1
$\begingroup$

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.

$\endgroup$
  • 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ "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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.