$\{a_{ij}\}_{i,j}\in \mathbf{C}$. Suppose $x=(x_1,x_2, . . . )$ be a sequence. Define a new sequence $Ax$ by $(Ax)_i = \sum_{j=1}^{\infty} a_{ij}x_j$ (if it makes sense). Consider the linear map $A:x\mapsto Ax$. Suppose $1<p<\infty$ and $q$ denotes the conjugate exponent of $p$, and the following three quantities are finite:
$\alpha_{p,q}=\sum_{i=1}^{\infty}(\sum_{j=1}^{\infty}|a_{ij}|^q)^\frac{p}{q}, \beta =sup_{i}\sum_{j=1}^{\infty}|a_{ij}|, \gamma=sup_{j}\sum_{i=1}^{\infty}|a_{ij}|$, then I need to show that A is a bounded linear operator from $l^p$ to itself and $||A||\leq min\{\alpha_{p,q} ^\frac{1}{p}, \beta^\frac{1}{q}\gamma^\frac{1}{p}\}$
I have shown that $||A||\leq \alpha_{p,q} ^\frac{1}{p}$ using Holder's inequality, need some help to show the other inequality.
For the second inequality, again using Holder's inequality, $$\forall i,\quad\bigg|\sum_ja_{ij}x_j\bigg|\le\sum_j|a_{ij}|^{1/q}|a_{ij}|^{1/p}|x_j|\le\left(\sum_j|a_{ij}|\right)^{1/q}\left(\sum_j|a_{ij}||x_j|^p\right)^{1/p}$$ The first term on the right is smaller than $\beta^{1/q}$.
Hence $$\|Ax\|_p^p=\sum_i\big|\sum_ja_{ij}x_j\big|^p\le\beta^{p/q}\sum_{i,j}|a_{ij}||x_j|^p\le\beta^{p/q}\gamma\|x\|_p^p$$ $$\therefore\quad\|Ax\|_p\le\beta^{1/q}\gamma^{1/p}\|x\|_p$$
This argument is originally due to Schur (for $p=q=2$), I believe.