Dear mathoverflowers.
Just wondering if the following inequality is true. For all $ p >1$ there is some $C$ such that
$ | |x+1|^p-|y+1|^p -p(x-y)| \le C ( |x|+|y| + |x|^{p-1} + |y|^{p-1} ) |x-y| $ for all real numbers $ x$ and $y$.
thanks..
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$\begingroup$ @Peter: So is the LHS along the diagonal. $\endgroup$– Connor MooneyCommented May 17, 2013 at 12:12
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$\begingroup$ Sorry, my mistake. $\endgroup$– Peter MichorCommented May 17, 2013 at 13:51
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2 Answers
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Yes, I think it's true. Say we follow the line $(x,\alpha x)$ for $x > 0$ and $0 < \alpha < 1$. Both sides of the desired inequality have no linear part at $0$, so we examine the second derivatives. Keeping only dependence of the coefficients on $\alpha$, the second derivative of the LHS goes like $$(1-\alpha)(1+x)^{p-2}$$ and for the RHS goes like $$(1-\alpha)(1+x^{p-2}).$$ It is clear that the left is controlled by a constant independent of $\alpha$ times the right for all $x > 0$. The other regions can probably be taken care of similarly.
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$\begingroup$ Thanks for your answer. I will attempt to follow through with the details. $\endgroup$– CraigCommented May 28, 2013 at 20:22
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If $p=2n$, a positive even integer, then your inequality is true -- use polar coordinates. In view of this fact, I think your inequality should be true for all $p>1$.
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$\begingroup$ I agree with you analysis. thanks. $\endgroup$– CraigCommented May 28, 2013 at 20:21