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I'm trying to find a relation between derivative and uniform continuity on $\mathbb{R}$. First, I found this property:

  1. If $f$ is differentiable on $\mathbb{R}$ and $\displaystyle \lim_{x\rightarrow +\infty} f'(x)=+\infty$, then $f$ is not uniformly continuous on $\mathbb{R}$.

Proving this is not so difficult. Indeed, if $f$ is uniformly continous on $\mathbb{R}$, then for every $\epsilon$, there exists $\delta$ such that for all $x,y$ satisfying $|x-y|<\delta$ then $|f(x)-f(y)|<\epsilon$. Since $f'\rightarrow +\infty$, there exists $X$ such that for all $x>X$, $f'(x)>\frac{2\epsilon}{\delta}$. Choose an arbitrary $x>X$ and consider Lagrange theorem: $$\left|f\left(x+\frac{\delta}{2} \right)-f(x)\right|=f'(c_x)\cdot \frac{\delta}{2}>\epsilon$$

since $c_x>x>X$. This is a contradiction.

Now I want to make the condition a little bit stricter.

  1. If $f$ is differentiable on $\mathbb{R}^+$ and $f'$ is not bounded above on $\mathbb{R}^+$, then $f$ is not uniformly continuous on $\mathbb{R}^+$.

($\mathbb{R}^+$ is just for convinience)

Since $f'$ is not bounded, there exists a sequence $(x_n)$ such that $x_n\rightarrow =\infty$ and $f'(x_n)\rightarrow +\infty$. Then there exists $N$ sucht that $f'(x_n)>\frac{2\epsilon}{\delta}$ for all $n>N$. Since $f$ is continuous, for all $n>N$, there exists a neighborhood $(x_n-\delta_n,x_n+\delta_n)$ such that $f'(x)>\frac{2\epsilon}{\delta}$ for all $x\in (x_n-\delta_n,x_n+\delta_n)$. If $\sup\delta_n>\frac{\delta}{2}$, there exists $k>N$ such that $\delta_k>\frac{\delta}{2}$. Applying Lagrange theorem for arbitrary $x,y\in(x_k-\delta_k,x_k+\delta_k)$ such that $|x-y|=\delta$: $$|f(x)-f(y)|=f'(c_{xy})|x-y|>2\epsilon$$

I'm stuck with the case $\sup \delta_n\le \frac{\delta}{2}$. Cound I continue, or the statement is just false itself?

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  • $\begingroup$ why cant the same arguements be repeated in 2 as in 1 $\endgroup$ – Learnmore Jan 14 '15 at 4:49
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    $\begingroup$ Since in the first statement, $f'\rightarrow+\infty$, while in the second one it is just unbounded. Notice that "unbounded" in not the same as "tending to infinity". Following the proof, if $\sup\delta_n\le \frac{\delta}{2}$, then we can not choose $x,y$ such that $|x-y|=\delta$. $\endgroup$ – Tien Kha Pham Jan 14 '15 at 4:54
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Take $$f(x)=\begin{cases}x^2\cos(1/x^2)&\text{ for }x\neq0\\0&\text{ for }x=0\end{cases}$$ Then $$f'(x)=\begin{cases}2x\cos(1/x^2)+2\frac{1}{x}\sin(1/x^2)&\text{ for }x\neq0\\0&\text{ for }x=0\end{cases}$$

Observe that $f(x)$ is continuous in the compact $[-1,1]$. Therefore $f(x)$ is uniformly continuous in $[-1,1]$. But we can see that $f'(x)$ is not bounded in $[-1,1]$.

This is an example where the problem is at the origin and we only define it in $[-1,1]$. To get a function on $\mathbb{R}^+$ you first ignore the part for $x<0$ and continue $f(x)$ to $[1,+\infty)$ by smoothly continuing the graph such that it bends into a constant function for $x$ large.


If we want the problem to be at $+\infty$ we can take

$$g(x)=\frac{1}{x}\cos(x^3)\text{ for }x\geq 1$$

which by a translation later you can assume defined in $\mathbb{R}^+$. Then

$$g'(x)=-\frac{1}{x^2}\cos(x^3)-2x\sin(x^3)$$

The derivative is unbounded. Let us prove $g(x)$ is uniformly continuous in $[1,+\infty)$.

Given $\epsilon>0$.

Choose $M>0$ such that if $x>M$ then $|g(x)|<\epsilon/2$. In particular if $x,y>M$ then $$|f(x)-f(y)|\leq |f(x)|+|f(y)|<\epsilon$$

On the compact $[1,M]$ the function $g$ is continuous and therefore uniformly continuous. Therefore there is $M/2>\delta_1>0$ such that if $x,y\in[1,M]$ with $|x-y|<\delta_1$ then $|f(x)-f(y)|<\epsilon$.

Finally, on the compact set $C:=[M-\delta_1,M+\delta_1]$ the function $g(x)$ is continuous and therefore there is a $\delta_2$ such that if $x,y\in C$ with $|x-y|<\delta_2$ then $|f(x)-f(y)|<\epsilon$.

Finally, take $\min(\delta_1,\delta_2)>\delta>0$, then if $x,y\geq1$ are such that $|x-y|<\delta$ then $x$ and $y$ lay both in one the the sets $[M,+\infty)$, $[1,M]$ or $C$. In each case $|f(x)-f(y)|<\epsilon$ and therefore $g(x)$ is uniformly continuous on $[1,+\infty)$.

Therefore $h(x)=g(x+1)$ does the job on $\mathbb{R}^+$.

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