$$ \left( \sum |z_i|^a \right)^b \geq \left( \sum |z_i|^b \right)^a, \quad \forall a\in [1,2], b\in [2,+\infty) $$

Actually this is equivalent to $$ \|z\|_a \geq \|z\|_b \quad \forall a\leq b $$

  • $\begingroup$ try it with just two terms and then use inductive reasoning and the associative property of addition. $\endgroup$ – robert bristow-johnson Dec 11 '16 at 3:08
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    $\begingroup$ BTW, there is a math stack exchange. one of them guys would be happy to take this on. $\endgroup$ – robert bristow-johnson Dec 11 '16 at 3:12
  • $\begingroup$ this hits pretty close to your question. $\endgroup$ – robert bristow-johnson Dec 11 '16 at 3:19
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    $\begingroup$ To be completely frank, this is off-topic here (as it's an equation, but you don't even try to explain what it has to do with signal processing), and it doesn't show any attempt of your own. Instead, this feels like you're asking us to do your homework... $\endgroup$ – Marcus Müller Dec 11 '16 at 3:59
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    $\begingroup$ @Pierre I didn't want to accuse you of being lazy – it's just that if someone "dumps" a formula on my desk, it feels like I'm doing their homework. As Laurent has very nicely expressed, norms are very important tools in DSP, and if you could expand the question to give a bit of background, I'm sure it's bound to be a good question :) $\endgroup$ – Marcus Müller Dec 11 '16 at 12:59

Although the question could belong to SE.math, mastering inequalities for $\ell_p$ norms (for $p\ge 1$) or quasinorms (for $0<p< 1$), and their norm ratios and powers, is quite important in modern signal/image processing: starting from Cauchy-Bunyakovsky-Schwarz, to power means, Rogers-Hölder, Young, Minkowski, Dresher-Gini etc.) There are scalar, additive and convolutive versions, and reverse forms, that are used in data restoration and deconvolution, especially when using signal sparsity.

If $a_n$ and $b_n$ are positive finite sequences of length $N$, $u,v,w>0$, and $1/u+1/v\le 1/w$, the generalized Rogers-Hölder's inequality gives:

\begin{equation} \left( \sum_{n=1}^{N} (a_n b_n)^w \right)^{1/w} \le \left( \sum_{n=1}^{N} (a_n )^u \right)^{1/u} \left( \sum_{n=1}^{N} (b_n )^v \right)^{1/v}\,. \label{eq_rogers-holder} \end{equation}

You can find a proof in P. S. Bullen, Handbook of Means and Their Inequalities, Kluwer Academic Publishers, 2nd edition, 2003, p. 187, Corollary 8. 22,

If you now set $a_n=|z_n|^p$, $b_n=1$, $w=p$, $v=q$ and $1/u=1/p-1/q$, you get \begin{equation} \ell_{q}{(|z|)} \leq \ell_{p}{(|z|)} \leq \ell_{0}^{\frac{1}{p}-\frac{1}{q}}\ell_{q}{(|z|)} \,. \end{equation}

$\ell_{0}$ here, not a norm, nor a quasinorm, is the count index, ie the number of nonzero terms in $z$, or the sparsity. In other words, there is an inclusion in $l_p$ spaces: if $p\le q$, $l_p\subset l_q$. Your answer is incorrect, because you should have $a\le b$ instead.

One of the basic block is alluded to by @robert bristow-johnson. If $0<c\le 1$ ($a_n$ positive as above):

$$ \left(\sum a_n\right)^c \le \sum a_n^c \,.$$

Using additivity, prove it first with two terms:

$$(a_1+a_2)^c \le a_1^c+a_2^c\,.$$

By factoring by the largest of $a_1$ or $a_2$, you end up to prove that for $0<t<1$:

$$(1+t)^c\le 1+t^c\,.$$

Just differentiate the function $t\mapsto 1+t^c- (1+t)^c$ as $c\left(t^{c-1}- (1+t)^{c-1}\right)$, and verify the latter is positive.

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