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I can't seem to work out the inequality $(\sum |x_n|^q)^{1/q} \leq (\sum |x_n|^p)^{1/p}$ for $p \leq q$ (which I'm assuming is the way to go about it).

Maximilian Janisch
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user1736
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5 Answers5

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You are right @user1736.

If $0<a\leq1$ then $$\left(\sum |a_n|\right)^a\leq \sum|a_n|^a.\tag{1}$$

Hence for $p\leq q$ we have $p/q\leq1$, and $$\left(\sum_n |x_n|^q\right)^{1/q}=\left(\sum_n |x_n|^q\right)^{p/qp}\leq \left(\sum_n |x_n|^{q(p/q)}\right)^{1/p}=\left(\sum|x_n|^p\right)^{1/p}$$


Edit: How do we prove (1) (for $0<a<1$)?

Step 1. It is sufficient to prove this for finite sequences because then we may take limits.

Step 2. To prove the statement for finite sequences it is sufficient to prove $$(x+y)^a\leq x^a+ y^a,\quad\text{ for $x,y>0$}\tag{2}$$ because the finite case is just iterations of (2).

Step 3. To prove (2) it suffice to prove $$(1+t)^a\leq 1+t^a,\quad\text{ where $0<t<1$} \tag{3}$$

Now, the derivative of the function $f(t)=1+t^a-(1+t)^a$ is given by $f'(t) = a(t^{a-1} - (1+t)^{a-1})$ and it is positive since $a>0$ and $t\mapsto t^b$ is decreasing for negative $b$. Hence, $f(t)\geq f(0)=0$ for $0<t<1$, which proves (3).

AD - Stop Putin -
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    Where did you learn the first inequality? – newbie Jun 13 '13 at 09:36
  • @newbie I really don't remember, but it is not that difficult to prove. Added some steps. – AD - Stop Putin - Jun 13 '13 at 12:40
  • @newbie If you are interested you might search for "locally bounded spaces" and "quasi-normed linear spaces". – AD - Stop Putin - Jun 13 '13 at 13:38
  • Thanks for your clarification. Do you mean that $\ell^p$ with $p\in(0,1)$ is quasi-normed? – newbie Jun 13 '13 at 13:48
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    @newbie Some references are Stefan Rolewicz *Metric Linear Spaces* and Wiesław Żelazko *Metric generalizations of Banach algebras*. An other common name is $p$-norm, while quasi-norm is something like $\|x + y\|\leq K(\|x\| +\|y\|)$ which is *equivalent* to a $p$-norm. So, "Yes" - is a sense - at least there is a strong connection. (Very interesting algebras, if I may say so, many things that works for $p\geq1$ are not true while others are but needs different kind of proofs). – AD - Stop Putin - Jun 13 '13 at 14:50
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    Can you explain why to prove (2) it suffices to prove (3)? I can't see it. – DrHAL Sep 10 '16 at 23:25
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    I also cant see that – Luiz Sep 11 '16 at 14:37
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    @DrHAL Suppose we knew (3), and suppose $x>y$, then $$(x+y)^a=x^a(1+(y/x))^a \leq x^a(1 + (y/x)^a) = x^a + y^a$$ because $y/x<1$. – AD - Stop Putin - Sep 12 '16 at 07:40
  • @LuizFernando Is this better? – AD - Stop Putin - Sep 12 '16 at 07:45
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    @AD. (1) could be proven *much more easily* by noting that $$\sum_i \frac{x_i^a}{\left(\sum_j x_j\right)^a}=\sum_i \left(\frac{x_i}{\sum_j x_j}\right)^a\geq \sum_i \frac{x_i}{\sum_j x_j}=1.$$ In my inequality I used that $x^a>x$ for $x\in]0,1]$ and $a\in]0,1]$. – Maximilian Janisch Nov 12 '19 at 15:10
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    @AD. Of course I meant $x^a\geq x$. – Maximilian Janisch Nov 12 '19 at 15:15
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    @MaximilianJanisch Indeed, nice of you to share that - short and simple. – AD - Stop Putin - Nov 19 '19 at 18:20
  • For step (2), just wanted to link this other proof https://math.stackexchange.com/questions/264156/prove-that-pqm-leq-pmqm?noredirect=1&lq=1 that seems even simpler than others already proposed in the comments. – D.R. Feb 09 '22 at 17:57
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I don't think you need to prove the inequality you have in the question; that's a bit too strong. Note that $\{x_n\}\in\ell_p$ if and only if $\left(\sum|x_n|^p\right)^{1/p}$ is finite, if and only if $\sum|x_n|^p\lt\infty$. So you really just need to show that if $\sum|x_n|^p$ is finite, then $\sum|x_n|^q$ is finite, assuming $p\leq q$.

You want to remember is two things:

  1. if $p\leq q$, then for $|x|>1$ you have $|x|^p\leq|x|^q$, but if $|x|<1$, then $|x|^p \geq |x|^q$.
  2. $\sum_{n=1}^{\infty}a_n$ converges if and only if for every $m\geq 1$, $\sum_{n=m}^{\infty}a_n$ converges.
Arturo Magidin
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The $\boldsymbol{\ell^{2^m}}$ Norm is Less Than the $\boldsymbol{\ell^1}$ Norm

Assume $b_k\ge0$, $$ \begin{align} \left(\sum_{k=1}^nb_k\right)^2 &=\sum_{k=1}^nb_k^2+2\!\!\!\sum_{\substack{j,k=1\\j\lt k}}^n\!\!b_jb_k\\ &\ge\sum_{k=1}^nb_k^2\tag1 \end{align} $$ Therefore, by induction, we have that $$ \left(\sum_{k=1}^nb_k\right)^{2^m}\ge\sum_{k=1}^nb_k^{2^m}\tag2 $$


Interpolate

For $1\lt r\lt2^m$, $$ \begin{align} \sum_{k=1}^nb_k^r &=\sum_{k=1}^n\left(b_k^{2^m}\right)^{\frac{r-1}{2^m-1}}b_k^{\frac{2^m-r}{2^m-1}}\tag3\\ &\le\left(\sum_{k=1}^nb_k^{2^m}\right)^{\frac{r-1}{2^m-1}}\left(\sum_{k=1}^nb_k\right)^{\frac{2^m-r}{2^m-1}}\tag4\\ &\le\left(\sum_{k=1}^nb_k\right)^{2^m\frac{r-1}{2^m-1}}\left(\sum_{k=1}^nb_k\right)^{\frac{2^m-r}{2^m-1}}\tag5\\ &=\left(\sum_{k=1}^nb_k\right)^r\tag6 \end{align} $$ Explanation:
$(3)$: $r=2^m\frac{r-1}{2^m-1}+\frac{2^m-r}{2^m-1}$
$(4)$: Hölder
$(5)$: Inequality $(2)$
$(6)$: $r=2^m\frac{r-1}{2^m-1}+\frac{2^m-r}{2^m-1}$


Apply to $\boldsymbol{\ell^p}$ and $\boldsymbol{\ell^q}$

Let $r=\frac qp$, and $b_k=a_k^p$, then $(6)$ says $$ \sum_{k=1}^na_k^q\le\left(\sum_{k=1}^na_k^p\right)^{q/p}\tag7 $$ which is equivalent to $$ \left(\sum_{k=1}^na_k^q\right)^{1/q}\le\left(\sum_{k=1}^na_k^p\right)^{1/p}\tag8 $$

robjohn
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Let $|x_i|^p=a_i$ and $\frac{q}{p}=k$.

Thus, $k\geq1$ and we need to prove that: $$(a_1+a_2+...+a_n)^k\geq a_1^k+a_2^k+...+a_n^k.$$ Now, let $a_1\geq a_2\geq...\geq a_n$.

Thus, since $f(x)=x^k$ is a convex function and $$(a_1+a_2+...+a_n,0,...,0)\succ(a_1,a_2,...,a_n),$$ by Karamata we obtain: $$(a_1+a_2+...+a_n)^k+0^k+...+0^k\geq a_1^k+a_2^k+...+a_n^k,$$ which is our inequality.

Michael Rozenberg
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For completeness I will add this as an answer (it is a slight adaptation of the argument from AD.):

For $a\in[0,1]$ and any $y_i\geq 0, i\in\mathbb N$, with at least one $y_i\neq0$ and the convention that $y^0=1$ for any $y\geq0$, \begin{equation}\label{*}\tag{*}\sum_{i=1}^\infty \frac{y_i^a}{\left(\sum_{j=1}^\infty y_j\right)^a}=\sum_{i=1}^\infty \left(\frac{y_i}{\sum_{j=1}^\infty y_j}\right)^a\geq \sum_{i=1}^\infty \frac{y_i}{\sum_{j=1}^\infty y_j}=1,\end{equation} where I have used $y^a\geq y$ whenever $y\in[0,1]$ and $a\in[0,1]$. (This can be derived for instance from the concavity of $y\mapsto y^a$.)

For $p=q$, there is nothing to prove. For $1\le p< q\le\infty$ and $x=(x_i)_{i\in\mathbb N}\in \ell^q$, set $a\overset{\text{Def.}}=\frac pq\in[0,1]$ and $y_i\overset{\text{Def.}}=\lvert x_i\rvert^q\ge0$. Then \eqref{*} yields \begin{equation*} \sum_{i=1}^\infty \lvert x_i\rvert^p\geq\left(\sum_{i=1}^\infty \lvert x_i\rvert^{q}\right)^{\frac pq}, \end{equation*} i.e. \begin{equation*} \lVert x\rVert_{\ell^q}\le\lVert x\rVert_{\ell^p}. \end{equation*}

Maximilian Janisch
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