# Prove that if $f$ is uniformly continuous on $[a,b]$ and $f$ is uniformly continuous on $[b,c]$, then $f$ is uniformly continuous on $[a,c]$.

Prove that if $f$ is uniformly continuous on $[a,b]$ and $f$ is uniformly continuous on $[b,c]$, then $f$ is uniformly continuous on $[a,c]$.

My attempt:

Let $\epsilon >0$ be given. Since $f$ is uniformly continuous on $[a,b]$ there $\exists \delta_1 >0$ such that if $x,y \in [a,b]$ and $|x-y|<\delta_1$, then $|f(x)-f(y)|<\epsilon$.

Since $f$ is uniformly continuous on $[b,c]$ there $\exists \delta_2 >0$ such that if $x,y \in b,c]$ and $|x-y|<\delta_2$, then $|f(x)-f(y)|<\epsilon$.

Now to show $f$ is continuous on $[a,c]$ how would i show this. Do i sort of add the two above relations?

• Use continuity at $b$ when $x\in[a,b]$ and $y\in[b,c]$. – hamam_Abdallah Dec 1 '16 at 19:34
• Hint: For $a \le x \le b \le y \le c$ follows $b - x, y-b \le y-x$. – user251257 Dec 1 '16 at 19:46
• Btw. Here is a probably unintended shortcut: $f$ is continuous at $b$ thus continuous on the compact interval $[a,c]$ thus uniformly continuous... – user251257 Dec 1 '16 at 20:14
• @jimm bo first of all function must be continuous on $[a,c].$ – neelkanth Feb 26 '18 at 7:59

Uniform continuity is not a pointwise, relative but an absolute attribute meaning "$f$ is uniformly continuous on $[a,b]$" may be put as "$f$ is uniformly continuous as function on $[a,b]$".

Let $\epsilon>0$ be arbitrary.

Since $f$ is uniformly continuous as function on $[a,b]$ there is a $\delta_1>0$ so that for all $x,y \in [a,b]$: $$|y-x|<\delta_1\implies |f(y)-f(x)|<\epsilon /2.$$

Since $f$ is uniformly continuous as function on $[b,c]$ there is a $\delta_2>0$ so that for all $x,y \in [b,c]$: $$|y-x|<\delta_2\implies |f(y)-f(x)|<\epsilon /2.$$

Set $\delta := \min\{\delta_1,\delta_2\}.$

Consider $x,y\in[a,c]$: If $x,y$ are both in either $[a,b]$ or $[b,c]$ we're done due to $\delta \leq \delta_1$ resp. $\delta \leq \delta_2$. So without loss of generality it suffices to consider $x\in [a,b]$ and $y\in[b,c]$ for the rest. We then have for $|y-x|<\delta$: $$|f(y)-f(x)|\leq|f(y)-f(b)|+|f(b)-f(x)| < \epsilon/2+\epsilon/2 = \epsilon.$$

Choose the minimum of the deltas from either interval corresponding to epsilon. This delta works.