Let $f$ and $g$ be uniformly continuous on A. Then given $\epsilon >0$ there exists a $\delta_{1} > 0$ such that if $|x-y| < \delta_{1}, \forall x,y \in A$, then $|f(x)-f(y)| < \frac{\epsilon}{2M}$. There also exists a $\delta_{2} > 0$ such that if $|x-y| < \delta_{2}, \forall x,y \in A$, then $|g(x)-g(y)| < \frac{\epsilon}{2M}$. Since $f$ and $g$ are both bounded on a, there exists $M_{1}>0$ such that $|f(x)| \leq M_{1}, \forall x,y \in A$ and there exists $M_{2}>0$ such that $|g(x)| \leq M_{2}, \forall x,y \in A$. Let $M=\left\{M_{1}, M_{2}\right\}$. Then $|f(x)| \leq M$ and $|g(x) < M$ for all $x \in A$. Let $\delta=\left\{\delta_{1}, \delta_{2}\right\}$. So if $|x-y|<\delta$ then $|f(x)-f(y)| < \frac{\epsilon}{2M}$ and $|g(x)-g(y)| < \frac{\epsilon}{2M}$. Now consider $|f(x)g(x) - f(y)g(y)| = |f(x)g(x) - g(x)f(y) + g(x)f(y)-f(y)g(y)|$. Then, $|g(x)||f(x)-f(y)| + |f(y)||g(x)-g(y)| \leq M|f(x)-f(y)|+M|g(x)-g(y)|< M\frac{\epsilon}{2M} + M\frac{\epsilon}{2M} = \epsilon$. Does this look right to anyone?

  • $\begingroup$ I am a student in analysis like you,but I think your proof is correct, except some minor fix like: $M = \text{max}(M_1,M_2), \delta = \text{min}(\delta_1,\delta_2)$. $\endgroup$ – DeepSea Mar 27 '16 at 2:42
  • $\begingroup$ A bit of reordering would be helpful. You are already using $M$ in the second sentence, but it is not defined until later. But the argument you are using is correct. $\endgroup$ – Bungo Mar 27 '16 at 5:11

And now for something slightly different:

Since $f,g$ are bounded on $A$, their ranges lie in some compact set $[-B,B]$. Since $[-B,B]^2$ is compact, we see that multiplication $\cdot : [-B,B]^2 \to \mathbb{R}$ is uniformly continuous.

Since $f,g$ are uniformly continuous separately, we see that the map $p(x) = (f(x),g(x))$ is also uniformly continuous.

Since the composition of uniformly continuous maps is again uniformly continuous (this follows almost immediately from the definition), we see that $\cdot \circ p$ is uniformly continuous.

  • $\begingroup$ @user251257: Thanks for catching that. A cut and paste laziness. $\endgroup$ – copper.hat Mar 27 '16 at 4:25

The following version of your proof works out the essentials:

For any sort of multiplication one has $$uv-u_0v_0=v(u-u_0)+u_0(v-v_0)\ .$$ If all quantities appearing here are bounded in absolute value by some constant $M>0$ it follows that $$|uv-u_0v_0|\leq |v||u-u_0|+|u_0||v-v_0|\leq M\bigl(|u-u_0|+|v-v_0|\bigr)\ .\tag{1}$$ Given an $\epsilon>0$ choose a $\delta>0$ that is sufficient for both $f$ and $g$ and tolerance $\epsilon':={\epsilon\over 2M}$. From $(1)$ it then follows that this $\delta$ is sufficient for $fg$ and tolerance $\epsilon$.


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