I'm trying to find a function $u: (0, \infty) \to \mathbb{R}$ which satisfies these conditions:

i) $u$ is bounded.

ii) $u^2$ increases over $(0, \infty)$.

iii) $\dot u(t)$ does not converge to $0$ as $t$ tends to infinity.

I think it's easier to choose first the derivative of $u^2$ and then integrate it. For example, I chose $$\frac{d}{dt} u^2 = \frac{\sin^2 t}{(t+1)^2},$$ so the function $u$ is $$u(t) = \sqrt{\int_0^t \frac{\sin^2 x}{(1+x)^2}dx}.$$ This function is bounded, but its derivative converges to $0$ as $t$ tends to infinity, which does not satisfy the iii) condition.

Those are my ideas for the problem. Thank you very much for any idea, hint or solution.

  • 2
    $\begingroup$ Just a thought. $u^2$ has limit at $+\infty$. If $u$ and $\dot u$ also do, then $\dot u$ must tend to zero at $+\infty$. $\endgroup$ – Stefano Mar 13 '17 at 10:34

A function that satisfies the conditions is given as an answer in the following question.

The following is a proof that we may replace condition ii), by the condition that $u$ is increasing.

Let's assume that you want $\dot{u}(t)$ to exists for all $t \in (0, \infty)$.

Suppose that $u(t_{0}) > 0$ for some$^{1}$ $t_{0} \in (0, \infty)$.

Let us show by contradiction that $u(t) > 0$ for all $t > t_{0}$. Suppose that $u(t) \leqslant 0$ for some $t > t_{0}$, then by the intermediate value theorem, there exists a $t' \in (t_{0},t]$ such that $u(t') = 0$. However, for this $t'$ we would have $u(t')^{2} = 0$, hence $u^{2}$ is seen to not be increasing.

Now we have that $u(t) > 0$ for all $t > t_{0}$, and it follows from the fact that $u^{2}$ is increasing that $u$ is increasing.

Now if we want $u$ to be bounded, it follows that $\lim_{t \rightarrow \infty} u(t) = c < \infty$.

1) This can be done without loss of generality, because the constant function $u \equiv 0$ does not satisfy your conditions. And if a function $u$ is non-positive for all $t$ and satisfies all the conditions, one sees that the function $-u$ satisfies all conditions as well.

  • $\begingroup$ Could you explain more specifically how we can conclude $\dot u(t) \to 0$? Thank you. $\endgroup$ – Tien Kha Pham Mar 13 '17 at 14:25
  • 1
    $\begingroup$ I can not, because it is false. My apologies, a counter-example (and thus an example of the function you are looking for) is given as an answer to another question [1]: math.stackexchange.com/questions/1038951/… $\endgroup$ – Peter Mar 13 '17 at 14:58
  • $\begingroup$ That's alright. Thank you very much for your counter-example :) $\endgroup$ – Tien Kha Pham Mar 13 '17 at 15:10

EDIT: The following only works if "increasing" means "strictly increasing" and not "nondecreasing"!

Are you sure such $u$ exists? Here is a proof that it does not. There may be a mistake I cannot see.

Suppose $u'(t_0)=0$ for some $t_0$. Then, $\frac{d}{dt}u^2(t_0)=2u(t_0)u'(t_0)=0$.

Because of condition (ii), we need $\frac{d}{dt}u^2(t)>0$ for all $t>0$, so the previous cannot be. Hence, $u'(t)\neq0$ for all $t>0$. Either $u'(t)>0$ or $u'(t)<0$ always. Without loss of generality, assume $u'(t)>0$ always (for the other case, the proof only changes replacing "increasing" for "decreasing"). Even when $t\rightarrow\infty$, the derivative doesn't tend to 0, so $u$ continues increasing.

Let $r$ be such that $u'(t)\geq r$ for all $t>0$. By the mean value theorem for derivatives, $u(t)-u(0)=f'(c)(t-0)$ for some $c\in(0,t)$. So, $u(t)-u(0)=f'(c)(t-0)\geq r(t-0)=rt$. So $u(t)\geq u(0)+rt$.

When $t\rightarrow \infty$, $u(t)\geq u(0)+rt\rightarrow u(0)+\infty=\infty$. This violates condition (i) ($u$ is bounded).

Therefore, such $u$ does not exist.

  • $\begingroup$ "Because of condition (ii), we need $\frac{d}{dt}u^2(t)>0$" That would be true if it was strictly increasing. If it is just increasing we can have points where the first derivative equals zero. $\endgroup$ – MathematicianByMistake Mar 13 '17 at 11:01
  • $\begingroup$ Doesn't "increasing" mean "strictly increasing"? Otherwise, it would be called "nondecreasing", wouldn't it? $\endgroup$ – Anna SdTC Mar 13 '17 at 11:03
  • $\begingroup$ Nope. There is a difference between increasing and strictly increasing. Actually we can have even the first derivative equal zero-but only at a stationary point-and the function be strictly increasing. Consider for example $f(x)=x^3$ in $x_0=0$ where $f'(x)=3x^2\ge 0$, $f'(0)=0$ but $f(x)$ is strictly increasing. $\endgroup$ – MathematicianByMistake Mar 13 '17 at 11:12
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    $\begingroup$ In your example, the derivative is anyway strictly positive in the open interval. But you are right, if we take "increasing" in the "weak" sense, then the $u$ function may exist and my proof is wrong. (But the ambiguity of what "increasing" means is known math.stackexchange.com/questions/115912/… .) $\endgroup$ – Anna SdTC Mar 13 '17 at 11:38

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