$\ell_p\subset\ell_q$ proof verification

Question

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?

Answer 1

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