# $| |t+1|^p - |t|^p - 1 | \leq k ( |t|^{p-1} + |t| )$ for some $k$

Let $1<p<\infty$. Also assume that $t \in [-1,1]$. Prove that there exists a constant $k \in \mathbb{R}$ that depends on $p$, such that: $$| |t+1|^p - |t|^p - 1 | \leq k ( |t|^{p-1} + |t| ).$$ Which would be equivalent to: $$\sup_{t\in[-1,1]} \left\{ \frac{||t+1|^p - |t|^p -1|}{|t|^{p-1} + |t|} \right\} < \infty.$$ Any hint would be appreciatied.

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Quick Note: The statement you are trying to prove and the "equivalent statement" are actually not equivalent for $t =0$. Therefore, you should exercise care if you are trying to prove the second statement. – JavaMan Oct 23 '12 at 6:27
I would try to use some integral like $$\int_0^t(x+1)^{p-1}-x^{p-1}dx=\frac{1}{p}\left((t+1)^p-t^p-1\right)$$ together with a max estimate of the integral. – AD. Oct 23 '12 at 8:02

For $t\ne0$, $$\frac{||t+1|^p - |t|^p -1|}{|t|^{p-1} + |t|} \leq\frac{||t+1|^p - 1|}{|t|^{p-1} + |t|} + \frac{|t|^p }{|t|^{p-1} + |t|} \leq\frac{||t+1|^p - 1|}{|t|} + \frac{|t|^p }{|t|^{p-1}} =\frac{||t+1|^p - 1|}{|t|} + |t|.$$ On the interval $[-1/2,3/2]$ the function $t\mapsto (t+1)^p$ is differentiable and $|t+1|=t+1$ (the choice is arbitrary, it only matters that it is $>-1$). By the Mean Value Theorem, there exist numbers $c_t$ between $t+1$ and $1$ (so $c_t\in[0,2]$) with $$(t+1)^p-1=pc_t^{p-1}\,t,$$ and so $$|\,|t+1|^p-1|=|(t+1)^p-1|=|pc_t^{p-1}t|\leq p2^{p-1}|t|,$$ i.e. $$\frac{|(t+1)^p-1|}{|t|}\leq p2^{p-1}.$$ For $t<-1/2$, $$\frac{|\,|t+1|^p-1|}{|t|}\leq\frac{|t+1|^p+1}{1/2}\leq 2(2^p+1).$$ Going back to the first inequality, we get $$\frac{||t+1|^p - |t|^p -1|}{|t|^{p-1} + |t|} \leq \max\{p2^{p-1},2(2^p+1)\} +1.$$