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Is it obvious that in $\mathbb R^n, \forall \ p,q \ \text{ s.t. } \ p \leq q$

$$ \vert \vert x\vert \vert_{p} \leq \vert \vert x\vert \vert_{q}, \text{where }\vert \vert x\vert \vert_{q} = (\sum_0^n x^q)^{\frac 1 q }$$ If yes, how to prove it ? I was thinking about the classical inequalities like Minkowski, Holder or Cauchy Schwartz but none seems to work here.

Moreover, is this inequality still true for any norm ? Like the norm $L_p $ and $L_q$ ? (I mean by this $ (\int \vert f(x)^q \vert dx ) ^{\frac 1 q } $ )

In other words, is it true for a space of infinite dimension ?

What about the infinite norm ? is it also always bigger than any other one ? (this one I know is false for the case $\mathbb R^n $, but it is true for functions spaces like the space of continuous functions...)


P.S. I'm sorry, I have not yet studied functionnal analysis. I'm asking this for my metric space class where we are proving that some spaces are complete, and I'm wondering when is it possible to use the very usefull inequality of norm I was talking about.

I'm asking this because intuitively, it seems right : enter image description here

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Actually, direction should be reversed: it holds that $$ \|x\|_p\ge \|x\|_q,\quad 1\le p< q\le\infty.\tag{*} $$ What your figure shows us is that $$ B_p\subset B_q,\quad p< q\tag{**} $$ where $B_r$ is a unit ball in $r$-norm. It is easy to see that $\text{(*)}$ is equivalent to $\text{(**)}$.

Now, we prove $\text{(*)}$. By the homogeneity of norms, it is sufficient to show that $$ \|x\|_q\le 1 $$ provided that $\|x\|_p= 1$. By the assumption, we have $$ |x_i|^p\le |x_1|^p+|x_2|^p+\cdots +|x_n|^p=1 $$ for all $i\le n$. Hence $|x_i|\le 1$ and it follows $$ |x_i|^q\le |x_i|^p,\quad i=1,2,\ldots,n $$ for $q<\infty$, and $\|x\|_\infty \le 1$ for $q=\infty$. Summing over $i$ yields $$ \|x\|_q^q =|x_1|^q+|x_2|^q+\cdots +|x_n|^q\le |x_1|^p+|x_2|^p+\cdots +|x_n|^p=1 $$ for $q<\infty$. This proves $\|x\|_q\le 1$ for all $1\le p< q\le \infty$ as desired.

Note: The inequality can be easily extended to $\Bbb R^{\Bbb N}$ equipped with $l^p$-norms, i.e. $l^p(\Bbb N)$. And reverse inequality is true if we are on a finite measure space, i.e. $$ \|f\|_p\le \mu(\Omega)^{1/p-1/q}\|f\|_q,\quad 1\le p<q\le\infty. $$ holds for $f\in L^q(\Omega)\subseteq L^p(\Omega)$. (It a direct consequence of Jensen's inequality.) But in a general measure space, in particular with infinite total measure such as $L^p(\Bbb R)$, none of the relationship $$ \|f\|_p\le C\|f\|_q $$ nor $$ \|f\|_q\le C\|f\|_p $$ is true for any $C>0$. (So there is no inclusion between $L^p(\Bbb R)$-spaces. I'll omit further explanation, but one maybe able to find examples of this.) Thus the fact that $$l^p(\Bbb N)\subseteq l^q(\Bbb N), \quad 1\le p<q\le \infty$$ is a special property, which can be attributed to certain properties of counting measure.

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  • $\begingroup$ Are you sure? In my lecture it s written : $$ abs(f_m(t) - f_n(t)) \leq \vert \vert f_m- f_n \vert \vert_{\infty} $$ and also for any element in the space of sequences $l^2_{ \mathbb F} $ : $$ abs( u_j^{(n) } - u_j^{(m) }) \leq \vert \vert u^{(n) } - u^{(m) } \vert \vert_{2} $$ $\endgroup$ Jan 23, 2019 at 19:20
  • $\begingroup$ And what about the other points I was talking about? Maybe it s linked to my other comment. Please be as precise as you can, i m baffled with all of this $\endgroup$ Jan 23, 2019 at 19:24
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    $\begingroup$ @MarineGalantin Yes, I'm so sure. If you are suspicious, why don't you calculate actual $p$-norms of a vector, say $(1,3,4)$? I think I fully explained why $\|x\|_p \ge \|x\|_q$ is true. And about those two facts, yes, they are true $\textbf{independently}$ of the inequality I showed. Those are just trivial pointwise bound obtained from the definition of sup-norm and $2$-norm. And they have nothing to do with the inequality discussed. $\endgroup$ Jan 23, 2019 at 19:43
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    $\begingroup$ @MarineGalantin The direction of the inequality is clear if you remember the obvious endpoint $$ \|x\|_{\ell^\infty} \le \|x\|_{\ell^1}$$ that cannot be reversed $\endgroup$ Jan 23, 2019 at 20:36
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    $\begingroup$ @MarineGalantin Yes, surely! To add some more explanation (using symbolic manipulations), let us denote $n\cdot B_p =\{n\cdot x\in l^p(\Bbb N)\;:\;x \in B_p\}$ (it is just a ball of radius $n$.) Then,$$ \forall n\ge 1,\quad n\cdot B_p\subset n\cdot B_q\subset l^q(\Bbb N).$$ Since $\bigcup_{n\ge 1} n\cdot B_p = l^p(\Bbb N)$ , it follows $$l^p(\Bbb N)\subset l^q(\Bbb N).$$ $\endgroup$ Jan 23, 2019 at 21:02

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