Are uniformly continuous function $f:\mathbb{R}^m \to \mathbb{R}^m$ bounded?

Question

I'm trying to understand how uniform continuity relates to the solutions of ODEs, and I have a hunch that because uniformly continuous functions are 'well-behaved' functions, the solutions to uniformly continuous ODEs do not escape to infinity in a finite time.

One attempt at proving this required the condition that the range of a function $f:\mathbb{R}^m \to \mathbb{R}^m$ given to be uniformly continuous is bounded (i.e. it takes on some maximum value $K$, and $f(x) \leq K$ for all $x$ in its domain); is this a valid assumption to make?

My current rationale for making this assumption is this: since $f$ is given to be uniformly continuous, it can't tend toward infinity on its range, because this would allow us to choose larger and larger $\epsilon$ values that contradict the uniform continuity hypothesis for a given $\delta$.

Any clarification would be much appreciated!

Answer 1

As pointed out by others, no, $f(x)=x$.

But it's worth analyzing

My current rationale for making this assumption is this: since $f$ is given to be uniformly continuous, it can't tend toward infinity on its range, because this would allow us to choose larger and larger $\epsilon$ values that contradict the uniform continuity hypothesis for a given $\delta$.

because you have the definition of [uniform] continuity sort of the wrong way round.

You specify an $\epsilon$. I have to tell you a $\delta$ such that "$f$ doesn't vary more than $\epsilon$ over regions of size $\delta$". In particular, it is always easier to satisfy this if you give me a bigger epsilon, because $\epsilon$ is an upper bound on something I have to tell you.

However if you're interested in showing a function cannot grow too big (or escape to infinity) in finite time then this is a true property of uniformly continuous functions, in the following precise sense:

Take any point $x$ in the domain of $f$. Then $\lVert f(y) - f(x)\rVert < C \lVert y-x\rVert$ is bounded by some constant multiplied by the distance from the starting point. Hence the function grows at most linearly.

Proof. Let $\delta$ correspond to $\epsilon = 1$. Now given $y,x$ we can choose a finite sequence of points $x_n$ between $x,y$ never more than $\delta$ apart, where there are at most $2\lVert x - y \rVert/\delta$ points in the sequence, say. Hence by the triangle inequality $\lVert f(y) - f(x) \rVert < \epsilon \times 2\lVert x - y \rVert/\delta$. Setting $C = 2/\delta$ gives the result.

Answer 2

If I understand your question correctly, the answer is extremely "no." Your sentence

since $f$ is given to be uniformly continuous, it can't tend toward infinity on its domain

is false; consider the identity function $f(\overline{x})=\overline{x}$. $f$ is uniformly continuous (just take $\delta=\epsilon$) , but clearly unbounded.

Or are you asking something else?

Answer 3

Firstly, it is not true that a uniformly continuous function is bounded. For instance, the identity function $f: \mathbb{R} \longrightarrow \mathbb{R}$ sending $f:x \mapsto x$ is uniformly continuous (one can always take $\delta = \epsilon$ in the classic $\delta$-$\epsilon$ formulation of uniform continuity), but is clearly not bounded.

Secondly, it does not make sense to talk about "uniform boundedness" of a single function. A function is bounded or it is not. But a family of functions can be uniformly bounded (this means that every function in the family is bounded by the same bounds).

Answer 4

This is not true. Let $f(x)=x$ and fix $\epsilon>0$. If $\vert x-y\vert<\epsilon$ then $\vert f(x)-f(y)\vert<\epsilon$. Thus $f$ is uniformly continuous but it's clearly not bounded.