I need to find an upper bound for $$\|x-y\|_2^\nu$$ in which $\nu \in [0,1]$. I was wondering if $\|x-y\|_2^2$ can be considered as an upper bound. or $\|x\|+\|y\|$?
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$\begingroup$ In the one-dimensional case with $\Vert\cdot\Vert =\vert\cdot\vert$ it's already the case that $\vert x-y \vert^\nu \leq \vert x-y\vert^2$ and $\vert x-y \vert^\nu \leq \vert x \vert + \vert y \vert$ can both be false. Do you have a reason to expect one of these expressions to be an upper bound? You'll get a better answer if you tell us the context of your question. $\endgroup$– Chris EagleCommented Aug 9, 2022 at 9:02
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$\begingroup$ If $0<\|x-y\|\le 1$ the upper bound is $1.$ If $ |x-y\|\ge 1$ the upper bound is $\|x-y\|.$ $\endgroup$– Ryszard SzwarcCommented Aug 9, 2022 at 13:52
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$\begingroup$ I would like to prove a theorem and I need an upper bound for $\|x-y\|^\nu$. $\endgroup$– Andreas deniroCommented Aug 9, 2022 at 14:37
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