Is sum, product and composition of uniformly continuous functions also uniformly continuous?

Question

I know that sum and composition are and product isn't but need prove of it or counter-example.

I have tried do it from definition but failed.

Answer 1

Hint: $x$ is uniformly continuous but $x^2$ is not.

Answer 2

I wonder if your question is more about how to use definitions and write proofs than about uniform continuity, since the proof for compositions is a quite straightforward application of the definition.

Suppose $g$ is uniformly continuous; then for every $\varepsilon>0$, no matter how small, there exists $\eta>0$ so small that whenever $|x-y|<\eta$ then $|g(x)-g(y)|<\varepsilon.$

But suppose $f$ is also uniformly continuous; then for every $\eta>0$, no matter how small, there exists $\delta>0$ so small that whenever $|s-t|<\delta$, then $|f(s)-f(t)|<\eta$.

So $|s-t|<\delta$ entails $|f(s)-f(t)|<\eta$ and $|f(s)-f(t)|<\eta$ entails $g(f(s))-g(f(t))|<\varepsilon.$

Given $\varepsilon>0$ there exists $\eta>0$ and given $\eta>0$ there exists $\delta>0$. Therefore, given $\varepsilon>0$ there exists $\delta>0$.

For sums, suppose we know that for every $\varepsilon>0$ there exists $\delta>0$ such that whenever $|x-y|<\delta$ then $|f(x)-f(y)|<\varepsilon$. This says something is true of EVERY positive number (in the role of $\varepsilon$). If it's true of EVERY positive number, then it's true of $\varepsilon/2$, since that is a positive number. So we can say there is some $\delta_1>0$ such that whenever $|x-y|<\delta_1$ then $|f(x)-f(y)|<\varepsilon/2$ (this number $\delta_1$ generally needs to be smaller than the value of $\delta$ mentioned earlier in this paragraph). Similarly, there is some number $\delta_2>0$ such that whenever $|x-y|<\delta_2$ then $|g(x) - g(y)|<\varepsilon/2.$ Now we can think about the sum $f+g.$ Let $\delta_3 = \min\{\delta_1,\delta_2\}.$ \begin{align} & \text{If }|x-y|<\delta_3 \text{ then } |x-y| < \delta_1 \text{ and so } |f(x)-f(y)|< \frac\varepsilon 2. \\[10pt] & \text{If }|x-y|<\delta_3 \text{ then } |x-y| < \delta_2 \text{ and so } |g(x)-g(y)| < \frac\varepsilon 2. \\[10pt] & \text{From this it follows that} \\[10pt] & | (f+g)(x) - (f+g)(y)| = \left| \Big(f(x) - f(y)\Big) + \Big(f(x)-g(y)\Big) \right| \\[10pt] \le {} & |f(x)-f(y)| + |g(x)-g(y)| < \frac\varepsilon 2 + \frac\varepsilon 2 = \varepsilon. \end{align}