0
$\begingroup$

Assume that $x(t)$ is nonnegative for $t\geq0$ and $\int_0^\infty{x(t)\;dt}<\infty$. If $x'(t)$ is bounded for $x\geq0$, then show that $\lim_{t\to\infty}{x(t)}=0$.


I started with contradiction taking the limit is finite and strictly positive. Then it is easy to derive the contradiction. But the limit may be oscillatory too or diverging to $+\infty$.

Any help would be apreciated. Thanks in advance.

$\endgroup$
11
  • 1
    $\begingroup$ Can you show us your contradiction proof? I think this question should be relatively straightforward, no? $\endgroup$
    – Jam
    Aug 19, 2018 at 19:05
  • 1
    $\begingroup$ $x(t)$ clearly can't diverge to $+\infty$ since $\int x(t)$ would not be less than $\infty$. $\endgroup$
    – Jam
    Aug 19, 2018 at 19:06
  • $\begingroup$ The contradiction is as follows : $\endgroup$
    – mathemagic
    Aug 19, 2018 at 19:06
  • $\begingroup$ Yes..I showed the exactly same thing but that works exactly when lim (t--> infinity) x(t)= d>0. $\endgroup$
    – mathemagic
    Aug 19, 2018 at 19:07
  • $\begingroup$ Can x(t) be oscillatory?? $\endgroup$
    – mathemagic
    Aug 19, 2018 at 19:08

3 Answers 3

6
$\begingroup$

If $x(t)\not\to 0$, then there exists a $\delta>0$ such that there exists a sequence $t_n\to\infty$ with $x(t_n)\ge \delta$. Taking a subsequence we can assume $t_{n+1}-t_n\ge \frac\delta M$. We have $|x'(t)|\le M$ so $x(t)\ge \frac\delta2$ for $|t-t_n|<\frac\delta{2M}$ and so $$\int_{t_n}^{t_{n+1}} x(t)dt \ge \frac{\delta^2}{4 M}.$$ Hence the integral must be infinite.

$\endgroup$
1
$\begingroup$

The hypothesis in the question seems weird to me, but here is a related theorem.

Lemma (Cauchy criterion for improper integrals) Let $f : [a, b[ \rightarrow \mathbb{R}$ be continue on $[a, b[$ where $-\infty<a<b\leq+\infty$. Then $\int_{a}^{b} f(t) dt$ converge iff $\forall\epsilon > 0$ $\exists A\in [a, b[$ $\forall (x, y)\in([A, b[)^2$, $|\int_{x}^{y} f(t) dt| < \epsilon$.

Proof Let $\epsilon > 0$ and $F : x \mapsto \int_{a}^{x} f(t)dt$. Suppose that $\int_{a}^{b} f(t) dt$ converge. Then $F(x)\xrightarrow[x\to b]{}\ell\in\mathbb{R}$. So there exists $A\in [a, b[$ such that for all $x\in [A, b[, |F(x)-\ell| < \frac{\epsilon}{2}$. Thus, for all $(x, y)\in([A, b[)^2$, $$|\int_{x}^{y} f(t) dt| = |F(y) - \ell +\ell - F(x)| < \epsilon.$$ Let's show the reciprocal. Let $(x_n)$ be sequence of $[a, b[$ converging to $b$. $(F(x_n))$ converge because it's a Cauchy sequence and $\mathbb{R}$ is complete. We note $\ell$ this limit. Let $(y_n)$ be any sequence of $[a, b[$ converging to b and $\ell'$ be the limite of (F(y_n)). Let $(z_n)$ be the sequence defined for all $n\in\mathbb{N}$ by $z_{2n} = x_n, z_{2n+1} = y_n$. Since $(z_n)$ converge to $b$, $(F(z_n))$ converge to a real $\ell''$. Since $(F(x_n))$ and $(F(y_n))$ are sub-sequences of $(F(z_n))$, $\ell = \ell'' = \ell'$. We have therefore show that for all sequence $(y_n)$ converging to b, $(F(y_n))$ converge to $\ell$. Thus $F(x)\xrightarrow[x\to b]{}\ell$. Hence $\int_{a}^{b} f(t)dt$ converge.

Theorem Let $a \in \mathbb {R}$ and $f$ be uniformly continuous over $[a, +\infty [$ such that $\int_{a}^{+\infty} f(t) dt$ converge. Then $ lim_{t\to+\infty} f(t) = 0 $.

Proof Let $\epsilon > 0$. Since $f$ is uniformly continuous on $[a, +\infty [$, there exists $\alpha > 0$ such that for all $(x, y)\in [a, +\infty[, |x-y|<\alpha \Rightarrow |f(x) - f(y)| < \frac{\epsilon}{2}$. On the other hand, since $\int_{a}^{+\infty} f(t) dt$ converge, via the criterion of cauchy, there exists $A\in [a, +\infty[$ such that for all $(x, y)\in([A, +\infty[)^2, |\int_{x}^{y} f(t)dt| < \frac{\alpha\epsilon}{2}$. Let $x>A$. $|f(x)| = \frac{1}{\alpha}|\int_{x}^{x+\alpha} f(x)dt| = \frac{1}{\alpha}|\int_{x}^{x+\alpha} (f(x)-f(t))dt + \int_{x}^{x+\alpha} f(t)dt| \leq \frac{1}{\alpha}|\int_{x}^{x+\alpha} (f(x)-f(t))dt| + \frac{1}{\alpha}|\int_{x}^{x+\alpha} f(t)dt| < \frac{1}{\alpha}|\int_{x}^{x+\alpha} (f(x)-f(t))dt| + \frac{\epsilon}{2} < \epsilon$. Thus $f(x)\xrightarrow[x\to+\infty]{}0$

In the context of this question, $ x'(t) $ is bounded on $ [0, +\infty [$, which equivalent to x being lipschitz continuous. A lipschitz function is uniformly continuous, so $ lim_{t\to+\infty} x(t) = 0 $.

$\endgroup$
-1
$\begingroup$

Incomplete proof (missing the case when $x(t)$ has no limit)

By contradiction, let $L \gt 0$ such that $\lim_{t \to \infty} x(t) = L$.

Then, by definition $\forall$ $\varepsilon \gt 0$ exists $k \in \mathbb{R^+}$ such that $$\mid x(t) - L \mid \lt \varepsilon \;\;\;\;\;\;\;\; \forall \;\;t \geq k$$ Let $\varepsilon \in (0, L)$, then $$0 \lt L - \varepsilon \lt x(t) \;\;\;\;\;\;\;\; \forall \;\; t \geq k$$ $$\Rightarrow 0 < \int_k^\infty (L - \varepsilon)dt < \int_k^\infty x(t)dt < \infty$$ But this is a contradiction since $L - \varepsilon$ is a positive constant so $\int_k^\infty (L - \varepsilon)dt$ is not bounded.

$\endgroup$
3
  • 2
    $\begingroup$ This might prove that the limit of $x(t) $ can't be greater than $0$ but doesn't examine the case of $x (t) $ having no limit. $\endgroup$
    – Jam
    Aug 20, 2018 at 16:29
  • 1
    $\begingroup$ Although I think you could extend this proof by using $\limsup $ in place of $\lim $, to cover the case of oscillatory functions with no limit. $\endgroup$
    – Jam
    Aug 20, 2018 at 16:33
  • $\begingroup$ @Jam I don't find a way to extending it. Could you please help? $\endgroup$
    – salvarico
    Aug 20, 2018 at 20:11

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.