Proof Explanation if $1 \leq p \leq q \lt \infty$ then $\|x\|_{q} \leq \|x\|_{p}$

Question

Proof taken from Inequality between $\ell^p$-norms

To see this, note that both sides of the inequality $\|x\|_{q} \leq \|x\|_{p}$ are homogeneous in $x$ (multiplying $x$ with a positive real number multiplies both sides with the same positive factor), so we may take without loss of generality an $x$ with $\|x\|_{p} = 1$. Then $\|x\|_{q}^{q} = \sum_{j = 1}^{\infty} |x_{j}|^{q} \leq \sum_{j = 1}^{\infty} |x_{j}|^{p} = 1$, and this is because for $t \leq 1$ and $p \leq q$ we have $t^{q} \leq t^{p}$.

The problem I have is that this proof completely ignores the roots in the norm expressions. For instance if $S = 0.001$ then $S^{\frac{1}{1000}} \approx 1 >> S$. So if $\sum_k |x_k|^q \leq \sum_k |x_k|^p$ then taking $p$ and $q$ roots could mean the inequality no longer holds.

Answer 1

Once you know $\sum_k |x_k|^q \leq \sum_k |x_k|^p$ just take the $q$-root on both sides. Notice that on the RHS will appear $1=\|x\|_p$.

Answer 2

It is sufficient to prove that $\| x \|_q \leqslant \| x \|_p$ for every $x \neq 0$. But then the inequality is equivalent to $$ \frac{\| x \|_q}{\| x \|_p} \leqslant 1 \quad \Longleftrightarrow \quad \left\| \frac{x}{\| x_{} \|_p} \right\|_q \leqslant 1. $$ Therefore it is sufficient to prove that $$ \| y \|_q^q \leqslant 1 \quad \forall y \,:\, \| y \|_p^p = 1. $$ For that you note that $$ 1 = \| y \|_p^p = \sum | y_i |^p \geqslant \sum | y_i |^q = \| y \|_q^q . $$ The previous inequality follows from the observation that since $\sum | y_i |^p = 1$, all $| y_i |$ are less than or equal to $1$. But if $| y_i | \leqslant 1$ then $| y_i |^q \leqslant | y_i |^p$ when $q \geqslant p$.