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I have proven that (and we are required to use this) $y^2\arctan y-x^2\arctan x\geq(y^2-x^2)\arctan x$ (the proof was near-trivial). I now have to use this to show that $f(x)=x^2\arctan x$ is not uniformly continuous in $\Bbb R$.

Here is what I have so far (I'm not used to writing mathematical proofs in English so bear with me):

Let there be $\delta>0$. Let $x=\frac{10}{\delta}, y=x+\frac{\delta}{2}$. It follows that $|x-y|<\delta$, and:

$\begin{align}|f(x)-f(y)| &=|x^2\arctan x-y^2\arctan y|\\ &\geq(y^2-x^2)\arctan x\\ &=\left(\left(\frac{10}{\delta}+\frac{\delta}{2}\right)^2-\left(\frac{10}{\delta}\right)^2\right)\arctan\frac{10}{\delta}\\ &=\left(\frac{100}{\delta^2}-\frac{100}{\delta^2}+2\frac{10\delta}{2\delta}+\frac{\delta^2}{4}\right)\arctan\frac{10}{\delta}\\ &=\left(10+\frac{\delta^2}{4}\right)\arctan\frac{10}{\delta}\end{align}$

And this is where I realized that my choice of $x$ and $y$ was probably not wise - I have failed to prove that the function is greater than some concrete number, as I can make no assumption about the value of $\arctan\frac{10}{\delta}$. I have failed to find better variables - I'd appreciate any help/guidance.

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    @DavidMitra Thank you very much!! You should consider posting all this as a full answer so I can choose it as the accepted answer :).2012-12-11

1 Answers 1

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Your argument can be slightly altered to work:

You need to use the fact that $\lim\limits_{x\rightarrow\infty}\arctan x=\pi/2$. Through the graces of this, given a $\delta>0$, we may (and do) choose a $0<\delta_1<\delta$ sufficiently small, so that $\arctan(10/\delta_1)>1$. Now define $x=10/\delta_1$ and $y=x+{\delta_1\over2}$. We have $\tag{1}|x-y|<\delta_1<\delta.$ Moreover, your computations carry over with minor changes to give $\tag{2} |f(x)-f(y)|\ge\bigl(10+\textstyle{\delta_1^2\over4}\bigr)\arctan(10/\delta_1)\ge10+\textstyle{\delta_1^2\over4}> 10. $ Since $\delta>0$ was arbitrary, it follows from $(1)$ and $(2)$ that $f(x)=x^2\arctan(x)$ is not uniformly continuous over $\Bbb R$.