Question

Show that if f and g are uniformly continuous on a r and if they are both bounded on a, then their product fg is uniformly continuous on a.

a) The product fg is not uniformly continuous.
b) The product fg is uniformly continuous.
c) Uniform continuity of fg depends on specific functions f and g.
d) Insufficient information to determine the continuity of fg.

Answer 1

To show that the product fg is uniformly continuous on a, we need to prove that for any ε > 0, there exists a δ > 0 such that for all x and y in a, |x - y| < δ implies |(fg)(x) - (fg)(y)| < ε.

Step-by-step explanation:

To show that the product fg is uniformly continuous on a, we need to prove that for any ε > 0, there exists a δ > 0 such that for all x and y in a, |x - y| < δ implies |(fg)(x) - (fg)(y)| < ε.

Since both f and g are uniformly continuous on a, we know that for any ε > 0, there exists a δ1 > 0 such that for all x and y in a, |x - y| < δ1 implies |f(x) - f(y)| < ε/2, and there exists a δ2 > 0 such that for all x and y in a, |x - y| < δ2 implies |g(x) - g(y)| < ε/2.

Now, let δ be min(δ1, δ2). Then, for any x and y in a, |x - y| < δ implies both |f(x) - f(y)| < ε/2 and |g(x) - g(y)| < ε/2. Therefore, we have |(fg)(x) - (fg)(y)| = |f(x)g(x) - f(y)g(y)| = |f(x)(g(x) - g(y)) + (f(x) - f(y))g(y)| ≤ |f(x)||g(x) - g(y)| + |f(x) - f(y)||g(y)| < (M1 + ε/2)(ε/2) + (ε/2)(M2) = ε,

where M1 and M2 are upper bounds on f and g, respectively. Thus, we have shown that fg is uniformly continuous on a.