Example of a continuous but not uniformly continuous function $f:Y\to Y$ on a complete convex bounded subset $Y$ of a metric space $X$.

Question

Does there exist a function on a complete convex bounded subset of a metric space (or normed linear space) which is continuous but not uniformly continuous?

I have tried to find such function but observed that first I need to consider non-compact closed bounded and convex subset. Since there is the convexity condition, I looked into Banach spaces. However, since the closed unit ball of a finite-dimensional Banach space is complete bounded convex but compact at the same time, there is no hope of getting continuous function on the closed unit ball which is not uniformly continuous.

On the other hand, in infinite-dimensional spaces the only functions I am familiar with, are bounded linear operators. However, every bounded linear operator is uniformly continuous.

Please help me with this problem. A detailed answer will be of very much help. Thanks in advance!

Answer 1

Take $(Y,d)=(\mathbb R,\delta )$ where $$\delta (x,y):=\min\{|x-y|,1\},$$ and $f(x)=x^2$. Then $f$ is continuous on $\mathbb R$ (which is bounded and convex) but it's not uniformly continuous on $\mathbb R$.

Answer 2

Let $(c_0,\|\cdot\|)$ be the Banach space of all real sequences $x=(\xi_k)_{k=1}^\infty$ with limit $0$ endowed with the maximum norm, $e_n =(0,\dots ,0,1_n,0,0,\dots)$ for $n \in \mathbb{N}$, $B$ the closed unit ball and $f:B \to B$ defined as $$ f(x)=(\xi_1,\xi_2^2,\xi_3^3,\xi_4^4\dots). $$ Then $f$ is continuous (check) but not uniformly continuous: Let $\varepsilon=1/2$, $\delta \in (0,1)$ and $n \in\mathbb{N}$. Then $\|e_n-(1-\delta)e_n\|=\delta$ and $\|f(e_n)-f((1-\delta)e_n)\|=1-(1-\delta)^n \ge 1/2$ for $n$ sufficiently big.