delta)$.

Since $f$ is bounded on any compact subinterval of $(0,2]$, we see that $\lim_{x \downarrow 0} \phi(x) = \infty$. Hence $f$ is not uniformly continuous.

Answer 2

Here a full answer (that i writte too to practice) but take into account that I am just a student so I hope it is correct.

1 - First let recall the definition of a non uniformly continuous function.
It exists at least one $\epsilon_0>0$ such that for every $\delta>0$ that we can choose it will always exists at least $x$ and $y$ that verifies $|x-y|<\delta$ but $|f(x)-f(y)|>\epsilon_0$.
More formally: $\exists \epsilon_0>0 \; , \forall \delta>0 \; : \; \exists |x-y|< \delta \Rightarrow |f(x)-f(y)| \geq \epsilon_0$

2 - Now let pay attention to the following inequalities:
(1): for any $\delta>0$ given it exists $N=Max(1; \left \lceil 1/ \delta \right \rceil)$ s.t. $1/N < \delta$ . Moreover all $n \geq N$ verifies too this inequality (by assumption we are in $(0; 2]$ ).
(2): $\forall n \in \mathbb{N} $ we have $|\frac{1}{(1/n)^2}-\frac{1}{(n+1/n)^2}| > 1/2$

3 - Now we can writte:
$\exists \epsilon_0 = \frac{1}{2}>0$ such that for any $\delta>0$ it will always exists ,with $n \geq N$ as define in (1), at least two points $x_n=n$ and $y_n=n+1/n$ that despite that verifying $|x_n-y_n|=|1/n| < \delta$ (by (1)) $ \Rightarrow|f(x_n)-f(y_n)|=|\frac{1}{(1/n)^2}-\frac{1}{(n+1/n)^2}|>1/2$.
More formally: $\exists \epsilon_0 = \frac{1}{2}>0 \; , \forall \delta>0 \; : \; \exists \; x_n = n, \; y_n=n+\frac{1}{n}$ with $n \geq max(1; \left \lceil 1/ \delta \right \rceil)$
By (1): $|x_n-y_n|< \delta$
By (2): $\Rightarrow |f(x_n)-f(y_n)|=|\frac{1}{(1/n)^2}-\frac{1}{(n+1/n)^2}| \geq \epsilon_0 = \frac{1}{2}$
Q.E.D.