Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 0 down vote accepted

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 $$ \frac{||t+1|^p - |t|^p -1|}{|t|^{p-1} + |t|} \leq \max\{p2^{p-1},2(2^p+1)\} +1. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.