A difficulty in understanding the boundedness of the linear operator associated with each $n \times n$ matrix 

The proof of the boundedness of this operators is given in the following picture but I did not understand from where the first inequality comes, could anyone clarify this for me please? Also I feel that the proof lacks many details, am I right? if so could anyone write the detailed proof for me please?

Thanks!  
 A: I don't know how to prove the first inequality, but you can prove that$$\|Ax\|^2\leqslant\left(\sum_{i,j=1}^n|a_{ij}|^2\right).\|x\|^2\tag{1}$$as follows: the absolute value of the $j^\text{th}$ component of $Ax$ is$$|a_{j1}\alpha_1+a_{j2}\alpha_2+\cdots+a_{jn}\alpha_n|=\bigl|(a_{j1},a_{j2},\ldots,a_{jn}).\left(\overline{\alpha_1},\ldots,\overline{\alpha_n}\right)\bigr|.$$By the Cauchy-Schwarz inequality, this number is smaller than or equal to$$\sqrt{\sum_{i=1}^n|a_{ji}|^2}.\sqrt{\sum_{i=1}^n|\alpha_i|^2}=\sqrt{\sum_{i=1}^n|a_{ji}|^2}.\|x\|^2.$$Since$$\|Ax\|^2=\sum_{j=1}^n|a_{j1}\alpha_1+a_{j2}\alpha_2+\cdots+a_{jn}\alpha_n|^2,$$ you have $(1)$.
A: By the first inequality, I assume you mean $$\|Ax\|^2\leq \sum_{i,j = 1}^n \lvert a_{ij}\rvert^2\lvert \alpha_j\rvert^2$$ Let $x = \alpha_1e_1+\cdots+\alpha_ne_n$. Then, $$\|Ax\|^2 = \sum_{j_1,j_2 = 1}^n \alpha_{j_1}\alpha_{j_2}\langle Ae_{j_1}, Ae_{j_2}\rangle = \sum_{j_1,j_2 = 1}^n \left[\alpha_{j_1}\alpha_{j_2}\sum_{i=1}^n a_{ij_1}a_{ij_2}\right] = \sum_{i,j_1,j_2 = 1}^n \alpha_{j_1}\alpha_{j_2}a_{ij_1}a_{ij_2}$$ The inequality is certainly not true if $\alpha_j$ and $a_{ij}$ are positive for each $1\leq i, j\leq n$, as we can break the sum up like $$\|Ax\|^2 = \sum_{i,j = 1}^n \alpha_j^2a_{ij}^2+\sum_{i=1}^n\sum_{1\leq j_1\neq j_2\leq n} \alpha_{j_1}\alpha_{j_2}a_{ij_1}a_{ij_2}$$ and the second sum will necessarily be positive.
A: First, I think this paper would gain in clarity using $x_i$ instead of $\alpha_i$ and $y_i$ instead of $\beta_i$ for $y=Ax$.
Also, it would be clearer to precise which norms are involved, here clearly we are trying to prove :
$$|||A|||_2\le ||A||_F$$



*

*Where $|||A|||_2=\sup\limits_{x\neq 0}\dfrac{||Ax||_2}{||x||_2}$ for $||x||_2=\sqrt{\sum\limits_{i=1}^n{x_i}^2}\quad$ is the subordinate $L^2$-norm 

*And $||A||_F=\sqrt{\operatorname{tr}(A^*A)}=\sqrt{\sum\limits_{i,j=1}^n |{a_{i,j}}|^2}\quad$ is the Frobenius norm.

So $y=Ax\quad$ with $y_i=\sum\limits_{j=1}^n a_{i,j}x_j$
By Cauchy-Scharz inequality we have $|y_i|=\left|\sum\limits_{j=1}^n a_{i,j}x_j\right|\le||a_{i,\bullet}||_2\,||x||_2$
Thus $\displaystyle||Ax||_2=||y||_2=\left(\sum\limits_{i=1}^n |y_i|^2\right)^{\frac 12}\le\left(\sum\limits_{i=1}^n {||a_{i,\bullet}||_2}^2\,{||x||_2}^2\right)^{\frac 12}\le \left(\sum\limits_{i=1}^n {||a_{i,\bullet}||_2}^2\right)^{\frac 12}||x||_2$
But note that $\sum\limits_{i=1}^n {||a_{i,\bullet}||_2}^2=\sum\limits_{i=1}^n\left(\sum\limits_{j=1}^n |a_{i,j}|^2\right)={||A||_F}^2$
So essentially the result was an application of Cauchy-Schwarz, the rest is just notations manipulation.
