An odd application of Jensen's inequality.

Question

Claim:

Suppose $p > 1$. Let $x,y \in \mathbb{R}$ and let $t \in (0,1)$. Then, $$ |x|^p - |x-y|^p \leq \frac{1}{t} \left( |x + ty|^p - |x|^p \right). $$

Attempt:

I know that $|x-y|^p$ is a convex function. So, by Jensen's inequality, $$ t |x- y|^p + (1-t) |k|^p \geq | tx - ty + k - kt |^p, $$ for any real number $k$. Thus, $$ t |x|^p - t|x-y|^p \leq t |x|^p + | tx - ty + k - kt |^p - (1-t) |k|^p. $$ If I can find a suitable $k$ so that $$ t |x|^p + | tx - ty + k - kt |^p - (1-t) |k|^p \leq \left( |x + ty|^p - |x|^p \right), $$ then I am done. However, I have been unable to find such a $k$.

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Any help would be greatly appreciated!

Answer 1

If $x=0$ the inequality is immediate. Otherwise, dividing through by $|x|^p$ and renaming $x:=y/x$ it is enough to show for $x\in\mathbb{R}$ that $1-|1-x|^p\le (1/t)(|1+tx|^p-1),$ i.e., $1+t\le |1+tx|^p+t|1-x|^p$, i.e., $2-|1+tx|^p\le 1-t+t|1-x|^p$. By Jensen's inequality for the function $x\mapsto |x|^p$, this follows if $2-|1+tx|^p\le |(1-t)1+t(1-x)|^p=|1-tx|^p,$ i.e., $1\le (1/2)|1+tx|^p+(1/2)|1-tx|^p,$ which follows by another application of Jensen's inequality.