Suppose $a,b,x,y \in \mathbb{R}$ and are non-zero. Suppose also that we know $$ |a-b| < \epsilon \tag{1} $$ Can (1) be used to provide an upper bound on the quantity, $$ |a|x| - b|y|| \tag{2} $$ tighter than an application of the triangle inequality? Or if not provably tighter, is there an upper bound on (2) that utilizes $|a-b|$, which is not simply $|a-b| \cdot \frac{|a||x| + |b||y|}{|a-b|}$?

  • $\begingroup$ upper bound in terms of which variables? $\endgroup$ – dezdichado Jul 7 '17 at 21:25
  • $\begingroup$ any or all of $a,b,x,y$, preferably something of the form $|a-b| \beta(x,y)$ $\endgroup$ – Chester Jul 7 '17 at 21:28

In general, it seems that there is no upper bound for the quantity $$|a|x|-b|y||$$ since, letting $y=x$, gives: $$|a|x|-b|y||=|a|x|-b|x||=|x|\cdot|a-b|$$ which, as $|x|\to+\infty$ tends to $+\infty$, as well, regardless how small is $|a-b|$, supposing $a\neq b$.

So, the quantity $|a|x|-b|y||$ cannot be uniformly bounded - this means that for every $x,y\in\mathbb{R}$ there is no $M>0$, independent of $x,y$, such that $|a|x|-b|y||\leq M$.

However, we can see that: $$\begin{align*}|a|x|-b|y||=&|a|x|-b|x|+b|x|-b|y|| =|(a-b)|x|+b(|x|-|y|)|\leq\\\leq&|x||a-b|+||x|-|y||<|x|\epsilon+||x|-|y||\end{align*}$$

And, due to symmetry over $x,y$, we also have: $$|a|x|-b|y||<|y|\epsilon+||x|-|y||$$ So, finally: $$|a|x|-b|y||<\min\{|x|,|y|\}\epsilon+||x|-|y||$$ or, in the requested for, one could write this as - altough that bound is slightly worse: $$|a|x|-b|y||<\max\{||x|-|y||,\min\{|x|,|y|\}\}(1+\epsilon)$$ However, the important is that some information about $x,y$ is needed, so as to have a uniform bound for this quantity.

Hope this helped! :)

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