$\ell_p\subset\ell_q$ proof verification I have a doubt. I early see a proof of the $\ell_p\subset\ell_q$, if $p<q$, how follows:
Suppose $x\in \ell_p$, then
\begin{align*}
    \|x\|_q^q 
    = \sum_i |x_i|^q 
    = \sum_i |x_i|^{q-p} |x_i|^p 
    \leqslant\left(\sup_i|x_i|\right)^{q-p} \sum_i |x_i|^p 
    &\leqslant \|x\|_p^{q-p} \sum_i |x_i|^p\\
    &= \|x\|_p^{q-p} \|x\|_p^p\\
    &= \|x||_p^q<\infty.
\end{align*}
Hence, $\ell_p\subset\ell_q$
But, using the same reasoning by otherwise, this is, if we take $x\in\ell_q$ we get
\begin{align*}
    \|x\|_p^p 
    = \sum_i |x_i|^p 
    = \sum_i |x_i|^{p-q} |x_i|^q 
    \leqslant\left(\sup_i|x_i|\right)^{p-q} \sum_i |x_i|^q 
    &\leqslant \|x\|_q^{p-q} \sum_i |x_i|^q\\
    &= \|x\|_q^{p-q} \|x\|_q^q\\
    &= \|x||_q^p<\infty.
\end{align*}
So, $\ell_q\subset\ell_p$?
But it's clear that something is wrong, 'cause there is a $x\in\ell_q$ such that $x\not\in\ell_p$, when $p<q$. So, this proof have any problem, or I'm missing anything?

Since $p-q<0$, we do not have $\sup_i\left(\lvert x_i\rvert^{p-q}\right)=\left(\sup_i\lvert x_i\rvert\right)^{p-q}$.
