Is $ f(x)=x^2 \sin x $ Uniformly Continuous on $\mathbb{R}$?

Question

Hello and good afternoon. I've tried to smash $f$ on 2 other functions and prove that they are u.c. on $\mathbb{R}$ , but $x^2$ isn't, & I have seen that it is not necessary that the product of 2 functions be uniformly continuous in $\mathbb{R}$. It will help me to find the derivative and follow the theory of uniform continuity ( if it is u.c. on $\mathbb{R}$ ). And if it isn't, is it good to prove that it is not by the method of inopportune induction?

Answer 1

Let $f(x)=x^{2}\sin(x)$ in $\mathbb{R}$. Assume that $f(x)$ is uniformly continuous in $\mathbb{R}$ and let $\varepsilon=1$, then there exists $\delta>0$ such that for $x,y\in\mathbb{R}$ with $|x-y|<\delta$ we have $|f(x)-f(y)|<1$. Let $y=x+\frac{\delta}{2}$ and $g(x)=x^{2}$. Thus, this means for $x\in\mathbb{R}$, we have : \begin{align*} \left|x^{2}\sin(x)-y^{2}\sin(y)\right|\leq&|x^{2}-y^{2}| \\ \implies &\left|g(x)-g\left(x+\frac{\delta}{2}\right)\right|<1 \\ \implies & \left|-\delta x -\frac{\delta^{2}}{4}\right|<1 \\ \implies & \delta x +\frac{\delta^{2}}{4}<1 \\ \implies &\text{$\forall x\in\mathbb{R}$ we have $x<\frac{1-\frac{\delta^{2}}{4}}{\delta}\implies$ Contradiction} \end{align*}