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Lets suppose$^1$ $x,y\in \mathbb R^3$. Then for which $s>1$ should I be expecting the following inequality to be true: $$|x|^s - |y|^s \leq C|x-y|^s \ ?$$ The constant $C$ should be independent of $x,y$.

For $s\leq 1$, its true with $C=1$, from the fact that $|\cdot|^s$ is sublinear. For $s>1$, convexity gives us $$ \left|\frac{x}2 \right|^s = \left|\frac{y}2 + \frac{x-y}2 \right|^s\leq \frac{|y|^s}2 +\frac{|x-y|^s}2$$ So that $$|x|^s \leq 2^{s-1}(|y|^s + |x-y|^s)$$ the factor of $2^{s-1}>1$ is not comforting, but I don't know if this can be improved.

$^1$ In my application, they have integer coordinates but I don't expect this to be relevant. I also don't expect the dimension 3 to play a significant role.

Calvin Khor
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1 Answers1

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No such constant $C$ exists if $s > 1$, in any dimension $n \ge 1$.

For $x = (x_0 + t, 0, \ldots, 0)$ and $y = (x_0, 0, \ldots, 0)$ with $x_0, t > 0$ we would have $$ (x_0+t)^s - x_0^s\le C t^s \Longrightarrow \frac{(x_0+t)^s - x_0^s}{t} \le C t^{s-1}\, . $$ Taking the limit for $t \to 0$ gives $$ s x_0^{s-1} \le 0 \, , $$ a contradiction.

Martin R
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