Let $g(x)=\sin(x)$ and $f(x)=x$,show that the function $fg$ defined by $(fg)(x)=f(x)g(x)$ is not uniformly continuous on $\mathbb R$.

Question

Let $g(x)=\sin(x)$ and $f(x)=x$,show that the function $fg$ defined by $(fg)(x)=f(x)g(x)$ is not uniformly continuous on $\mathbb R$.

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A function $f:D \to \mathbb R$ is not uniformly continuous on $D$ if and only if $$\exists (x_n),(y_n) \subset D: \lim_{n\to \infty} (x_n-y_n)=0 \;\;\text{But} \;\;\lim_{n\to \infty} (f(x_n)-f(y_n))\ne 0$$

Take $\left\{x_n\right\}=\left\{2 \pi n\right\},\left\{y_n\right\}=\left\{2\pi n +1/n\right\} \subset \mathbb R$,now $$\lim_{n\to \infty} (x_n-y_n)=\lim_{n\to \infty}(2 \pi n-2 \pi n-1/n)=0$$

But $$\lim_{n\to \infty} (f(x_n)g(x_n)-f(y_n)g(y_n))$$$$=\lim_{n\to \infty} (2\pi n \sin(2\pi n)-(2\pi n +1/n)\sin(2\pi n +1/n))$$ $$=-\lim_{n\to \infty}(2\pi n +1/n)\sin(1/n)=-2\pi \ne 0$$

And so the function is not uniformly continuous on $\mathbb R$.

I would kike to know if my proof is correct.

Answer 1

The if and only if statement in your question is false. Uniform continuity implies Cauchy continuity but Cauchy continuity does not imply uniform continuity. Both of the definitions I quote below.

Uniform continuity according to wikipedia:

Given metric spaces $\left(X, d_{1}\right)$ and $\left(Y, d_{2}\right),$ a function $f: X \rightarrow Y$ is called uniformly continuous if for every real number $\varepsilon>0$ there exists real $\delta>0$ such that for every $x, y \in X$ with $d_{1}(x, y)<\delta,$ we have that $d_{2}(f(x), f(y))<\varepsilon$ If $\mathrm{X}$ and $\mathrm{Y}$ are subsets of the real line, $d_{1}$ and $d_{2}$ can be the standard one-dimensional Euclidean distance, yielding the definition: for all $\varepsilon>0$ there exists a $\delta>0$ such that for all $x, y \in X,|x-y|<\delta \Longrightarrow|f(x)-f(y)|<\varepsilon$

and Cauchy continuity:

For a function between metric spaces, uniform continuity implies Cauchy continuity (Fitzpatrick 2006). More specifically, let A be a subset of $\mathbf{R}^{n}$. If a function $f: A \rightarrow \mathbf{R}^{m}$ is uniformly continuous then for every pair of sequences $x_{n}$ and $y_{n}$ such that $$ \lim _{n \rightarrow \infty}\left|x_{n}-y_{n}\right|=0 $$ we have $$ \lim _{n \rightarrow \infty}\left|f\left(x_{n}\right)-f\left(y_{n}\right)\right|=0 $$

The negation of Cauchy continuity in your comment is false but correct in question statement with the exception of absolute values.

All in all, your proof would be correct if you said: suppose $fg$ is uniform continuous which implies Cauchy continuous which implies contradiction (essentially, not Cauchy cont. implies not uniform cont.) while justifying each step including your limits.