Let $f:\mathbb R \to \mathbb R$. Prove that if $f$ is uniformly continuous on $[a,b]$ and $[b,c]$, then $f$ is uniformly continuous on $[a,c]$.

My attempt at a solution:

I've came up with a solution but I am having doubts if it is correct, so I would like to check that:

Let $\epsilon>0$, we know that there exist $\delta_1, \delta_2$ such that if $x,y \in [a,b] |x-y|<\delta_1 \implies |f(x)-f(y)|<\dfrac{\epsilon}{2}$

and if

$x,y \in [b,c], |x-y|<\delta_2 \implies |f(x)-f(y)|<\dfrac{\epsilon}{2}$

Let $\delta=\min\{2\delta_1,2\delta_2\}$, if $|x-y|=|x-b+b-y|\leq |x-b|+|b-y|<\delta \implies |f(x)-f(y)|\leq |f(x)-f(b)|+|f(b)-f(y)|\leq \dfrac{\epsilon}{2}+\dfrac{\epsilon}{2}=\epsilon$. This proves that $f$ is uniformly continuous.

I would appreciate if anyone could tell me if my proof is ok or if I've made any mistakes.
(S.A. Understanding Analysis. pp 119 question 4.4.7)


Your proof never uses any of the properties of $\delta$, so it's incomplete. I think you want to set $\delta = \frac{1}{2} \min(\delta_1, \delta_2)$ rather than doubling the deltas. I'd suggest a slight rewrite:

Given $\epsilon > 0$, suppose that $|x - y| < \delta$. Then either (a) $x$ and $y$ are both greater than $b$, or (b) they're both less than $b$ or (c) one is less and one is greater. Without loss of generality, assume $x \le c \le y$. We'll proceed case by case to show that $|f(x) - f(y)| < \epsilon$.

(a) Since $|x - y| < \delta$, we know $|x - y| < \delta_2$; that means (by the unif. continuity assumption for $[b, c]$ that $|f(x) - f(y)| < \epsilon$.

(b) similar to case a.

(c) [your proof, somewhat modified and cleaned up, goes here. You'll need to use the fact that $|x - b| + |b - y| = |x - y|$ because $b$ is between $x$ and $y$. Then you'll put bounds on each of the two terms, and you're on your way.]

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  • $\begingroup$ Extremely useful, thanks. My proof didn't convince me but I couldn't see where my mistake was. $\endgroup$ – user100106 Nov 26 '13 at 23:51
  • $\begingroup$ I couldn't understand which is the correct $\delta$ for the third point: I know that if for $x,y\in [a,b]$, then there's a $\delta_1$ such that $|x-y|<\delta_1$ implies... Same for $w,z\in [b,c]$. Now, if $x\in[a,b], z\in [b,c]$ which $\delta$ works? $\endgroup$ – YoTengoUnLCD Feb 28 '16 at 21:25

Very close. When you have chosen your $2\delta = \min\{\delta_1,\delta_2\}$, I suggest the following: Take $x, y \in [a,c]$ with $|x-y|<\delta$. Now, break this into 3 cases. Case 1: $x, y \in [a,b]$, Case 2: $x, y \in [b,c]$, Case 3: $x \in [a,b]$ and $y \in (b,c]$. For Case 3, use the triangle inequality you suggested. Otherwise, good job!

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  • $\begingroup$ Thanks for the help!, but be careful with $\delta$, see what John wrote above. $\endgroup$ – user100106 Nov 26 '13 at 23:50
  • $\begingroup$ @user100106 Didn't I choose the same $\delta$ as John? I just wrote $2 \delta = \min(\delta_1, \delta_2)$ versus $\delta = \frac{1}{2} \min(\delta_1, \delta_2)$. $\endgroup$ – Tom Nov 26 '13 at 23:51
  • $\begingroup$ … although admittedly, when I first wrote the solution, I did write $\delta = \min(2\delta_1, 2\delta_2)$, but immediately edited for a fix on realization of the error… $\endgroup$ – Tom Nov 26 '13 at 23:57
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    $\begingroup$ @user100106 I think both John and I made an overkill with the adjusted $\delta$. You should be able to get away with your $\epsilon/2$ proof with simply $\delta = \min(\delta_1, \delta_2)$. Another good point. $\endgroup$ – Tom Nov 27 '13 at 0:04
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    $\begingroup$ @user100106 No Problem! The point is the following: You won't need to do a triangle inequality with $\delta$ in mind. The only triangle inequality you will need is $|f(x)-f(y)| \leq |f(x)-f(b)|+|f(b)-f(y)|$ and therefore only needed to adjust the $\epsilon$. $\endgroup$ – Tom Nov 27 '13 at 0:09

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