Suppose $f : X \to Y$ is a (continuous) bounded map.Does this implies that $f$ is uniformly continuous?

Question

It's well known that if $ f : \bf (X,d) \to \bf (Y,e) $ is a uniformly continuous function then $f$ maps bounded set to bounded set.Does the converse hold ? More Precisely,

Suppose $f : X \to Y$ is a (continuous) bounded map (i.e. $f$ maps bounded set to bounded set).Does this implies that $f$ is uniformly continuous?

Answer 1

The classic function $f:(0,1) \to [0,1]$ $$f(x) = \sin(1/x)$$ is a counterexample.

Answer 2

There is a well known lemma saying that if a function $f$ defined on $\mathbb{R}$ such that $\int_0^{+\infty}f(x)dx$ converge and $f$ is uniformly continuous then $lim_{x\rightarrow +\infty} f(x)dx = 0$.

Consider the function $f:\mathbb{R}\rightarrow \mathbb{R}$ defined by $f(x)=\sin(x^2)$. It satisfies your hypothesis and the integral of $f$ converges and its limit at $+\infty$ is not $0$ hence it's not uniformly continuous.