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I'm interested in a simple way to estimate $|\sqrt[n]{x}-\sqrt[n]{y}|$ from $|x-y|$.

For $n=2$ and $n=3$, we have $|\sqrt{x}-\sqrt{y}|\le \sqrt{x-y}$ for $x,y\ge 0$ and $|\sqrt[3]{x}-\sqrt[3]{y}|\le \sqrt[3]{4(x-y)}$ for $x,y\in \mathbb{R}$.

When $n$ is even, $|\sqrt[n]{x}-\sqrt[n]{y}|\le \sqrt[n]{|x-y|}$ holds since $x,y$ are nonnegative.

My question is: for an arbitrary odd number $n$, is there a constant $C_n$ such that $$|\sqrt[n]{x}-\sqrt[n]{y}|\le \sqrt[n]{C_n|x-y|}$$ for any $x,y\in \mathbb{R}$?

user107678
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1 Answers1

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Let $x, y$ be real numbers and $n$ an odd integer.

If $x$ and $y$ have the same sign then $$ |\sqrt[n]{x}-\sqrt[n]{y}|\le \sqrt[n]{|x-y|} \, , $$ see for example proving that $f(x) = x^s$ is holder continuous with holder exponent s or How do you prove the follwing $|\sqrt[n]{x}-\sqrt[n]{y}| \le \sqrt[n]{|x-y|}$.

And if $x < 0 < y$ then, proceeding as in Prove a function is Hölder continuous but not Lipschitz continuous, $$ \left( \frac{|x|}{|x| + y}\right)^{1/n} + \left( \frac{y}{|x| + y}\right)^{1/n} \le \max \{ u^{1/n} + (1-u)^{1/n} \mid 0 < u < 1 \} = 2^{(n-1)/n} $$ so that $$ \sqrt[n]{y}-\sqrt[n]{x} = |x|^{1/n} + y^{1/n} \le 2^{(n-1)/n} (|x| + y)^{1/n} = 2^{(n-1)/n}\sqrt[n]{y-x} \, . $$

Therefore $$ |\sqrt[n]{x}-\sqrt[n]{y}|\le \sqrt[n]{2^{n-1}|x-y|} $$ for all $x, y \in \Bbb R$ and odd integers $n$. Choosing $x=-y \ne 0$ shows that $C_n = 2^{n-1}$ is the best possible constant.

Martin R
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