Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Suppose that we have two vectors, $x,y\in\mathbb{R}^n$. Let $z\in\mathbb{R}^n$ be the vector given by $z_i=\sqrt{x_i y_i}$. With an abuse of notation, I may write $z=\sqrt{xy}$. Consider the quantity $$X(t)=2\|z\|_{1}\|z\|_{t}^{t}-\left(\|x\|_{1}\|y\|_{t}^{t}+\|y\|_{1}\|x\|_{t}^{t}\right)$$ where $\|x\|_t=\left(x_1^t +\cdots+x_n^t\right)^\frac{1}{t}$ refers to the $t$-norm.

I only care about when $t$ is very very close to $t=1$. Notice that $$X(1)=2\|z\|_1^2-\|x\|_1\|y\|_1=2\left(\sum_{i=1}^n \sqrt{x_i y_i}\right)^2-2\left(\sum_{i=1}^n x_i\right)\left(\sum_{i=1}^n y_i\right).$$ We can then show that $X(1)\leq 0$ by expanding and using the AM_GM.

Here is my question:

Is it possible to show that for very small $\epsilon$ we have, $$X(1-\epsilon)\leq X(1)\leq X(1+\epsilon)?$$

share|cite|improve this question
Did you mean $\leq$? Otherwise, it's false for $x = y$. – user7530 Oct 26 '11 at 14:12
@user7530: Yes! sorry about that, corrected. – Eric Naslund Oct 26 '11 at 14:36
up vote 3 down vote accepted

This doesn't work already for $n=1$. In that case we have



$$ \begin{eqnarray} X'(t) &=& (xy)^{(t+1)/2}\log(xy)-xy^t\log y-yx^t\log x\;, \\ X'(1) &=& xy\log(xy)-xy\log y-yx\log x=0 \end{eqnarray}$$


$$ \begin{eqnarray} X''(t) &=& \frac12(xy)^{(t+1)/2}(\log(xy))^2-xy^t(\log y)^2-yx^t(\log x)^2\;, \\ X''(1) &=& \frac12xy(\log(xy))^2-xy(\log y)^2-yx(\log x)^2=-\frac12xy(\log x-\log y)^2\;, \end{eqnarray}$$

which will generally not vanish, so $X$ has a local extremum at $t=1$.

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.