Uniform continuity proof problem

Let $f$ be continuous on $\mathbb R$ and assume that there is a constant $C$ such that

$$|f(x)| \leq C$$

for all $x \in \mathbb R$.

Let $g$ be uniformly continuous on $\mathbb R$. Show that $fg$ is uniformly continuous on $\mathbb R$.

I just need a hint on how to start this! I'm drawing a blank.

• Start with the definition of uniform continuity! – Emily Nov 27 '13 at 20:42
• Hint: It's wrong. $f$ needs to be uniformly continuous too. And even that isn't sufficient. – Daniel Fischer Nov 27 '13 at 20:46
• @DanielFischer, we are now given that |g(x)| is less than or equal to a constant M. Does this make it solvable? – Pineapple Tacos Nov 29 '13 at 0:02
• You still need the uniform continuity of $f$. Consider $f(x) = \sin (x^2)$ for example. – Daniel Fischer Nov 29 '13 at 0:06

It is not true: you need more. In particular, let us understand what is missing. $fg$ is uniformly continuous on whole $\mathbb R$ if

$$\forall \epsilon>0 ~\text{there exists} ~\delta=\delta(\epsilon)~ \text{s.t.}~ \forall x_1,x_2\in\mathbb R : |x_1-x_2|<\delta\Rightarrow |f(x_1)g(x_1)-f(x_2)g(x_2)|<\epsilon~~(*)$$

Let us study conditions on $f$ and $g$ to ensure $(*)$. We can write

$$|f(x_1)g(x_1)-f(x_2)g(x_2)|=|f(x_1)g(x_1)+f(x_1)g(x_2) - f(x_1)g(x_2) -f(x_2)g(x_2)|\leq |f(x_1)g(x_1)-f(x_1)g(x_2) |+ |f(x_1)g(x_2)-f(x_2)g(x_2)|=\\|f(x_1)||g(x_1)-g(x_2)|+ |g(x_2)||f(x_1)-f(x_2)|;$$

Now, in the first term in the r.h.s. of the last equality, i.e. $|f(x_1)||g(x_1)-g(x_2)|$ we can use that $f$ is bounded and $g$ is uniformly continuous. What about the second term, i.e. $$|g(x_2)||f(x_1)-f(x_2)| ~~(**)$$

? Can you find additional hypothesis on $f$ and $g$ to estimate $(**)$ and arrive at the thesis, i.e. $(*)$?

• Okay. That makes sense to me. So there's absolutely no way to estimate (**)? |f(x_1) - f(x_2)| would be less than 2C, right? but I guess we can't estimate |g(x_2)|? – Pineapple Tacos Nov 27 '13 at 21:32
• if $f$ is also uniformly continuous and $g$ is bounded, then...ps: btw, do you see in $(*)$ where "uniformly" lies? – Avitus Nov 27 '13 at 21:34
• I'm not sure what you're asking? – Pineapple Tacos Nov 27 '13 at 21:44
• what is the difference between "continuity" and "uniform continuity"? It is a small detail in $(*)$, i.e. the fact that $\delta$ is a function only of $\epsilon$, and not of points $x\in\mathbb R$. This is an important fact, that makes uniform continuity on a set stronger than continuity – Avitus Nov 27 '13 at 21:48
• @Avitus: I gave the start for the OP to work his problem out and see what conditions he needs to impose on the functions. – Mhenni Benghorbal Nov 29 '13 at 19:50

Here is a start,

$$|g(x)f(x)-g(y)f(y)| = |g(x)f(x)-g(y)f(x)+g(y)f(x)-g(y)f(y)|$$

$$\leq |g(x)f(x)-g(y)f(x)| + |g(y)f(x)-g(y)f(y)| .$$

• What's the downvote for? This is a start for OP to work his problem out and see what kind of conditions he needs to impose on his function. Instead, @Avitus did the job. – Mhenni Benghorbal Nov 29 '13 at 19:51