0

For a $f:X \rightarrow Y$
Where $(X, d)$ and $(Y, \rho)$ are metric spaces.
Standard $\epsilon \text{-} \delta$ definition of uniform continuity:
Suppose $\emptyset \ne A \subset X$, $f$ is said to be uniformly continuous on $A$ if $\forall \epsilon \gt0 \exists \delta \gt0$ s.t $\forall x \in A, y \in X$ $d(y,x) \lt \delta \implies \rho(f(y), f(x)) \lt \epsilon$.

While perusing through a problem in math StackExchange, I came to this result :-
If for every $h>0$ we have that $|f(x+h)−f(x)|$ is unbounded on $I$, then $f$ is not uniformly continuous on $I$.

How to prove the above statement.
Thanks in advance!!

Saptarshi Sahoo
  • 93
  • The statement doesn't apply in arbitrary metric spaces, but only to normed vector spaces. What context are you working in? Note also that it is not an alternative definition: the converse statement is false. – Rob Arthan Apr 07 '21 at 22:31
  • That is only a necessary condition for uniform continuity. You cannot call it an alteranative definition of uniform continuity. – Kavi Rama Murthy Apr 07 '21 at 23:24
  • @RobArthan I'm aware of the notion that this is applicable in normed vector space only. – Saptarshi Sahoo Apr 08 '21 at 06:04

2 Answers2

1

Suppose $f$ is u.c. Then for small $h > 0$ we have $|f(x + h) - f(x)| < \frac{1}{n}$.

But for $h > 0$ and $n \ge 1$ there is a $x_n \in I$ such that $n < |f(x_n + h) - f(x_n)|$.

It follows that if $h>0$ is small, then $n < |f(x_n + h) - f(x_n)| < \frac{1}{n}$.

Lázaro Albuquerque
  • 2,464
0

A bounded continuous function can fail to be uniformly continuous. Let $X=\Bbb R^+$ and $Y=\Bbb R$ and $f(x)=\cos 1/x.$ Then $|f(x)|\le 1$ and $|f(x+h)-f(x)|\le |f(x+h)|+|f(x)|\le 2.$ But $f$ is not uniformly continuous. For if $0<\epsilon<2$ then for any $\delta >0,$ take $n\in \Bbb N$ with $2n\pi >1/\delta,$ and we have

$|1/2n\pi -1/(2n+1)\pi|<\delta$

but $|f(1/2n\pi)-f(1/(2n+1)\pi)|=2>\epsilon.$

DanielWainfleet
  • 59,529