# Relations between p norms

The $$pp$$-norm on $$Rn\mathbb R^n$$ is given by $$‖\|x\|_{p}=\big(\sum_{k=1}^n |x_{k}|^p\big)^{1/p}$$. For $$0 < p < q0 < p < q$$ it can be shown that $$\|x\|_p\geq\|x\|_q\|x\|_p\geq\|x\|_q$$ (1, 2). It appears that in $$\mathbb{R}^n\mathbb{R}^n$$ 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: $$\|x\|_{1} \leq\sqrt n\,\|x\|_{2}\|x\|_{1} \leq\sqrt n\,\|x\|_{2}$$(3), $$\|x\|_{2} \leq \sqrt n\,\|x\|_\infty\|x\|_{2} \leq \sqrt n\,\|x\|_\infty$$ (4). Geometrically, it is easy to see that opposite inequalities must hold in $$\mathbb R^n\mathbb R^n$$. For instance, for $$n=2n=2$$ and $$n=3n=3$$ one can see that for $$0 < p < q0 < p < q$$, the spheres with radius $$\sqrt n\sqrt n$$ with $$\|\cdot\|_p\|\cdot\|_p$$ inscribe spheres with radius $$11$$ with $$\|\cdot\|_q\|\cdot\|_q$$.

It is not hard to prove the inequality (4). According to Wikipedia, inequality (3) follows directly from Cauchy-Schwarz, but I don't see how. For $$n=2n=2$$ it is easily proven (see below), but not for $$n>2n>2$$. So my questions are:

1. How can relation (3) be proven for arbitrary $$n\,n\,$$?
2. Can this be generalized into something of the form $$\|x\|_{p} \leq C \|x\|_{q}\|x\|_{p} \leq C \|x\|_{q}$$ for arbitrary $$0?
3. Do any of the relations also hold for infinite-dimensional spaces, i.e. in $$l^pl^p$$ spaces?

Notes:

$$\|x\|_{1}^{2} = |x_{1}|^2 + |x_{2}|^2 + 2|x_{1}||x_{2}| \leq |x_{1}|^2 + |x_{2}|^2 + \big(|x_{1}|^2 + |x_{2}|^2\big) = 2|x_{1}|^2 + 2|x_{2}|^2\|x\|_{1}^{2} = |x_{1}|^2 + |x_{2}|^2 + 2|x_{1}||x_{2}| \leq |x_{1}|^2 + |x_{2}|^2 + \big(|x_{1}|^2 + |x_{2}|^2\big) = 2|x_{1}|^2 + 2|x_{2}|^2$$, hence $$=2\|x\|_{2}^{2}=2\|x\|_{2}^{2}$$
$$\|x\|_{1} \leq \sqrt 2 \|x\|_{2}\|x\|_{1} \leq \sqrt 2 \|x\|_{2}$$.
This works because $$|x_{1}|^2 + |x_{2}|^2 \geq 2|x_{1}\|x_{2}||x_{1}|^2 + |x_{2}|^2 \geq 2|x_{1}\|x_{2}|$$, but only because $$(|x_{1}| - |x_{2}|)^2 \geq 0(|x_{1}| - |x_{2}|)^2 \geq 0$$, while for more than two terms $$\big(|x_{1}| \pm |x_{2}| \pm \dotsb \pm |x_{n}|\big)^2 \geq 0\big(|x_{1}| \pm |x_{2}| \pm \dotsb \pm |x_{n}|\big)^2 \geq 0$$ gives an inequality that never gives the right signs for the cross terms.

1. Using Cauchy–Schwarz inequality we get for all $x\in\mathbb{R}^n$
2. Such a bound does exist. Recall Hölder's inequality

Apply it to the case $|a_i|=|x_i|^p$, $|b_i|=1$ and $r=q/p>1$

Then

In fact $C=n^{1/p-1/q}$ is the best possible constant.
3. For infinite dimensional case such inequality doesn't hold. For explanation see this answer.