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Let $1<p<\infty$ be fixed. Suppose $L=\{(x_1,\dots,x_n):x_i\ge 0, \sum_i x_i=1, \sum_i a_i x_i=b\}$ for some real numbers $a_i$ and $b$. I am wondering whether the following would be true.

Suppose that $\bf{x}$ $ =(x_1,\dots,x_n)$, $\bf{y}$ $ =(y_1,\dots,y_n)\in L$ such that $x_i=0$ whenever $y_i=0$ and $y_j\neq 0$ but $x_j=0$ for some $j$. Then there exists $\bf{z}$ $ =(z_1,\dots,z_n)\in L$ such that $\bf{z}$ has properties as $\bf{y}$ (i.e., $x_i=0$ whenever $z_i=0$ and $z_j\neq 0$ but $x_j=0$ for some $j$) and $$\sum_i \frac{x_i}{\|\bf{x}\|_p} k_i < \sum_i \frac{z_i}{\|\bf{z}\|_p} k_i$$ where $k_i>0$ for all $i$ are some fixed real numbers such that $\|\bf{k}\|_q=1$, $\frac{1}{p}+\frac{1}{q}=1$.

Any help is appreciated.

Update:

This claim is essentially saying that when the quantity $\sum_i \frac{x_i}{\|\bf{x}\|_p} k_i$ is maximized over $\bf{x}$ $ \in L$, the maximum is attained at a point $\bf{x}^*$ which has maximal support among the points in $L$. (I have already proved that there is a unique maximizer)

So I wanted to prove the following equivalent one.

If I know that $\bf{x}$ and $\bf{y}$ are in $L$, and $\bf{y}$ has more support than $\bf{x}$, then $\bf{x}$ cannot be a maximizer.

I thought in the following way.

By Holder's inequality, $$\sum_i\frac{x_i}{\|\bf{x}\|_p}k_i\le \frac{1}{\|\bf{x}\|_p}\cdot \|\bf{k}\|_{\infty}$$ Instead of the left side quantity, if I look at the right side quantity, its more for vectors in the interior than the vectors on the boundary in the probability simplex, atleast in 2 or 3 dimensions. But, after all, even this intutive idea is for the right side quantity only.

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Remark: cross-posted to MO. –  Willie Wong May 29 '12 at 12:06
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