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.
First noted that you may type
\circin $\LaTeX$ 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.