# Intermediate values of $p$-norms

Let $x,y \in [0,1]^n$ be $n$ dimensional vectors with elements in the continuous set $[0,1]$.

Suppose we have the following conditions:

1) $\Vert x\Vert_1 = \Vert y \Vert_1=1$ ,

2) $\Vert x\Vert_2 \geq \Vert y \Vert_2$ ,

3) $\Vert x\Vert_\infty \geq \Vert y \Vert_\infty$,

where $\Vert x\Vert_\alpha := \left( \sum x_i^\alpha\right)^{1/\alpha}.$

Can we deduce the following: $\Vert x\Vert_\alpha \geq \Vert y \Vert_\alpha$ for all $\alpha>1$?

• Note, we can prove this for $n=2$ by noting that $x$ majorizes $y$ and that the $p$-norm is schur convex. – David Simmons Jan 2 '17 at 16:26

To see that this is a bit too optimistic, imagine two vectors $x$ and $y$ that have the same 1-norm, same 2-norm, and same $\infty$-norm. If your conjecture were true, than they would have the same $\alpha$-norm for every $\alpha$, which is impossible without them being the same. For a concrete example, take

$$x=(10,2,2,0),\qquad y=(10,1,(3+\sqrt{5})/2, (3-\sqrt{5})/2)$$ which satisfy

• $\|x\|_1=\|y\|_1 = 14$,
• $\|x\|_2=\|y\|_2 = \sqrt{108}$,
• $\|x\|_\infty=\|y\|_\infty = 10$

These will have different $3$-norms, for example ($1016^{1/3}$ and $1019^{1/3}$, to be specific). They can be normalized to have 1-norm equal to $1$, just divide by 14 if desired.

• Why would my conjecture imply that they have the same $\alpha$-norm for all $\alpha$? – David Simmons Jan 2 '17 at 20:18
• Because if the assumptions hold with equality, then one can interchange $x$ and $y$. – user357151 Jan 2 '17 at 20:28

No.

A counter example is:

$$x = \left[0.354019 \quad 0.162424 \quad 0.302226 \quad 0.145733 \quad 0.035598\right]$$ $$y = \left[0.137069\quad 0.065747\quad 0.327494 \quad 0.346246\quad 0.123444\right]$$ $$p = 8 .$$

I wrote the following Matlab script to prepare vectors in the required form. I then loop over different values of $p$ at the end to check if I can find vectors that do not conform to $\Vert x \Vert_p \geq \Vert y \Vert_p$.

clear all n = 5; p = 2; count = 0;trial = 1000; for i = 1:trial x(i,:) = exprnd(1,1,n); x(i,:) = x(i,:)/norm(x(i,:),1); y(i,:) = exprnd(1,1,n); y(i,:) = y(i,:)/norm(y(i,:),1); if max(x(i,:)) < max(y(i,:)) temp = y(i,:); y(i,:) = x(i,:); x(i,:) = temp; end if norm(x(i,:),2) > norm(y(i,:),2) count = count +1; x1(count,:) = x(i,:); y1(count,:) = y(i,:); end end for p = 1.1:0.1:10 for i = 1:count if norm(x1(i,:),p) < norm(y1(i,:),p) norm(x1(i,:),p) < norm(y1(i,:),p) p end end end