Show that $f$ is not uniformly continuous

Question

let $f:\mathbb{R}\rightarrow\mathbb{R}$

There exists a real number $c$ and two sequences $x_n, y_n$ such that

• $\forall$ $n$: $|x_n-y_n|=c$

• $|f(x_n)-f(y_n)|\xrightarrow[n \to \infty]{}\infty$

Prove that $f$ is not uniformly continuous on $\mathbb{R}$

Hi, i have been trying to solve this question but i keep hitting a brick wall.

What i tried to prove:

there exists an $\varepsilon_0$ such that for all $\delta>0$ there exists $x,y$ such that $|x-y|<\delta$

but: $|f(x)-f(y)|\geq\varepsilon_0$

So i chose $\varepsilon_0=1$, let $\delta>0$.

If $\delta>c$:

There exists $N$ such that for all $n>N$ : $|f(x_n)-f(y_n)|>1$

let $n>N$:

$|x_n-y_n|=c<\delta$

and $|f(x_n)-f(y_n)|>1$

so $f$ is not uniformly continuous

If $\delta\leq c$: this is where i got stuck.

Answer 1

Hint: Suppose for every $\epsilon >0$ there exists $ \delta >0$ such that $|f(x)-f(y)| <\epsilon$ whenver $|x-y| <\delta$. Take $\epsilon=1$ in this.

W.l.o.g. we may suppose $x_n <y_n$ for each $n$. There exists a partition $(t_{i,n})$ of $[x_n,y_n]$ such that $|t_{i,n}-t_{i-1,n}| <\delta$ for each $i$ and the number of subintervals in the partiton is bounded w.r.t. $n$. [This is where $|x_n-y_n|=c$ is used]. Now use triangle inequality and the fact that $|f(t_{i,n})-f(t_{i-1,n})|<\epsilon$ to show that $|f(x_n)-f(y_n)|$ is bounded.

Answer 2

Suppose $f$ is uniformly continuous. Let $\delta' > 0$ be as given in the definition of uniform continuity, corresponding to $\varepsilon = 1$. Put $\delta := \delta'/2$.
Define $N$ be the smallest positive integer such that $N\delta \geqslant c$.

Claim. If $|y - x| = c$, then $|f(y) - f(x)| < N$.

Proof. Let $x, y \in \Bbb R$ be arbitrary such that $y = x + c$. Consider the points $$x, x + \delta, \ldots, x + (N - 1)\delta, y.$$ The consecutive points are $< \delta'$ away in distance and thus, applying triangle inequality repeatedly gives \begin{align} |f(y) - f(x)| &\leqslant |f(y) - f(x + (N - 1)\delta)| + \cdots + |f(x + \delta) - f(x)| \\ &<1 + \cdots + 1 = N. \qquad \Box \end{align}

From this, it follows that a sequence like yours cannot exist.