# Neccesary condition for uniform continuity under norm induced metric spaces.

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!!

• 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 7 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 7 at 23:24
• @RobArthan I'm aware of the notion that this is applicable in normed vector space only. – Saptarshi Sahoo Apr 8 at 6:04

## 2 Answers

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 $$n < |f(x_n + h) - f(x_n)| < \frac{1}{n}$$.

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