Find bounded function satisfying given conditions

Question

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.

Answer 1

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.

Answer 2

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 $\lim_{t \rightarrow \infty} u(t) = c < \infty$.

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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.