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Suppose that $f$: $[0, infty) rightarrow mathbb{R}$ is continuous on $[0, infty)$ and differentiable on $(1, infty)$ with bounded derivative. Show that $f$ is uniformly continuous. (HINT: split $[0, infty)$ into two pieces, on each of which $f$ is uniformly continuous, then explain why this implies that $f$ is uniformly continuous on the whole $[0, infty)$.)

Question: I have solved the majority of the problem, but wanted a little clarification with regards to the explanation in the second part. Specifically there is the situation where:

$ x in [0, 1)$ and $y > 1$ (or vice versa). In this scenario I still have to show $|x - y| < delta$. Given the conditions that I have can I do the following: $$|x - y| leq |x - 1| + |y - 1| < frac{delta}{2} + frac{delta}{2} = delta$$

And if this is allowed is the reason that I should choose $1$ because the two intervals split at that specific point so it allows me to relate $x,y$ to each other?

Assuming the previous step is true then it follows:

$$ |f(x) - f(y)| leq |f(x) - f(1) + |f(1) - f(y)| < frac{epsilon}{2} + frac{epsilon}{2} = epsilon$$

Use MVT to show $|f(x)-f(y)|le M|x-y|$ on $(1,infty)$. Since $f$ is uniformly continuous on $[0,1]$, for any $epsilon>0$ there exists $delta>0$ so that $|f(x)-f(y)|<epsilon$ when $|x-y|<delta$ and $x,yin[0,1]$. If $xle 1le y$, because $f$ is continuous at $1$, there is $eta$ so that if $|x-y|<eta$ implies $|x-1|,|y-1|<eta$ and
$$|f(x)-f(y)|le |f(x)-f(1)|+|f(1)-f(y)|<epsilon $$
Now if $|x-y|<min{epsilon/M,delta,eta}$, we get $|f(x)-f(y)|<epsilon$.

The mean value theorem gives Lipschitz and therefore uniform continuity 
of $f$ in $[1,infty)$ including $1$. That is because you need 
differentiablity only in the interior. Moreover, $f$ is uniformly 
continuous in $[0,1]$ since every continuous function on compact sets 
are automatically uniformly continuous. 

Now, let be $varepsilon > 0$. Then,
there exist $delta_{1/2}$ such that $$ lvert f(x_i) - f(1) rvert < 
frac{varepsilon}{2} $$ for $lvert x_i - 1 rvert < delta_i$ where $0 leq x_1 leq 1 leq x_2$. Of course, $delta_i$ does not depend on $1$ nor $x_i$ due to the uniform continuity; this is important. Hence, 
if we choose $delta = min{ delta_1,delta_2 }$, we obtain $$ lvert 
f(x_1) - f(x_2) rvert leq lvert f(x_1) - f(1) rvert + lvert f(1) - 
f(x_2) rvert < frac{varepsilon}{2} + frac{varepsilon}{2} = 
varepsilon$$ since $$lvert x_i - 1 rvert leq lvert x_1 - x_2 rvert < delta leq delta_i$$
for $0 leq x_1 leq 1 leq x_2$. For arbitrary $x_1,x_2$ both entirely in either $[0,1]$ or $[1,infty)$, we can use the same $delta$ due to the choices of each $delta_i$, which proves the statement.