Assume that $g$ is defined on an open interval $(a, c)$ and it is known to be uniformly continuous on $(a, b]$ and $[b, c)$, where $a < b < c$. Prove that g is uniformly continuous on $(a, c)$.

My attempt at proof:

By hypothesis, we have: $$\forall \epsilon>0, \exists \delta_1>0, \forall x,y \in (a,b]: |x-y|<\delta_1 \implies |f(x)-f(y)|<\epsilon. $$ and $$\forall \epsilon>0, \exists \delta_2>0, \forall x,y \in [b,c): |x-y|<\delta_2 \implies |f(x)-f(y)|<\epsilon. $$ If we pick $\delta = \min \{\delta_1,\delta_2 \}$, then we have $$\forall \epsilon>0, \exists \delta>0: |x-y|<\delta \implies |f(x)-f(y)|<\epsilon, $$ whenever $x,y$ is in $(a,b]$ or $[b,c)$, or equivalently, whenever $x,y$ is in $(a,c)$.

Is this correct?

  • $\begingroup$ I am still not sure why we have to be explicit on the cases about where $x$ or $y$ will fall in $(a,b]$ or $[b,c)$. $\endgroup$ – Kurome Jul 12 '15 at 15:40
  • $\begingroup$ I updated my answer with (hopefully) all of the correct details filled in. Please review it carefully, and if you are still confused about why we have to worry about the cases, let me know and I will comment back. $\endgroup$ – layman Jul 12 '15 at 15:46

I think the idea of your proof is right. Basically, when proving continuity (or uniform continuity), given $\epsilon > 0$, if you can find a $\delta > 0$, then any smaller $\delta '$ will also work. I just think you need to be a bit more careful about the details.

So we know given $\epsilon > 0$, we can find $\delta_{1} > 0$ such that for all $x, y \in (a,b]$ (even if one of them equals $b$), $|x - y| < \delta_{1} \implies |f(x)- f(y)| < \frac{\epsilon}{2}$, right?

Also, for the same $\epsilon > 0$, we can find $\delta_{2} > 0$ such that for all $w, z \in [b,c)$ (in particular, also if one of them equals $b$), $|x - y| < \delta_{2} \implies |f(x) - f(y)| < \frac{\epsilon}{2}$.

Now, if both of our inputs are in $(a,b]$, then letting $\delta = \min \{\delta_{1}, \delta_{2} \}$ is OK because any smaller $\delta$ still works. Similarly, if they are both in $[b,c)$, it still works. We only have to worry about if $x \in (a,b]$ and $y \in [b,c)$ (since we want $x,y \in (a,c)$ to be arbitrary, so we have to consider all cases). So let's assume this case since the other cases are taken care of.

So, let's assume $x \in (a,b]$ and $y \in [b,c)$. Then it should be clear that the distance between $x$ and $b$ is smaller than the distance between $x$ and $y$, right? i.e., $|x - b| \leq |x - y|$. Similarly, the distance between $y$ and $b$ is smaller than the distance between $x$ and $y$, i.e., $|y - b| \leq |x - y|$. Then if $|x - y| < \delta$, we have $|x - b|$, $|y - b| < \delta$, right? Then the first one is smaller than $\delta_{1}$, and the second one is smaller than $\delta_{2}$, right? So $|f(x) - f(b)| < \frac{\epsilon}{2}$ and $|f(y) - f(b)| < \frac{\epsilon}{2}$.

So, if $|x - y| < \delta$, the above inequalities hold. Finally, we want to use the trick of adding $0$ in the form of $-f(b) + f(b)$ to show that $ |f(x) - f(y)| < \epsilon$ if $|x - y| < \delta$.

So, given $\epsilon > 0$, let $\delta = \min \{ \delta_{1}, \delta_{2} \}$. Then $|x - y| < \delta$ implies $ |f(x) - f(y)| = |f(x) - f(b) + f(b) - f(y)| \leq |f(x) - f(b)| + |f(b) - f(y)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$, and we are done.

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  • $\begingroup$ So my proof only succeeded to cover for the case when $x,y \in (a,b]$, or $x,y \in [b,c)$ but not for the case $x\in (a,b]$ and $y \in [b,c)$? $\endgroup$ – Kurome Jul 12 '15 at 17:21
  • $\begingroup$ @Kurome That's right. If you think about the XY-plane, and imagine a function drawn in it over some interval $(a,c)$, then uniform continuity says for every $\epsilon > 0$, you can find some $\delta$ such that every interval of length $\delta$ in $(a,c)$ satisfies that for each two points in it, the distance between their images is $< \epsilon$. So you can imagine drawing the interval $(a,c)$, and picking some $b$ in between the end points. Then make your index finger and thumb into a fixed width, and run it along the interval $(a,c)$ on your paper... $\endgroup$ – layman Jul 12 '15 at 17:27
  • $\begingroup$ ...pretend this width has length $\delta$. Then as you run your fingers back and forth along this interval, you are basically going through all of the intervals in $(a,c)$ of length $\delta$. Each of these intervals better have the property that any two points in the interval have images less than $\epsilon$ apart. But when your fingers run over $b$ so that $b$ is between you two fingers (i.e., in the interval), you still need to show that for any two points in this interval, their images are $< \epsilon$ apart. It's easy (and follows by assumption) if the two points are also in $(a,b]$... $\endgroup$ – layman Jul 12 '15 at 17:29
  • $\begingroup$ ...or if they are both in $[b,c)$. But you need any two points in this interval to satisfy their images are less than $\epsilon$ apart, so you need to verify the last case where one point is in $(a,b]$ and one point is in $[b,c)$. I hope this is clear and that I didn't confuse you. $\endgroup$ – layman Jul 12 '15 at 17:30
  • $\begingroup$ I'll read it next time, but I think I already understood the proof for the latter case. $\endgroup$ – Kurome Jul 12 '15 at 17:35

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