# Uniform continuity on $[a,b]$ and $[b,c]$ $\implies$ uniform continuity on $[a,c]$.

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)

• – Analysis JD Feb 16 '14 at 10:18

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.]

• Extremely useful, thanks. My proof didn't convince me but I couldn't see where my mistake was. – user100106 Nov 26 '13 at 23:51
• 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? – 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!

• Thanks for the help!, but be careful with $\delta$, see what John wrote above. – user100106 Nov 26 '13 at 23:50
• @user100106 Didn't I choose the same $\delta$ as John? I just wrote $2 \delta = \min(\delta_1, \delta_2)$ versus $\delta = \frac{1}{2} \min(\delta_1, \delta_2)$. – Tom Nov 26 '13 at 23:51
• … 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… – Tom Nov 26 '13 at 23:57
• @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. – Tom Nov 27 '13 at 0:04
• @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 $\epsilon$. – Tom Nov 27 '13 at 0:09