# An inequality with $\ell_p$ norm

I encounter the following claim in my research for which I couldn't get a solution for a long time. I asked a more general version of the question at math.stackexchange which did not attract much attention there. Following is a particular version of the question.

Let $p>1$ and let

$$L = \{ \ \textbf{x} \in \mathbb{R}^n:\ \ x _i\ge 0, \sum_i x_i=1, \sum_i a_i x_i =b\ \}$$ for some real numbers $a_i$ and $b$.

Suppose that $\textbf{x}, \textbf{y}\in L$ with the following property: $x_i=0$ wherever $y_i=0$, and $y_j\neq 0$ but $x_j=0$ for some $j$, i.e., $\bf{y}$ has more support than $\bf{x}$. Then $\|\textbf{x}\|_p>\|\textbf{y}\|_p$ or there exists $\textbf{z}\in L$ such that $\textbf{z}$ has more support than $\textbf{x}$, i.e., $x_i=0$ wherever $z_i=0$ and $z_k\neq 0$ but $x_k=0$ for some $k$ and $\|\textbf{x}\|_p>\|\textbf{z}\|_p$.

The claim is roughly that the $\ell_p$-norm of points of $L$ which are more closer to the boundary of the probability simplex is larger than the ones which are interior enough. In 2-dimension, the claim is very obvious. The only possibility of $L$ there is $\{ \(x_1,x_2):\ x_i\ge 0, x_1+x_2=1\ \}$ and the boundary points $(1,0),(0,1)$ have the maximum $\ell_p$-norm and as we go deeper interior the norm is lesser and lesser. I am wondering whether the same would be the case even in the higher dimensions.

• What is $a_i$ (in the definition of $L$)? – Sergei Ivanov May 29 '12 at 11:55
• $a_i$ and $b$ are some real numbers. – Ashok May 29 '12 at 12:20
• The logic of the question looks strange as the inequality $\|x\|_p>\|y\|_p$ implies the property "there exists $z$ ..." (because you can take $y$ for $z$). So you are asking for "A or B" when A in fact implies B and hence "A or B" is equivalent to just B. Is this indeed what you intend to ask? – Sergei Ivanov May 29 '12 at 12:45
• Yes Sergei Ivanov, you are right. Actually I want to prove B: There exists $\textbf{z}$ such that... I know for sure that $\textbf{x}, \textbf{y}\in L$ with the mentioned properties. If $\textbf{y}$ itself works as $\textbf{z}$ its fine. I am claiming, otherwise, there exists $\textbf{z}$... – Ashok May 29 '12 at 13:00

The answer is no, for any $n\ge 3$ and any $C^1$-smooth strictly convex norm on $\mathbb R^n$. (Here "strictly convex" means that the triangle inequality is strict for any two non-collinear vectors. This is equivalent to the following: the norm restricted to any affine subspace not containing 0 is a strictly convex function on that subspace.)
The claim you are asking for is equivalent to the following: if $L$ is the intersection of the probability simplex and a hyperplane, then the norm restricted to $L$ cannot attain its minimum at a point of the relative boundary of $L$.
To construct a counter-example, consider the set $F$ of points in the simplex where the coordinate $x_1$ is zero and all other coordinates are positive. (Thus $F$ is the relative interior of one of the $(n-2)$-dimensional faces of the simplex.) Consider the norm as a function on this $(n-2)$-dimensional convex set. Since this function is strictly convex, there is at most one point where its derivative vanishes. Choose $\mathbf x=(x_i)\in F$ so that this restricted derivative does not vanish at $\mathbf x$ (here I use the assumption that $n\ge 3$). Let $(a_i)$ be the vector of partial derivatives of the norm at $\mathbf x$ and let $b=\sum a_ix_i$. It is easy to see that $b\ne 0$.
Consider the hyperplane $Y=\{\mathbf y=(y_i):\sum a_iy_i=b\}$ and let $L$ be the intersection of $Y$ and the simplex. By the choice of $\mathbf x$, this hyperplane intersects the relative interior of the simplex, so $L$ contains vectors all whose coordinates are nonzero and $\mathbf x$ belongs to the relative boundary of $L$. On the other hand, the norm restricted on $Y$ is a strictly convex function and by construction it has a critical point at $\mathbf x$. Therefore $\mathbf x$ is the minimum point of the norm on $Y$ and hence on $L$.