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Uniform continuous

Let f:[0, +\infty) \rightarrow \mathbb{R} be a continuous function such that for each h>0 it holds

    \[\lim_{x \rightarrow +\infty} \left( f(x+h) - f(x) \right)=0\]

Prove that f is uniformly continuous.


Firstly, let us define the following set:

    \[S_{\varepsilon,X} = \{t \in \mathbb R_{\geq 0} : \text{if }x\geq X\text{ then }|f(x+t)-f(x)| \leq \varepsilon\}\]

By hypothesis, for any fixed \varepsilon, we have

    \[\bigcup_{X\in\mathbb R_{\geq 0}}S_{\varepsilon,X} = \mathbb R_{\geq 0}\]

Note that each set S_{\varepsilon,X} is closed because for a fixed x, the set of values of t such that |f(x+t)-f(x)|\leq \varepsilon is closed and S_{\varepsilon,X} is an intersection of these closed sets over all x\geq X.

Note that we could also say that

    \[\bigcup_{X\in\mathbb Z_{\geq 0}}S_{\varepsilon,X} = \mathbb R_{\geq 0}\]

since the sets S_{\varepsilon,X} increase with X – giving a countable union of closed sets whose union is the whole space.

We can then apply the Baire Category Theorem to say that since a countable union of closed sets has non-empty interior, some element of the union must have an interior! In particular, for any \varepsilon>0, there must be some X such that some interval (a,b) is a subset of S_{\varepsilon,X}. However, then if we have that x,y \geq X + a and |x-y| < |b-a| we could choose some pair a',b'\in (a,b) with x-a' = y-b' and then observe that

    \[f(x)=f(x-a' + a')\]

    \[f(y)=f(y-b' + b')\]

Then the distance from f(x) to f(x-a') is at most \varepsilon as is the distance from f(y) to f(y-b') since a',b'\in \subseteq S_{\varepsilon,X}. Thus we find that if x,y \geq X+a and |x-y| < |b-a| we have |f(x)-f(y)| \leq 2\varepsilon – and this works out for any choice of \varepsilon>0. This fact suffices to establish that f is uniformly continuous with a small bit of further work.

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