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 $\alpha>1$?
 A: 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.
A: 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
