Non-Uniform Continuity Criteria

Let $A\subseteq\mathbb{R}$, $f:A\rightarrow \mathbb{R}$. Then, TFAE:

1. $f$ is not uniformly continuous on $A$
2. There exists an $\varepsilon _0>0$ such that for every $\delta >0$ there are points $x_\delta, u_\delta\in A$ such that

$\left| x_{s}-u_{s}\right| <\delta$ and $\left| f\left( x_\delta\right) -f\left( u_\delta\right) \right| \geq\varepsilon _0$

1. There are $\varepsilon >0$ and two sequence $x_n,y_n\in A$ such that $lim (x_n-y_n)=0$ and $\left| f\left( x_{n}\right) -f\left( y_n\right) \right| \geq \varepsilon _{0}$ for all $n\in\mathbb{N}$

Firstly, I want to show that $1)$ $\implies$ $2)$ but I couldn't do anything because I think by the definition it should be clear. Can you give a hint/help?

• 2. Is just the negation of the definition of uniform continuity. There's nothing more to it. – JH vd Walt Dec 28 '17 at 10:22

I think you third clause should be:

There are $\epsilon>0$ and two sequences $(x_n),(y_n)\subseteq A$ such that $\lim\limits_{n\to\infty}(x_n-y_n)=0$ and $|f(x_n)-f(y_n)|\geq\epsilon_0$ for all $n\in\mathbb{N}$.

Now, let us prove it.

• $1\Rightarrow 2$. Well, let us rememeber the definition of uniform continuity on a set $A\subseteq\mathbb{R}$:

Uniform Continuity: A function $f$ is called uniformly coniunuous on a set $A\subseteq\mathbb{R}$ if: $$\forall\ \epsilon>0\ \exists\ \delta>0:\ \forall\ x\ \forall\ y\ (x,y\in A\text{ and }|x-y|<\delta)\Rightarrow|f(x)-f(y)|<\epsilon$$

Now, let us consider the logical negation of the above proposition (so every $\exists$ becomes a $\forall$ and vice versa and we also negate the final clause):

Negation of the definition: A function $f$ is not uniformly continuous on a set $A\subseteq\mathbb{R}$ if: $$\exists\ \epsilon_0>0\ \forall\ \delta>0\ \exists\ x_\delta\ \exists\ y_\delta\ (x_\delta,y_\delta\in A\text{ and }|x_\delta-y_\delta|<\delta)\text{ and }|f(x_\delta)-f(y_\delta)|\geq\epsilon_0$$ Which is the wanted result.

• $2\Rightarrow 3$. Well, given the clause 2, we have that there exists an $\epsilon_0>0$ such that for every $\delta>0$ there exist $x_\delta,y_\delta\in A$ such that: $$|x_\delta-y_\delta|<\delta\text{ and }|f(x_\delta)-f(y_\delta)|\geq\epsilon_0$$ Now, consider $\delta=\frac{1}{n}$, $n\in\mathbb{N}$. Then there exist $x_n,y_n\in A$ such that: $$|x_n-y_n|<\frac{1}{n}\text{ and }|f(x_n)-f(y_n)|\geq\epsilon_0$$ So, we have two sequences $(x_n),(y_n)\subseteq A$ such that: $$\lim_{n\to\infty}|x_n-y_n|=0\Leftrightarrow\lim_{n\to\infty}(x_n-y_n)=0$$ and $$|f(x_n)-f(y_n)|\geq\epsilon_0$$ which is the requested.
• $3\Rightarrow 1$. Considering that clause 3 is true, let us suppose that clause 1 is false, so let us suppose that $f$ is uniformly continuous. Then for $\epsilon=\epsilon_0$ (the one given in the third clause) there exists a $\delta>0$ such that for every $x,y\in A$ we have: $$|x-y|<\delta\Rightarrow|f(x)-f(y)|<\epsilon_0$$ From our hypothesis there exist two sequences $(x_n),(y_n)\subseteq A$ such that: $$x_n-y_n\to0\text{ and }|f(x_n)-f(y_n)|\geq\epsilon_0$$ For the found $\delta>0$ there exists a $n_0\in\mathbb{N}$ such that: $$|x_n-y_n|<\delta\ \forall\ n\geq n_0$$ Now, for every $n\geq n_0$, from clause 3 we have that: $$|f(x_n)-f(y_n)|\geq\epsilon_0$$ From our hypothesis about unifrom continuity we have that: $$|f(x_n)-f(y_n)|<\epsilon_0$$ which is a contradiction. So $f$ is not uniformly continuous on $A$.

Having proved $1\Rightarrow2\Rightarrow3\Rightarrow1$ the proof is complete.