Given $p>1$, I would like to show that there exists constants $C>0$ and $\epsilon_0 > 0$ such that
$$
\left\lvert \left\lvert x+ y\right\rvert^{p-1}(x+y) – \left\lvert x \right\rvert^{p-1}x – \left\lvert y \right\rvert^{p-1}y \right\rvert \le C\left(\left\lvert x\right\rvert^\epsilon \left\lvert y\right\rvert^{p-\epsilon} + \left\lvert x\right\rvert^{p-\epsilon}\left\lvert y\right\rvert^\epsilon\right)
$$
for all $x,y\in \mathbb{R}^n$ and $0<\epsilon < \epsilon_0$.
While I hope this inequality is true, I am unable to prove it and therefore not 100% sure. Any suggestions on how to approach it would be appreciated.
What I have tried:
- Some calculus methods. For instance, I tried defining $f(x,y,\epsilon)$ to be the difference between the right-hand and left hand side (where I will figure out later how large $C$ needs to be). Then, I was hoping to take the derivative with respect to $\epsilon$ and somehow show that $f(x,y,\epsilon)$ approximates $f(x,y, 0)$ in an appropriate way.
- I also tried going by contradiction and constructing a sequence $(x_m), (y_m), (\epsilon_m)$ such that $0 < \epsilon_m < 1/m$ and
$$
\left\lvert \left\lvert x_m + y_m\right\rvert^{p-1}(x_m+y_m) – \left\lvert x_m \right\rvert^{p-1}(x_m) – \left\lvert y_m \right\rvert^{p-1}y_m \right\rvert > m\left(\left\lvert x_m\right\rvert^{\epsilon_m} \left\lvert y_m\right\rvert^{p-\epsilon_m} + \left\lvert x_m\right\rvert^{p-\epsilon_m}\left\lvert y_m\right\rvert^{\epsilon_m}\right)
$$
for each $m\in\mathbb{N}$. Unfortunately, this led me nowhere.
Best Answer
Put $z=-x-y$. Then $x+y+z=0$ and $$\left\| \left\| x+ y\right\|^{p-1}(x+y) - \left\| x \right\|^{p-1}x - \left\| y \right\|^{p-1}y \right\|=$$ $$\left\| \left\| x \right\|^{p-1}(-x) + \left\| y \right\|^{p-1}(-y)+\left\| z \right\|^{p-1}(-z) \right\|=$$ Let $(u,v,w)$ be a permutation of $(x,y,z)$ such that $\|u\|\le \|w\|$ and $\|v\|\le \|w\|$. Then the triangle inequality implies that $$\left\| \left\| x \right\|^{p-1}(-x) + \left\| y \right\|^{p-1}(-y)+\left\| z \right\|^{p-1}(-z) \right\|=$$ $$\left\| \left\| u+v \right\|^{p-1}(u+v) - \left\| u \right\|^{p-1}u-\left\| v \right\|^{p-1}v \right\|=$$ $$\left\| (\left\| u+ v\right\|^{p-1} - \left\| u \right\|^{p-1})u +(\left\| u+ v\right\|^{p-1}- \left\| v \right\|^{p-1})v \right\|\le $$
$$\left(\left\| u+ v\right\|^{p-1} - \left\| u \right\|^{p-1}\right)\|u\| +\left(\left\| u+ v\right\|^{p-1}- \left\| v \right\|^{p-1}\right)\|v\|\le $$
$$\left(\left(\left\|u\right\|+ \left|v\right\|\right)^{p-1} - \left\| u \right\|^{p-1}\right)\|u\|+\left(\left(\left\|u\right\|+ \left|v\right\|\right)^{p-1} - \left\| v \right\|^{p-1}\right)\|v\|=$$ $$\left(\left\|u\right\|+ \left|v\right\|\right)^p-\|u\|^p-\|v\|^p.$$
Since $(u,v,w)$ be a permutation of $(x,y,z)$ such that $\|u\|\le \|w\|$ and $\|v\|\le \|w\|$, we have that $$C(\|u\|^{p-\epsilon}\|v\|^\epsilon+\|u\|^\epsilon \|v\|^{p-\epsilon})\le C(\|x\|^{p-\epsilon}\|y\|^\epsilon+\|x\|^\epsilon \|y\|^{p-\epsilon}).$$ Thus to provide the required inequality it suffices to provide $$\left(\left\|u\right\|+ \left|v\right\|\right)^p-\|u\|^p-\|v\|^p\le C(\|u\|^{p-\epsilon}\|v\|^\epsilon+\|u\|^\epsilon \|v\|^{p-\epsilon})$$
Put $a=\|u\|$ and $b=\|v\|$. Thus it suffices to provide $(a+b)^p-a^p-b^p\le C(a^{p-\epsilon}b^\epsilon+a^\epsilon b^{p-\epsilon})$.
Put $\epsilon_0=1$. Swapping $a$ and $b$, if needed, we can suppose that $a\le b$. If $a=0$ then the left hand side is zero, so the inequality holds. Suppose that $a>0$ and put $t=b/a\ge 1$. Thus we have to provide $(1+t)^p-1-t^p\le C(t^{p-\epsilon}+t^\epsilon).$
Note that if $\epsilon\le p-1$ then since $(t^{p-\epsilon}+t^\epsilon)-(t^{p-1}+t)=(t^{p-1}-t^\epsilon)(t^{1-\epsilon}-1) \ge 0$ it suffices to provide $(1+t)^p-1-t^p\le C(t^{p-1}+t)$.
By Lagrange's theorem, there exists $s\in (t,t+1)$ such that $(1+t)^p-t^p=ps^{p-1}<p(t+1)^{p-1}$. Then $$(1+t)^p-1-t^p<p(t+1)^{p-1}\le p(2t)^{p-1}\le 2^{p-1}pt^{p-\epsilon},$$ so we can put $C=2^{p-1}p$.
Let $2\le p\le 3$. Note that in this case $\epsilon_0\le p-1$. By Taylor's formula with the remainder in Lagrange's form, there exists $s\in (t,t+1)$ such that
$$(t+1)^p=t^p+pt^{p-1}+\tfrac{p(p-1)}2t^{p-2}+\tfrac{p(p-1)(p-2)}6s^{p-3} \le t^p+pt^{p-1}+pt+1,$$ so we can put $C=p$.
I shall try to decrease the value of $C$ for the remaining values of $p$.