If $a,b>0$ is there a function $f:\Bbb R^2\rightarrow \Bbb R$ such that $$|ax-by|\leq f(a,b)|x-y|$$ holds at all $x,y\in\Bbb R$ (take $a=\frac32$ and $b=\frac12$ for instance)?
Is there any condition on $a,b$ that could give a function $f(x)$?
On bounding a linear form.
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$\begingroup$
inequality
linear-transformations
1 Answers
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Fix $x=1$ and look at $\sup_{y \neq 1} | {a-by \over 1-y} | $. If $a \neq b$, then $\lim_{y \to 1} | {a-by \over 1-y} | = \infty $, so there is no upper bound. If $a=b$, then $f(a,b) = 1$ will do.
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0what if $|\cdot|$ is ultrametric valuation? – 2017-02-23
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0Same general idea, show that ${ \max (|a|,|by|) \over |1-y| }$ is unbounded as $y \to 1$. – 2017-02-23