# uniformly continuous functions in [0,∞)

Help me please to solve this. Prove or disprove that:

1. If $f,g$ are uniformly continuous functions in $[0,\infty )$ then also $fg$ (product) is uniformly continuous in $[0,\infty )$.

2. Same question, but about the composition of $f$ and $g$ Thanks beforehand.

-

## 1 Answer

For the first one let $f(x) = g(x) = x$. Then $f$ and $g$ are uniformly continuous since for $\varepsilon > 0$ you can choose $\delta := \varepsilon$ so that for $| x - x^\prime| < \delta = \varepsilon$ you get $|f(x) - f(x^\prime)| = |x - x^\prime| < \varepsilon$.

On the other hand, $h(x) := f(x)g(x) = x^2$ is not uniformly continuous on $[0, \infty)$ (Proof taken from here):

Let $\varepsilon := 1$. Let $\delta > 0$ be arbitrary but fixed. For $n \in \mathbb N$ let $x_n := \sqrt{ n + 1}$ , $y_n := \sqrt{n}$. Then $|h(x_n) - h(y_n)| = |x_n^2 - y_n^2| = 1$ and $$0 < x_n - y_n < \frac{(\sqrt{n + 1} - \sqrt{n})(\sqrt{n + 1} + \sqrt{n})}{\sqrt{n + 1} + \sqrt{n}} = \frac{1}{\sqrt{n + 1} + \sqrt{n}} < \frac{1}{\sqrt{n} + \sqrt{n}} < \delta$$

for $n$ large enough. Hence you have found a $\varepsilon > 0$ such that for every $\delta > 0$ you have $|x - y| < \delta$ but $| f(x) - f(y)| \geq \varepsilon$.

-
If one of the functions is bounded for example then the product will still be uniformly continuous. The composition of uniformly continuous functions is of course uniformly continuous: For $\varepsilon \gt 0$ find $\delta_{g}$ such that $|x - y| \lt \delta_g$ implies $|g(x) - g(y)| \lt \varepsilon$ and now choose $\delta_f$ such that $|x-y| \lt \delta_f$ implies $|f(x) - f(y)| \lt \delta_g$ then $|g(f(x)) - g(f(y))| \lt \varepsilon$. – t.b. Mar 20 '12 at 9:45
@t.b. Nice, thank you for providing the missing parts to my answer. : ) – Rudy the Reindeer Mar 20 '12 at 20:28