# If $x'(t)$ is bounded for $x\geq0$ show that $\lim_{t\to\infty}{x(t)}=0$

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.

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

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.

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 $lim_{t\to+\infty} x(t) = 0$.

# 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.

• 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.
– Jam
Aug 20, 2018 at 16:29
• 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.
– Jam
Aug 20, 2018 at 16:33
• @Jam I don't find a way to extending it. Could you please help? Aug 20, 2018 at 20:11