Let $p \in (1,\infty)$ and $q$ the corresponding Hölder conjugate $T: \ell^{p} \to \ell^{p}$ where $(Tx)_{j}=\sum\limits_{k \in \mathbb N}c_{j,k}x_{k}$ and $a:=\sup\limits_{k \in \mathbb N}\sum\limits_{j \in \mathbb N}\vert c_{j,k}\vert<\infty$ and $b:=\sup\limits_{j \in \mathbb N}\sum\limits_{k \in \mathbb N}\vert c_{j,k}\vert<\infty$
Show that $\vert \vert T \vert \vert\leq a^{\frac{1}{p}}b^{\frac{1}{q}}$
Hint:
$\vert \vert T \vert \vert =\sup\limits_{0\neq y \in \ell^{q},0\neq x \in \ell^{p}}\frac{\vert\langle y, Tx\rangle\vert}{\vert\vert y\vert\vert_{q}\vert\vert x\vert\vert_{p}}$ for every bounded operator $T: \ell^{p} \to \ell^{p}$
I have gotten to the following point, using the above hint:
$\vert\langle y, Tx\rangle\vert\leq \vert \sum\limits_{n \in \mathbb N}y_{n}(Tx)_{n}\vert=\vert\sum\limits_{n \in \mathbb N}y_{n}\sum\limits_{k \in \mathbb N}c_{n,k}x_{k}\vert\leq\sum\limits_{n \in \mathbb N}\vert y_{n}\vert\sum\limits_{k \in \mathbb N}\vert c_{n,k}x_{k}\vert=\sum\limits_{n \in \mathbb N}\vert y_{n}\vert \sum\limits_{k \in \mathbb N}\vert c_{n,k}\vert^{\frac{1}{p}}\vert c_{n,k}\vert^{\frac{1}{q}}\vert x_{k}\vert=\sum\limits_{n \in \mathbb N}\vert y_{n}\vert \sum\limits_{k \in \mathbb N}\vert c_{n,k}\vert^{\frac{1}{p}}\vert x_{k}\vert\vert c_{n,k}\vert^{\frac{1}{q}}\leq \sum\limits_{n \in \mathbb N}\vert y_{n}\vert(\sum\limits_{k\in \mathbb N}(\vert c_{n,k}\vert^{\frac{1}{p}}\vert x_{k}\vert)^{p})^{\frac{1}{p}}(\sum\limits_{k \in \mathbb N}\vert c_{n,k}\vert)^{\frac{1}{q}}\leq b^{\frac{1}{q}}\sum\limits_{n \in \mathbb N}\vert y_{n}\vert(\sum\limits_{k\in \mathbb N}(\vert c_{n,k}\vert^{\frac{1}{p}}\vert x_{k}\vert)^{p})^{\frac{1}{p}}$
I get to this point, but then in the solutions (with the justification Tonelli and Hölder), it continues:
$b^{\frac{1}{q}}\sum\limits_{n \in \mathbb N}\vert y_{n}\vert(\sum\limits_{k\in \mathbb N}(\vert c_{n,k}\vert^{\frac{1}{p}}\vert x_{k}\vert)^{p})^{\frac{1}{p}}\leq b^{\frac{1}{q}} \vert \vert y\vert \vert_{q}(\sum\limits_{n \in \mathbb N}\sum\limits_{k \in \mathbb N}\vert c_{j,k}\vert \vert x_{k}\vert^{p})^{\frac{1}{p}}\leq b^{\frac{1}{q}}\vert \vert y\vert \vert_{q}a^{\frac{1}{p}}\vert\vert x\vert\vert_{p}$
I do not understand the last two inequalities. Essentially, how do I get, using Hölder and Tonelli:
$\sum\limits_{n \in \mathbb N}\vert y_{n}\vert(\sum\limits_{k\in \mathbb N}(\vert c_{n,k}\vert^{\frac{1}{p}}\vert x_{k}\vert)^{p})^{\frac{1}{p}}\leq \vert \vert y\vert \vert_{q}(\sum\limits_{n \in \mathbb N}\sum\limits_{k \in \mathbb N}\vert c_{j,k}\vert \vert x_{k}\vert^{p})^{\frac{1}{p}}\leq \vert \vert y\vert \vert_{q}a^{\frac{1}{p}}\vert\vert x\vert\vert_{p}$
It's a little late, but in case you still needed an answer:
Let $z_n = \sum_{k \in \mathbb{N}}|c_{n,k}||x_k|^p$. Then the leftmost term in your last line is $\sum_{n}|y_n||z_n|^{\frac{1}{p}}$. Applying Hölder immediately, we get: \begin{align*} \sum_{n}|y_n|(|z_n|^{\frac{1}{p}}) &\le \|y\|_q \|\{|z_n|^{\frac{1}{p}}\}_{n\in \mathbb{N}}\|_p = \|y\|_q (\sum_{n} |z_n|^{\frac{1}{p}p})^\frac{1}{p}\\ &\le \|y\|_q (\sum_{n} |z_n|)^\frac{1}{p} = \|y\|_q (\sum_n \sum_k|c_{n,k}||x_k|^p)^\frac{1}{p}. \end{align*}
So the first inequality is just Hölder. For the second inequality, we reverse the order of summation using Tonelli to get:
$$\|y\|_q (\sum_n \sum_k|c_{n,k}||x_k|^p)^\frac{1}{p} = \|y\|_q (\sum_k (\sum_n|c_{n,k}|)|x_k|^p)^\frac{1}{p} \le \|y\|_q (\sum_k a|x_k|^p)^\frac{1}{p} = a^\frac{1}{p}\|y\|_q\|x\|_p.$$