Question

Uniform continuity:

Prove that if f and g are uniformly continuous on R and are bound, then f*g is uniformly continuous on R.

Answer 1

Conditions

Uniform continuity:

Prove that if $f$ and $g$ are uniformly continuous on $R$ and are bound, then $f*g$ is uniformly continuous on $R$.

Solution

Consider:

$$f: M \in R \to R$$

Function $f$ is uniformly continued in $M$, if:

$$\forall \varepsilon > 0 \ \exists \delta = \delta(\varepsilon) > 0: \forall x_1, x_2 \in M \ |x_1 - x_2| < \delta \ |f(x_1) - f(x_2)| < \varepsilon$$

For our case:

As functions are bounded, so

$$\exists M_1, M_2: \forall x \in R \ |f(x)| < M_1 \ |g(x)| < M_2$$ $$\forall \varepsilon > 0 \ \exists \delta_1 = \delta_1(\varepsilon) > 0: \forall x_1, x_2 \in R \ |x_1 - x_2| < \delta_1 \ |f(x_1) - f(x_2)| < \frac{\varepsilon}{2M_1}$$ $$\forall \varepsilon > 0 \ \exists \delta_2 = \delta_2(\varepsilon) > 0: \forall x_1, x_2 \in R \ |x_1 - x_2| < \delta_2 \ |g(x_1) - g(x_2)| < \frac{\varepsilon}{2M_2}$$

Fix $\varepsilon > 0$, $\exists \delta = \min(\delta_1, \delta_2)$. Consider:

$$\begin{array}{l} |f(x_1)g(x_1) - f(x_2)g(x_2)| = |f(x_1)g(x_1) - f(x_1)g(x_2) + f(x_1)g(x_2) - f(x_2)g(x_2)| \leq \\ \leq |f(x_1)g(x_1) - f(x_1)g(x_2)| + |f(x_1)g(x_2) - f(x_2)g(x_2)| < \\ < M_1 |g(x_1) - g(x_2)| + M_2 |f(x_1) - f(x_2)| \leq \varepsilon \end{array}$$

Q.E.D.