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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, $$X(1-\epsilon)\leq X(1)\leq X(1+\epsilon)?$$

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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

1 Answer 1

up vote 3 down vote accepted

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

$$X(t)=2zz^t-xy^t-yx^t=2(xy)^{(t+1)/2}-xy^t-yx^t\;,$$

so

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

and

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

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

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