I have been reading this paper and in the beginning of section 2 it is written something that I am failing to see why it holds. It is stated that if $A$ is a $dxd$ bounded, symmetric matrix such that there are constants $0<\Lambda_m<\Lambda_M<\infty$ that satisfy the following inequalities for all $z \in \mathbb{R}^d$:
$$\Lambda_m\sum_{i=1}^dz_i^2 \leq \sum_{i,j=1}^da_{ij}z_iz_j \leq \Lambda_M\sum_{i=1}^dz_i^2$$
then:
$$\sup\limits_{j \in \{1,...,d\}}\sqrt{\sum_{i=1}^d a_{ij}^2} \leq \Lambda_M$$
Can someone give me some hints on why this is true? I am thinking that some functional analysis results may come handy, but I have not come up with anything so far. Thanks in advance!
Interpreting $z$ as a column vector, we can write the inequalities as $$ 0 < \Lambda_m \leq z^TAz \leq \Lambda_M. $$ The Rayleigh-Ritz theorem tells us that the maximal possible value of $z^TAz$ is equal to the largest eigenvalue of $A$, and because $A$ is symmetric this is also equal to $$ \|A\| := \max_{\|x\| = 1} \|Ax\|, $$ and this is at most equal to $\Lambda_M$. Thus, if $e_i$ denotes the $i$th standard basis vector, then $$ \|Ae_j\| \leq \|A\| \leq \Lambda_M, $$ hence $\max_{j \in 1,\dots,d} \|Ae_j\| \leq \Lambda_M$, which is the deisred result.
For a direct proof, we could do the following. Select $j \in \{1,\dots,d\}$, let $y = Ae_j/\|Ae_j\|$. With the help of the polarization identity and Cauchy-Schwarz inequality, we have \begin{align} 4\sqrt{\sum_{i=1}^d a_{ij}^2} &= 4y^T Ae_j \\ & = (e_j + y)^TA(e_j + y) - (e_j - y)^TA(e_j - y) \\ & \leq \Lambda_M \|e_j + y\|^2 - \Lambda_M\|e_j - y\|^2 \\ & = \Lambda_M(\|e_j + y\|^2 - \|e_j - y\|^2) \\&= 4\Lambda_M e_j^Ty \leq 4 \Lambda_M \|e_j\|\cdot\|y\| = 4 \Lambda_M \end{align}