# Limit of function variation over finite interval

I am working on a proof and I have stumbled in the following question. Consider a continuously differentiable bounded function $f(t)$ with bounded, uniformly continuous derivative defined in $[0,\infty)$ with the property that $$\lim_{t\rightarrow\infty}\left[f(t)\dot{f}(t)\right]=0.$$

Is the claim $$\lim_{t\rightarrow\infty}|f(t+T)-f(t)|=0\qquad\qquad (1)$$ true for all finite $T>0$ ?

My attempt: Since $$\lim_{t\rightarrow\infty}\frac{d}{dt}\left[f^2(t)\right]=2\lim_{t\rightarrow\infty}f(t)\dot{f}(t)=0$$ we directly obtain from the mean value theorem that $$\lim_{t\rightarrow\infty}\left[f^2(t+T)-f^2(t)\right]=0$$ for all finite $T>0$. I believe that now using continuity of $f$ and the fact that the above limit holds for all $T>0$ will maybe suffice to prove (1) but I am somehow missing the final link.

• This is bad... where is the "uniformness" applied in here? There is only one function, not a set of functions in where to apply uniform boundness. – Brethlosze Sep 5 '17 at 21:04
• @hyprfrcb My bad..bounded, will edit! – RTJ Sep 5 '17 at 21:13

## 1 Answer

Ok this appears to be true. It follows directly from the claim $$\lim_{t\rightarrow\infty}\dot{f}(t)=0$$ We can prove this by contradiction. If this is not true, then there exists some $\epsilon>0$ and a sequence of times $\{t_i\}_{i=1}^{\infty}$ with $\lim_{i\rightarrow\infty}t_i=+\infty$ such that $$\dot{f}(t)>\epsilon \qquad \forall t\in[t_i,t_i+\delta(\epsilon)]\qquad \qquad (1)$$ or $$\dot{f}(t)<-\epsilon \qquad \forall t\in[t_i,t_i+\delta(\epsilon)] \qquad \qquad (2)$$ Without loss of generality we proceed with the first case (1). Then, due to $\lim_{t\rightarrow\infty}f(t)\dot{f}(t)=0$ there exists $i_0$ sufficiently large such that $$|f(t)\dot{f}(t)|<\frac{\delta(\epsilon)\epsilon^2}{2} \quad \forall t\in[t_i,t_i+\delta(\epsilon)], \: \forall i\geq i_0$$ or equivalently $$|f(t)|< \frac{\delta(\epsilon)\epsilon^2}{2\dot{f}(t)}<\frac{\delta(\epsilon)\epsilon}{2}\quad \forall t\in[t_i,t_i+\delta(\epsilon)], \: \forall i\geq i_0 \qquad \qquad (3)$$

Using (1), (3) we obtain $$f(t_i+\delta(\epsilon))>f(t_i)+\epsilon\delta(\epsilon)>-\frac{\delta(\epsilon)\epsilon}{2}+\epsilon\delta(\epsilon)=\frac{\delta(\epsilon)\epsilon}{2} \quad \forall i\geq i_0$$ which is a contradiction to (3). This completes the proof.

• $f(t)f'(t)\to 0$ imply $f'(t)\to 0$ or $f(t)\to 0$. A trivial case is missing... – Brethlosze Sep 6 '17 at 11:06
• @hyprfrcb How is this directly obtained? I can't see it. If $f_1f_2\rightarrow 0$ then that does not mean that $f_1\rightarrow 0$ or $f_2\rightarrow 0$. – RTJ Sep 6 '17 at 11:15
• Both functions are bounded, so the proposition should hold; it is easy to figure a function with $f'(t)=f_0$ and $f(t) \to 0$, which requires that case to be covered. – Brethlosze Sep 7 '17 at 14:26