Exercise 4.4.10

Question

Assume that $f$ and $g$ are uniformly continuous functions defined on a common domain $A$. Which of the following combinations are necessarily uniformly continuous on $A$:
  $$
  f(x)+g(x), \quad f(x) g(x), \quad \frac{f(x)}{g(x)}, \quad f(g(x)) ?
  $$
  (Assume that the quotient and the composition are properly defined and thus at least continuous.)

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\roman*)] \item $f(x) + g(x)$ is clearly uniformly continuous \item $f(x) = g(x) = x$ are individually uniformly continuous over $\mathbf{R}$ but $f(x)g(x) = x^2$ is not. Note this counterexample only works when $A$ is unbounded. If $A$ is bounded you can prove $f(x)g(x)$ must be uniformly continuous the same way you prove the product rule. \item False, consider $f(x)=1$ and $g(x) = x$ over $(0,1)$. Both are uniformly continuous on $(0,1)$ but $f/g = 1/x$ is not. \item Want $|f(g(x)) - f(g(y))|<\epsilon$. Since $f$ is uniformly continuous we can find an $\alpha > 0$ so that $$|g(x) - g(y)| < \alpha \implies |f(g(x))-f(g(y))| < \epsilon$$ Now since $g$ is uniformly continuous we can find a $\delta > 0$ so that $$|x - y| < \delta \implies |g(x) - g(y)| < \alpha$$ Combine the two to get $$|x-y| < \delta \implies |f(g(x)) - f(g(y))| < \epsilon$$ as desired, which proves that $f(g(x))$ is uniformly continuous. \end{enumerate}

Exercise 4.4.10

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Exercise 4.4.10

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