Bounded derivative and uniform continuity

[Kudasai]

Let $f:[0,\infty)\to\mathbb R$ be a differentiable function such that for all $a>0$ exists a constant $M_a$ such that $|f'(t)|\le M_a$ for all $t\in[0,a]$ and $f(t)\xrightarrow[n\to\infty]{}0.$ Show that $f$ is uniformly continuous.

Basically, I need to prove that $f$ is uniformly continuous for all $t\ge0,$ however, I know that $f$ is uniformly continuous for $t\in[0,a]$ since $f$ is bounded there, but I don't know how to use the fact of the limit to prove uniform continuity for all $t\ge0.$

 

[Opalg]

Let $f:[0,\infty)\to\mathbb R$ be a differentiable function such that for all $a>0$ exists a constant $M_a$ such that $|f'(t)|\le M_a$ for all $t\in[0,a]$ and $f(t)\xrightarrow[n\to\infty]{}0.$ Show that $f$ is uniformly continuous.

Basically, I need to prove that $f$ is uniformly continuous for all $t\ge0,$ however, I know that $f$ is uniformly continuous for $t\in[0,a]$ since $f$ is bounded there, but I don't know how to use the fact of the limit to prove uniform continuity for all $t\ge0.$

Let $\varepsilon>0$. The fact that $f(t)\to0$ as $t\to\infty$ tells you that there exists $a$ such that $|f(t)| < \frac12\varepsilon$ for all $t>a$. The function $f$ is uniformly continuous on $[0,a]$ because its derivative is bounded there. Therefore there exists $\delta>0$ such that if $s,t\in [0,a]$ and $|s-t|<\delta$ then $|f(s)-f(t)| < \varepsilon.$ On the other hand, if $s,t>a$ then $|f(s) - f(t)| \leqslant |f(s)| + |f(t)| <\frac12\varepsilon + \frac12\varepsilon <\varepsilon$.

Putting those results together, you see that $|s-t|<\delta$ implies $|f(s) - f(t)| < \varepsilon$, regardless of whether $s$ and $t$ are less than $a$ or greater than $a$. Since that implication holds for all $\varepsilon>0$, it follows that $f$ is uniformly continuous on $[0,\infty)$.

You may have spotted that there is a gap in that argument, in that I have not dealt with the possibility that one of $s,t$ is less than $a$ and the other one is greater than $a$. I will leave you to think about how to deal with that case.