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Here is the question I'm having trouble with:

Suppose that $g$ is uniformly continuous on an interval $I$ and that $f$ is uniformly continuous on an interval $J$ that contains $g(I)$. Prove that the composite function $f g$ is uniformly continuous on $I$.

First I'm taking the definition of uniform continuity as:

The function $f$ is uniformly continuous on $I$ if for every $ε > 0$, there exists a $δ > 0$ such that: $|x − y| < δ$ implies $|f(x) − f(y)| < ε$.

I am able to prove this if the function $f$ and $g$ are defined as uniformly continuous and bounded on the same interval $I$. Here is my proof in that case:

Because $f$ and $g$ are bounded, there is an $M > 0$ such that $|f(x)| < M$ and $|g(x)| < M$ for all $x$ in $I$. Because $f$ and $g$ are uniformly continuous on $I$, given $ε > 0$, there is a $δ > 0$ such that if $x$ and $y$ are in $I$ and $|x − y| < δ$, then $|f(x) − f(y)| < ε/2M$ and $|g(x) − g(y)| < ε/2M$. Then:

\begin{align*} |f(x)g(x) − f(y)g(y)| &= |f(x)g(x) − f(y)g(x) + f(y)g(x) − f(y)g(y)|\\ &\le |f(x)g(x) − f(y)g(x)| + |f(y)g(x) − f(y)g(y)| & \text{(Triangle Inequality)}\\ &\le |g(x)||f(x) − f(y)| + |f(y)||g(x) − g(y)| \\ &\le M|f(x) − f(y)| + M|g(x) − g(y)| & \text{(f and g bounded by M)}\\ & < M (ε/2M) + M(ε/2M) & \text{(Uniform continuity)}\\ &= ε. \end{align*}

However, I am having trouble expanding this proof to include the case where $f$ is uniformly continuous on an interval $J$ that contains $g(I)$. Please help!

Thanks a lot.

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    $\begingroup$ Welcome to Math.SE! You'll find that you are much more likely to get a positive, helpful response to your question if you modify it. Rather than just stating the fact you are trying to prove, tell us what you've tried and where you're getting stuck. It makes it easier to answer, AND more likely that you'll actually learn something from it! $\endgroup$ – Nick Peterson Feb 5 '17 at 7:27
  • $\begingroup$ Can you show us how you've applied the definition of uniform continuity and where you've got stuck? $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 5 '17 at 7:27
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    $\begingroup$ Thanks for typing your work. (-1) withdrawn. It'd be even better if you use $\rm \LaTeX$. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Feb 5 '17 at 9:07
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    $\begingroup$ Do you mean the composite function $f \circ g$, or the product function $fg$ ? The question says the former, but you seem to be working on the latter. It would also help others write answers more at your level if you include more context about where you encountered the problem: what level of course? What textbook? $\endgroup$ – Carl Mummert Feb 5 '17 at 13:26
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    $\begingroup$ If you're dealing with the composite function, then you don't want to look at expressions like $|f(x)g(x) - f(y)(g(y)|$. You want to look at $|f(g(x)) - f(g(y))|$ instead. $\endgroup$ – Carl Mummert Feb 5 '17 at 21:44
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Suppose that $g$ is uniformly continuous on an interval $I$ and that $f$ is uniformly continuous on an interval $J$ that contains $g(I)$. Prove that the composite function $f \circ g$ is uniformly continuous on $I$.

First noted that you may type \circ in $\LaTeX$ for the composition operator $\circ$.


Proof.

Since $g$ is uniformly continuous on $I$, by definition,

Given any $\varepsilon' > 0$, we have for some $\delta'(\epsilon') >0$, $x', x'' \in I $ and $\lvert x' - x'' \rvert < \delta' \Rightarrow \lvert g(x') - g(x'') \rvert < \varepsilon' $.

Now we observe $g(I) \subset J$. It is clear that the restriction of $f$ on $g(I) \subset J$ is also uniform continuous, thus applying the definition, we have:

Given any $\varepsilon > 0$, $\exists \delta(\epsilon) > 0$, such that $y', y'' \in g(I)$ and $\vert y' - y'' \rvert < \delta \Rightarrow \lvert f(y') - f(y'') \rvert < \epsilon$.

Hence if we choose $\varepsilon' = \delta(\epsilon)$ in the first application, we have $x', x'' \in I$ and $\lvert x' - x'' \rvert < \delta'(\delta(\varepsilon)) \Rightarrow \lvert g(x') - g(x'') \rvert < \delta(\varepsilon)$. Noted $g(x'), g(x'') \in g(I) \subset J$, hence it further implies that $\lvert f(g(x')) - f(g(x'')) \rvert < \varepsilon$.

In other words, we have shown that given any $\varepsilon > 0$, we may find some $\delta > 0$ such that $x' ,x'' \in I$ and $\lvert x' - x'' \rvert < \delta \Rightarrow \lvert f \circ g(x') - f \circ g(x'') \rvert < \varepsilon$. Hence by definition $f \circ g$ is uniform continuous on $I$, as desired.


Noted in this proof we do not need to use the fact that $f$ and $g$ are bounded.

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