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.
 A: 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.
