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 :
- 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 $\mathbb R$.
- 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 $x\geq 0$.
- 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 $\varepsilon=1$
$|f(x_1+w)|- |f(x_2+w)|\leq K(x+1)<1$ .... ..... .....