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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 $\varepsilon=1$

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

  • 0
    It is the *exact* duplicate indeed... May I suggest to close the other thread, for it has no answer?2011-12-14

1 Answers 1

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The problem can be solved as follows.

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 $K=1/\delta$.

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    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 $\Delta =1$.2011-12-16