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I'm trying to prove that if $f$ and $g$ are uniformly continuous on an interval $(a,b)$ then so is $f+g$. I have a picture in my head but I can't seem to make it into a formal proof.

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  • $\begingroup$ $\epsilon = \min(\epsilon_f, \epsilon_g)/2$. $\endgroup$ – marty cohen Feb 18 '15 at 2:40
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It is the same proof that the limit of the sum is the sum of the limits, the same proof that the sum of (just) continuous functions is continuous, etc: take the minimum $\delta$, and use the triangle inequality.

Let $\epsilon > 0$. Then exists $\delta_1,\delta_2 > 0$ such that: $$\begin{cases} |x-y|<\delta_1 \implies |f(x)-f(y)|<\epsilon/2 \\ |x-y|<\delta_2 \implies |g(x)-g(y)|<\epsilon/2\end{cases}$$ Choose $\delta = \min\{\delta_1,\delta_2\}$. So if $|x-y|<\delta$ all the above holds, and we have: $$|(f+g)(x)-(f+g)(y)| \leq |f(x)-f(y)|+|g(x)-g(y)|\leq \frac{\epsilon}{2}+\frac{\epsilon}{2} = \epsilon.$$

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Claim: If $f$ and $g$ are uniformly continuous on their common domain $D$, then $f+g$ is uniformly continuous on $D$.

Proof. Let $\epsilon>0$ be given. To show $f+g$ is uniformly continuous on $D$, we need to find $\delta>0$ so that $|(f+g)(x)-(f+g)(t)|<\epsilon$, for all $x,t\in D$ satisfying $|x-t|<\delta$. To this end, observe that $f$ uniformly continuous implies that there exists $\delta_1>0$ such that $|f(x)-f(t)|<\frac{\epsilon}{2}$ whenever $|x-t|<\delta_1$ and $x,t\in D$. Also, $g$ uniformly continuous implies that there exists $\delta_2>0$ such that $|g(x)-g(t)|<\frac{\epsilon}{2}$ whenever $|x-t|<\delta_2$ and $x,t\in D$. Thus, choose $\delta=\min\{\delta_1,\delta_2\}$. Then for $x,t\in D$ with $|x-t|<\delta$, we have $$ |(f+g)(x)-(f+g)(t)|\leq|f(x)-f(t)|+|g(x)-g(t)|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon. $$ Hence, $f+g$ is uniformly continuous on $D$.

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Let $\{f_n \}, \{g_n\}$ be sequences that converge uniformly to $f$ and $g$, respectively. Then given $\frac{\epsilon}{2}>0$ there exists $M, N \in \Bbb{N}$ such that $$\left|f_n(x)-f(x)\right|<\frac{\varepsilon}{2} \quad \text{and} \quad \left|g_m(x)-g(x)\right|<\frac{\varepsilon}{2}$$ for all $x \in (a,b)$ and all $n \geq N, \space m \geq M$. WLOG let $N = \max\{N,M\}$. Then $\left|g_n(x)-g(x)\right|<\frac{\varepsilon}{2}$ still holds. By adding up the inequalities and using the triangle inequality, we have $$\frac{\varepsilon}{2}+\frac{\varepsilon}{2} > \left|g_n(x)-g(x)\right|+\left|f_n(x)-f(x)\right| \\ \geq \left|(g_n(x)-g(x))+(f_n(x)-f(x))\right| \\ = \left|(g_n(x)+f_n(x))-(f(x)+g(x))\right|$$ Hence, $$\varepsilon > \left|(g_n(x)+f_n(x))-(f(x)+g(x))\right|$$ for all $\varepsilon>0$, all $x \in (a,b)$ and all $n \geq N$.

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