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Proving uniform continuity on an interval
If $f$ is uniformly continuous on $(-\infty,-a]$ and $[-a,0]$, then it is uniformly continuous on $(-\infty,0]$ where $a\in(0,\infty)$.
Proof:
From the definition of uniform continuity we have that:
$\forall \epsilon > 0 \exists \delta_1: \forall x,y \in(-\infty,-a] : |x-y|\lt \delta_1: |f(x)-f(y)| \lt \frac{\epsilon}{2}$
and
$ \forall \epsilon >0 \exists \delta_2: \forall x,y \in [-a,0]: |x-y|< \delta_2 : |f(x)-f(y)| \lt \frac{\epsilon}{2}.$
If we take $\delta = \min\{\delta_1,\delta_2\}$ then we have that:
If $x,y\in(-\infty,-a]$ then $|x-y|<\delta \rightarrow |f(x)-f(y)|<\epsilon$.
If $x,y\in[-a,0]$ then $|x-y|< \delta\rightarrow |f(x)-f(y)|<\epsilon$.
If $x\in(-\infty,-a]$ and $y\in[-a,0]$ then we have that as $\delta>|x-y|>|x-(-a)|$ and $\delta>|x-y|>|y-(-a)|$
from the uniform continuity of $f$ on $(-\infty,-a]$ and $[-a,0]$ we have that $ \begin{align*} |f(x)-f(y)| &= |f(x)-f(-a)+f(-a)-f(y)| \\ &< |f(x)-f(-a)|+|f(y)-f(-a)| < \frac{\epsilon}{2} + \frac{\epsilon}{2}=\epsilon . \end{align*} $
Is the above correct? Thanks for any help!
So we have that $f$ is uniformly continuous on $(-\infty,0]$.