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).


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).

  • 1
    $\begingroup$ Where did you learn the first inequality? $\endgroup$
    – newbie
    Jun 13 '13 at 9:36
  • 1
    $\begingroup$ @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). $\endgroup$
    – AD.
    Jun 13 '13 at 14:50
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    $\begingroup$ @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 $y/x<1$. $\endgroup$
    – AD.
    Sep 12 '16 at 7:40
  • 4
    $\begingroup$ @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 $a\in]0,1]$. $\endgroup$ Nov 12 '19 at 15:10
  • 2
    $\begingroup$ @AD. Of course I meant $x^a\geq x$. $\endgroup$ Nov 12 '19 at 15:15

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.
  • $\begingroup$ That makes sense. Thanks a lot! Is the inequality that I wrote down true though? $\endgroup$
    – user1736
    Sep 5 '10 at 20:24
  • 2
    $\begingroup$ Arturo, it is not too strong, as you can see in my answer. $\endgroup$
    – AD.
    Jan 21 '11 at 12:40
  • $\begingroup$ This logic is easy to think, so useful. $\endgroup$
    – Supriyo
    Feb 9 '14 at 11:29
  • $\begingroup$ @Dutta Could you explain how to proceed using these hints? I do not see how to prove it using this. $\endgroup$
    – Soap
    Mar 29 '17 at 20:03
  • $\begingroup$ @Simoes I am out of touch with functional analysis nowadays. You may go through a standard undergraduate textbook on functional analysis. $\endgroup$
    – Supriyo
    Mar 30 '17 at 5:25

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 $$ \left(\sum_{k=1}^nb_k\right)^{2^m}\ge\sum_{k=1}^nb_k^{2^m}\tag2 $$


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 $$ \left(\sum_{k=1}^na_k^q\right)^{1/q}\le\left(\sum_{k=1}^na_k^p\right)^{1/p}\tag8 $$


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.


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*}


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