Let $f$ be uniformly continuous ,$f(0)=0$ show $\exists M>0$ st $|f(x)|\le 1+M|x|$

Question

Consider a uniformly continuous function $f:\mathbb{R}\to \mathbb{R}$ such that $f(0)=0$, Prove that there exists $M>0$ such that $|f(x)|\le 1+M|x|$ $\forall x \in \mathbb{R}$

Answer 1

Let $\epsilon=1$. Then there exists $\delta>0$ such that $|x-y|<\delta$ implies $|f(x)-f(y)|<\epsilon=1$. Let $M=\frac1 \delta$.

Now if $x>0$ let $n$ be minimal with $n\delta>x$. Let $x_k=\frac {kx}{n}$, for $0\le k\le n$. Then $x_0=0$, $x_n=n$ and $x_{k+1}-x_k=\frac x{n}<\delta$, hence $|f(x_{ĸ+1})-f(x_k)|<1$ and finally $|f(x_n)-f(x_0)|<n$, i.e. $|f(x)|<n$. Since $(n-1)\delta\le x$, we see that $|f(x)|<\frac x\delta+1=1+M|x|$.

The case $x<0$ follows similarly.

Answer 2

Suppose that $f$ is uniformly continuous. That is: for any $\epsilon>0:$ for any $x,y\in\mathbb R$, there is a $\delta_\epsilon$ such that $$ |x-y|<\delta_\epsilon \implies |f(x)-f(y)|<\epsilon $$ Setting $\epsilon=1$, there is a $\delta_1$ such that $$ |x-y|<\delta_1 \implies |f(x)-f(y)|<1 $$ Thus, selecting some $\Delta x<\delta_1$, we have $$ \left|f(x+\Delta x)-f(x)\right|<1 $$ Or, perhaps more intuitively, $$ \left|\frac{f(x+\Delta x)-f(x)}{\Delta x}\right|<\frac{1}{\Delta x} $$ That is, over any interval of length $\Delta x$, $f$ has a maximum average rate of change of $\frac{1}{\Delta x}$.

Now, I claim that setting $M=\frac1{\Delta x}$ meets the requirements of this theorem, by the following proof:

Consider any $x_0\in\mathbb R$. Let $n=\lfloor x_0/\Delta x \rfloor$, $\Delta x_0= x_0-n\Delta x$, and $M=\frac1{\Delta x}$. Then we have (in the general case that $n>2$) $$ \begin{align} f(x_0)&=f(n\Delta x+\Delta x_0)\\ &<1+f((n-1)\Delta x+\Delta x_0)\\ &<2+f((n-2)\Delta x+\Delta x_0)\\ &<\cdots\\ &<n+f(0+\Delta x_0)\\ &<n+1 = 1+ \lfloor x_0/\Delta x \rfloor\\ &< 1 + x_0/\Delta x = 1 + Mx_0 \end{align} $$ Similarly, $$ \begin{align} f(x_0)&=f(n\Delta x+\Delta x_0)\\ &>-1+f((n-1)\Delta x+\Delta x_0)\\ &>-2+f((n-2)\Delta x+\Delta x_0)\\ &>\cdots\\ &>-n+f(0+\Delta x_0)\\ &>-n-1 = -\left(1+ \lfloor x_0/\Delta x \rfloor\right)\\ &> \left(1 + x_0/\Delta x\right) = -\left(1 + Mx_0\right) \end{align} $$ The statement follows.