Equivalence of norm Am I right, when I state the following:
Let $p\in[1,\infty) \text{ then }\exists c>0 \text{ s.t. } (\sum_{i=1}^n |a_i| )^p\leq c \sum_{i=1}^n |a_i|^p $ where c is independent of the vector $a=(a_1,\dots,a_n)$.
This follows trivially by the fact that all norms of finite vectorspaces are equivalent, right?
Cheers
 A: 
This follows trivially by the fact that all norms of finite vector spaces are equivalent, right?

Yes, but it may be a too sophisticated tool. An alternative way is to use the convexity of the map $t\geqslant 0\mapsto t^p$. Notice that $c$ depends on the dimension $n$.
A: This is correct.
Since we are only dealing with absolute values of the terms, let's assume that $a_k\ge0$.
It also follows from Jensen's Inequality, and the fact that $x\mapsto x^{q/p}$ is convex when $p\le q$, that
$$
\left(\frac1n\sum_{k=1}^ na_k^p\right)^{q/p}
\le\frac1n\sum_{k=1}^ n\left(a_k^p\right)^{q/p}\tag{1}
$$
Taking $q^\text{th}$ roots, gives
$$
\left(\frac1n\sum_{k=1}^ na_k^p\right)^{1/p}
\le\left(\frac1n\sum_{k=1}^ na_k^q\right)^{1/q}\tag{2}
$$
which is
$$
\|a_k\|_p\le n^{\frac{q-p}{qp}}\|a_k\|_q\tag{3}
$$
In the other direction, find an integer $m\ge\frac qp$, then by simply tossing out terms we have
$$
\left(\sum_{k=1}^na_k^p\right)^m
\ge\sum_{k=1}^na_k^{pm}\tag{4}
$$
Taking $pm^\text{th}$ roots and using $(3)$,
$$
\begin{align}
\|a_k\|_p
&=\left(\sum_{k=1}^na_k^p\right)^{\frac1p}\\
&\ge\left(\sum_{k=1}^na_k^{pm}\right)^{\frac1{pm}}\\
&=\|a_k\|_{pm}\\
&\ge n^{\frac{q-pm}{qpm}}\|a_k\|_q\tag{5}
\end{align}
$$
That is,
$$
n^{\frac{q-pm}{qpm}}\|a_k\|_q\le\|a_k\|_p\le n^{\frac{q-p}{qp}}\|a_k\|_q\tag{6}
$$
