The p-norm is given by $||x||_{p} = (\sum_{n=1}^{\infty }|x_{n}|^{p})^{1/p}$. For $0 < p < q$, it can be shown that $||x||_{p} \geq ||x||_{q}$ (1, 2). It appears that in $R^{n}$ a number of opposite inequalities can also be obtained. In fact, since all norms in a finite-dimensional vector space are equivalent, this must be the case. So far, I only found the following: $||x||_{1} \leq \sqrt n ||x||_{2}$(3), $||x||_{2} \leq \sqrt n ||x||_{\infty}$ (4). Geometrically, it is easy to see that opposite inequalities must hold in $R^{n}$. For instance, for n=2 and n=3 one can see that for $0 < p < q$, the spheres with radius $\sqrt n$ with $||\cdot||_{p}$ inscribe spheres with radius 1 with $||\cdot||_{q}$.
It is not hard to prove inequality (4). According to Wikipedia, inequality (3) follows directly from Cauchy-Schwarz, but I don't see how. For n=2 it is easily proven (see below), but not for n>2. So my question is:
- How can relation (3) be proven for arbitrary n?
- Can this be generalized into something of the form $||x||_{p} \leq C ||x||_{q}$ for arbitrary $0
?
- Do any of the relations also hold for infinite dimensional spaces, i.e. in $l^{p}$ spaces?
Notes:
$||x||_{1}^{2} = |x_{1}|^2 + |x_{2}|^2 + 2|x_{1}||x_{2}| \leq |x_{1}|^2 + |x_{2}|^2 + (|x_{1}|^2 + |x_{2}|^2) = 2|x_{1}|^2 + 2|x_{2}|^2 = 2 ||x||_{2}^{2}$, hence $||x||_{1} \leq \sqrt 2 ||x||_{2}$. This works because $|x_{1}|^2 + |x_{2}|^2 \leq 2|x_{1}||x_{2}|$, but only because $(|x_{1}| - |x_{2}|)^2 \geq 0$, while for more than two terms $(|x_{1}| \pm |x_{2}| \pm \dotsb \pm |x_{n}|)^2 \geq 0$ gives an inequality that never gives the right signs for the cross terms.