Cauchy property and one-sided double limit

Question

Let $u,\phi:[0,\infty)\to\mathbb{R}$ and assume that $\lim_{t\to\infty}\phi(t)$ exists. Suppose that for any $s>0$, we have \begin{equation} \lvert u(t)-u(s)\rvert\leq\phi(t)-\phi(s),\quad t\in[0,s]. \end{equation} Given the above, I would like to understand the (one-sided double limit) expression \begin{equation} \lim_{\substack{t,s\to+\infty\\s>t}}\lvert u(t)-u(s)\rvert=0 \end{equation} and why this ''Cauchy property'' suffices for $\lim_{t\to\infty}u(t)$ to exist.

Answer 1

The definition of $$\lim_{\substack{t,s\to+\infty\\s>t}}f(s,t)=\ell$$ is: $$\forall\varepsilon>0\quad\exists M\quad\forall s>t>M\quad|f(s,t)-\ell|<\varepsilon. $$ It should then become obvious that

• if $\lim_{+\infty}\phi$ exists then $\lim_{\substack{t,s\to+\infty\\s>t}}(\phi(t)-\phi(s))=0,$ and that
• if $\forall t\in[0,s]\quad\lvert f(s,t)\rvert\le g(s,t)$ and $\lim_{\substack{t,s\to+\infty\\s>t}}g(s,t)=0,$ then $\lim_{\substack{t,s\to+\infty\\s>t}}f(s,t)=0.$

Now comes the more interesting: why the ''Cauchy property'' $$\lim_{\substack{t,s\to+\infty\\s>t}}\lvert u(t)-u(s)\rvert=0 $$suffices for $\lim_{+\infty}u$ to exist.

For any real sequence $t_n\to+\infty,$ the Cauchy property implies that $(u(t_n))$ is a Cauchy sequence, hence converges. Moreover, the limit does not depend on the choice of $(t_n)$ since if $t'_n\to+\infty,$ the sequence $(t_1,t'_1,t_2,t'_2,\dots)$ also tends to infinity hence its image by $u$ converges, which proves that the two subsequences $(u(t_n))$ and $(u(t'_n))$ have the same limit. This ends the proof.

Answer 2

The main idea of the Cauchy Property is apply the triangle inequality with the limit.

Let $L=\lim_{t \to \infty}\phi(t)$. Then, $$|u(t)-u(s)|\leq \phi(t)-\phi(s) \leq|\phi(t)-\phi(s)|\leq |\phi(t)-L|+|L-\phi(s)|$$ For any $t \in [0,s]$.

Since $s>t$, if $t \to \infty$, then $s \to \infty$. Now if u take the limit as $t \to \infty$, we get that $s\to \infty$ and by the sandwich theorem, $$0\leq \lim_{t,s\to \infty, s>t}|u(t)-u(s)|\leq 0$$ Thus, $$ \lim_{t,s\to \infty, s>t}|u(t)-u(s)|=0$$