# $f\colon \mathbb{R}_{+} \to \mathbb{R}$ uniformly continuous : problem

a few days ago, I tried to solve this problem

Let $f\colon \mathbb{R}_{+} \to \mathbb{R}$ uniformly continuous. Prove that exists $K>0$ such that for each $x\in \mathbb{R}_{+},$ $$\sup_{w>0}\{ |f(x+w) -f(w)|\}\le K \,\, ( x + 1).$$

and some members of this community have suggested that road :

1. Let $F\colon\mathbb R_+\to\mathbb R$ defined by $\displaystyle F(x)=\sup_{t>0}|f(x+t)-f(x)|$. Show that $F$ is uniformly continuous on $\mathbb R$.
2. Let $h\colon \mathbb R_+\to\mathbb R_+$ an uniformly continuous function on $\mathbb R_+$. Prove that we can find a constant $K>0$ such that $h(x)\leq K(x+1)$ for all $x\geq 0$.
3. Conclude.

I tried, only step 1,but I do not know how to continue

$$|f(x_1+w) -f(w)- f(x_2+w) +f(w)|=$$ $|f(x_1+w)|- |f(x_2+w)|\leq|f(x_1+w)- f(x_2+w)|\leq K(x+1)<\varepsilon$

if I take $\varepsilon=1$

$|f(x_1+w)|- |f(x_2+w)|\leq K(x+1)<1$ .... ..... .....

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possible duplicate of Problem with this question : uniformly continuous – Did Dec 14 '11 at 14:38
It is the exact duplicate indeed... May I suggest to close the other thread, for it has no answer? – Pacciu Dec 14 '11 at 16:27

If you choose $\varepsilon =1$ in the definition of uniform continuity, you gain a $\delta \in ]0,1]$ s.t. for all $a,b >0$: $$\tag{1} |a-b|\leq \delta \quad \Rightarrow \quad |f(a)-f(b)|\leq 1\; .$$ Now, fix $x>0$: since $\mathbb{R}$ has the Archimedean property, there exists $N\in \mathbb{N}$ s.t.: $$\tag{2} N\delta \leq x\leq (N+1)\delta \; ;$$ hence split the interval $[w,x+w]$ into $N+1$ subintervals using the $N+2$ points: $$w_k:=w+k\ \delta \qquad \text{for } k=0,\ldots ,N;\quad w_{N+1}:=x+w$$ and write: $$\tag{3} |f(x+w)-f(w)|\leq \sum_{k=0}^N|f(w_k)-f(w_{k+1})|\; ;$$ at this stage, (1) can be used to increase the RHside of (3): $$|f(x+w)-f(w)| \leq N+1$$ and (2) yields: $$\tag{4} |f(x+w)-f(w)| \leq \frac{1}{\delta}\ x+1\; ;$$ finally from $0<\delta \leq 1$ and (4) you get: $$|f(x+w)-f(w)| \leq \frac{1}{\delta}\ (x+1)\; ,$$ which is your claim with $K=1/\delta$.
A remark. For $h>0$, let $u^h:]0,\infty [\ni x\mapsto u(x+h)\in \mathbb{R}$ be the $h$-translate of $u$; hence $\sup_{x>0} |u(x+h)-u(x)|\leq M\ (h+1)$ recasts as an estimate of the $\infty$-distance from $u^h$ to $u$ in terms of the length of the translation via a constant $M$ depending only on the modulus of continuity of $u$, i.e. $\lVert u^h-u\rVert_\infty \leq M_u\ (h+1)$. Therefore, if $\mathfrak{F}\subset C(]0,\infty[)$ is an equi-uniformly continuous family, then there exists $M=M_\mathfrak{F}$ s.t. $\lVert u^h-u\rVert_\infty \leq M\ (h+1)$ for all $u\in \mathfrak{F}$ and $h>0$. – Pacciu Dec 14 '11 at 17:03
why you write: "If you choose $\varepsilon =1$ in the definition of uniform continuity, you gain a $\delta \in ]0,1]$ s.t. for all $a,b >0$" i know only that $\delta>0$ not $\delta \in ]0,1]$ ... – FrConnection Dec 15 '11 at 15:59
You can always take $\delta$ less than a positive number $\Delta$ in the definition of uniform continuity. In fact, if you call $\delta_1 >0$ the value you gain from the definition of uniform continuity with $\varepsilon =1$, then also $\delta :=\min \{ \Delta ,\delta_1\}$ makes the definition work because: $$|x-y|\leq \delta\ \Rightarrow \ |x-y|\leq \delta_1\ \Rightarrow\ |f(x)-f(y)|\leq 1\; ;$$ moreover such a $\delta$ does not exceed $\Delta$. In particular, in the previous proof I chose $\Delta =1$. – Pacciu Dec 16 '11 at 0:33