3

Let $f_{1}(x),f_{2}(x),\dots,f_{n}(x)$ be $n$ bounded and uniformly continuous functions on $\mathbb{R}$. Prove that their product $f_1(x)f_2(x)\cdots f_n(x)$ is also a uniformly continuous function on $\mathbb{R}$

My approach is to use mathematical induction. At the last step, I need to prove that the product of two uniformly continuos functions (which are the product of the first $n-1$ uniformly continuous functions by inductive hypothesis and the $n^{\text{th}}$ function), which doesn't require me to use the fact the functions are bounded.

So, I sense that this approach is not correct (but I'm not sure why) as I didn't use the fact that the functions are bounded. If it's really wrong, how do I make use of the fact that the functions are bounded? Thank you :)

Indrik Wijaya
  • 85

3 Answers3

2

The product of two uniformly continuous functions might not be uniformly continuous.

Take $f_1(x)=f_2(x)=x$. $f_1,f_2$ are uniformly continuous but $f(x)=f_1(x)f_2(x)=x^2$ is not. If you were able to prove that the product of two real functions defined on $\mathbb R$ is uniformly continuous... there is something wrong in your proof.

So indeed, to prove that $f_1f_2$ is uniformly continuous, you have to use that $f_1,f_2$ are bounded.

mathcounterexamples.net
  • 70,018
1

For case $n=2$ Let $|f(x)|<M$ and $|g(x)|<N$ for all $x$ and be uniformly continues. We should prove: $$\forall \epsilon>0,\exists \delta >0,\forall x,y(|x-y|<\delta \rightarrow |f(x)g(x)-f(y)g(y)|<\epsilon)$$ We know $$|f(x)g(x)-f(y)g(y)|=|f(x)g(x)-f(y)g(x)+f(y)g(x)-f(y)g(y)|\leq$$ $$|g(x)||f(x)-f(y)|+|f(y)||g(x)-g(y)|\leq N|f(x)-f(y)|+M|g(x)-g(y)|$$, so if we choose $\epsilon '=\min(\frac{\epsilon}{2N},\frac{\epsilon}{2M})$, there exists a $\delta_1>0$ and $\delta_2>0$ such that $$\forall x,y(|x-y|<\delta_1 \rightarrow |f(x)-f(y)|<\epsilon')$$ $$\forall x,y(|x-y|<\delta_2 \rightarrow |g(x)-g(y)|<\epsilon')$$ Therefore if we choose $\delta=\min(\delta_1,\delta_2)$, for all $x,y$ such that $|x-y|<\delta$, $$|f(x)g(x)-f(y)g(y)|\leq N|f(x)-f(y)|+M|g(x)-g(y)|\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ So it completes your induction proof.

Erfan Khaniki
  • 717
0

It is well known that all continuous function from a metric and compact space to a metric space is uniformly continuous. The product function $g(x)=f_1(x)f_2(x)\cdot\cdot\cdot f_n(x)$ is clearly continuous on all compact of $\mathbb R$ hence uniformly continuous on all compact of $\mathbb R$. Since all of the $f_i$ is bounded one has $g(x)$ is also bounded. This finish the proof.

Piquito
  • 29,594