The p-norm on Rn is given by ‖. 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:

- 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<p < q\,?
- 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.

**Answer**

- 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

- 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. - For infinite dimensional case such inequality doesn't hold. For explanation see this answer.

**Attribution***Source : Link , Question Author : PianoEntropy , Answer Author : Community*