# Relations between p norms

The $$p$$-norm on $$\mathbb R^n$$ is given by $$\|x\|_{p}=\big(\sum_{k=1}^n |x_{k}|^p\big)^{1/p}$$. For $$0 < p < q$$ it can be shown that $$\|x\|_p\geq\|x\|_q$$ (1, 2). It appears that in $$\mathbb{R}^n$$ 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: $$\|x\|_{1} \leq\sqrt n\,\|x\|_{2}$$(3), $$\|x\|_{2} \leq \sqrt n\,\|x\|_\infty$$ (4). Geometrically, it is easy to see that opposite inequalities must hold in $$\mathbb R^n$$. 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 the 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 questions are:

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

Notes:

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

• Your first link (1) has exactly the equation you seek. Actually a version of Norberts excellent answer, generalized to measure spaces. – Thomas Ahle Mar 31 '16 at 20:57
• Probably a dead thread, but norms in finite dimensional spaces are equivalent, so one can always find c ad C such that $c\|x\|_p\leq \|x\|_q\leq C\|x\|_p$. – BigM Dec 1 '19 at 20:23

1. Using Cauchy–Schwarz inequality we get for all $x\in\mathbb{R}^n$ $$\Vert x\Vert_1= \sum\limits_{i=1}^n|x_i|= \sum\limits_{i=1}^n|x_i|\cdot 1\leq \left(\sum\limits_{i=1}^n|x_i|^2\right)^{1/2}\left(\sum\limits_{i=1}^n 1^2\right)^{1/2}= \sqrt{n}\Vert x\Vert_2$$
2. Such a bound does exist. Recall Hölder's inequality $$\sum\limits_{i=1}^n |a_i||b_i|\leq \left(\sum\limits_{i=1}^n|a_i|^r\right)^{\frac{1}{r}}\left(\sum\limits_{i=1}^n|b_i|^{\frac{r}{r-1}}\right)^{1-\frac{1}{r}}$$ Apply it to the case $|a_i|=|x_i|^p$, $|b_i|=1$ and $r=q/p>1$ $$\sum\limits_{i=1}^n |x_i|^p= \sum\limits_{i=1}^n |x_i|^p\cdot 1\leq \left(\sum\limits_{i=1}^n (|x_i|^p)^{\frac{q}{p}}\right)^{\frac{p}{q}} \left(\sum\limits_{i=1}^n 1^{\frac{q}{q-p}}\right)^{1-\frac{p}{q}}= \left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{p}{q}} n^{1-\frac{p}{q}}$$ Then $$\Vert x\Vert_p= \left(\sum\limits_{i=1}^n |x_i|^p\right)^{1/p}\leq \left(\left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{p}{q}} n^{1-\frac{p}{q}}\right)^{1/p}= \left(\sum\limits_{i=1}^n |x_i|^q\right)^{\frac{1}{q}} n^{\frac{1}{p}-\frac{1}{q}}=\\= n^{1/p-1/q}\Vert x\Vert_q$$ In fact $C=n^{1/p-1/q}$ is the best possible constant.
• @Arun, because for $x=(1,1,1,\ldots,1)$ this bound is attained – Norbert Jun 10 '15 at 14:27
• @Norbert: I'm trying to derive a similar bound for the case $p>q$. Any ideas about how to proceed ? – pikachuchameleon Jun 22 '15 at 12:19
• @AshokVardhan what does "similar" mean? Do you want to prove something like $\Vert x\Vert_p \leq C\Vert x\Vert_q$ for $p>q$ ? – Norbert Jun 23 '15 at 14:24