log norm inequality for lower triangular part of matrix Suppose $L$ is the lower triangular part of a matrix $A \in \mathbb{C}^{n\times n}$. Prove that $||L||_2 \leq ||A||_2 \log_2(2n)$. Here $||\cdot||_2$ is the matrix norm induced by the $p=2$  vector norm. This problem is from the book A Brief Introduction to Numerical Analysis by E. Tyrtyshnikov, Problem 16, Lecture 2.
Hints:


*

*If $B$ is a sub-matrix of $A$, $||B||_2 \leq ||A||_2$

*$L$ can be represented as sum of sub-matrices of A, each having  one non-zero block of $A$.


Despite the hints, am unable to solve this. Please help.
Similar question here: Bounding lower triangular perturbation , this  answer probably solves that too.
 A: EDIT. That follows is the key of the proof.
Proposition. If $||L_n||_2\leq \log_2(2n)||A_n||_2$, then  $||L_{2n}||_2\leq \log_2(4n)||A_{2n}||_2$.
Proof. $L_{2n}=\begin{pmatrix}L'_n&0\\C_n&L''_n\end{pmatrix}$ is the lower submatrix of $A_{2n}=\begin{pmatrix}A'_n&B_n\\C_n&A''_n\end{pmatrix}$. Note that $L_{2n}=\begin{pmatrix}L'_n&0\\0&L''_n\end{pmatrix}+\begin{pmatrix}0&0\\C_n&0\end{pmatrix}$. Clearly $||L'_n||_2\leq \log_2(2n)||A'_n||_2\leq \log_2(2n)||A_{2n}||_2$ and $||L''_n||_2\leq \log_2(2n)||A_{2n}||_2$; consequently $||diag(L'_n,L''_n)||_2\leq \log_2(2n)||A_{2n}||_2$.
Finally $||L_{2n}||_2\leq (\log_2(2n)+1)||A_{2n}||_2$ and we are done.
EDIT. Do as follows. One has, for every $n$,  $||L||_2\leq \sqrt{n}||A||_2$, that is, for $n=4$, $||L||_2\leq 2||A||_2$. According to the reasoning above, the factors of $||A||_2$ are 
$n=4 \rightarrow 2,n=8 \rightarrow 3,\cdots,n=256 \rightarrow 8,\cdots$. Assume that $n\in [257,512]$. Then the reasoning above gives the factor $8+1=9$ (increase the matrix with 1's on the diagonal until the dimension $512$); we are done because $2^9=512<2\times 257=514$.
