Matrix-vector norm inequalities in tridiagonal matrix Let's consider the linear system $A\vec{x}=\vec{b}$, with $A$ being a diagonally dominant by columns and tridiagonal matrix, that is:
$$A=\begin{pmatrix} d_1&& c_2    &&        &&        &&      \\
                    a_1&& d_2    && c_3    &&        &&      \\
                       && \ddots && \ddots && \ddots &&      \\
                       &&        && a_{n-2}&& d_{n-1}&& c_{n}\\
                       &&        &&        && a_{n-1}&& d_n \end{pmatrix}$$
with $a_n=c_1=0$. Prove that $K\|\vec{x}\|_{\infty} \leq \|\vec{b}\|_{\infty}$ is satisfied, where $$K = \min\limits_{1 \leq i \leq n} \left\{|d_i|-|a_i|-|c_i| \right\}$$

I've tried using the property $$\|\vec{b}\|_{\infty} = \|A\vec{x}\|_{\infty} \leq \|A\|_{\infty} \|\vec{x}\|_{\infty}$$ but I don't get anywhere near the answer. Any suggestions?
 A: I think there's something wrong in the description.
As it is, the statement is false and a counter example is provided by
$$
A=\begin{pmatrix}1 & 2 & 0\\ 0 & 3 & 0\\ 0 & 0 & 1\end{pmatrix},
\quad
\vec{x} = \begin{pmatrix}2\\ -\frac12\\ 0 \end{pmatrix},
$$
which gives $K=1$, $\Vert\vec{x}\Vert_{\infty} = 2$ and $\Vert\vec{b}\Vert_{\infty} = \frac32$.
Conversely, if one considers $A^{\top}\vec{x}=\vec{b}$, then one can prove the inequality as follows. If $K=0$, the inequality is trivially satisfied. We can therefore assume $K>0$ in the following, which means that $A^\top$ is strictly diagonally dominant (by row) and therefore invertible [1]. Hence we can write
$$
\Vert\vec{x}\Vert_{\infty} = \Vert A^{-\top}\vec{b}\Vert_{\infty}
\le
\Vert A^{-\top}\Vert_{\infty} \, \Vert\vec{b}\Vert_{\infty},
$$
where the inequality follows from the standard property of matrix norms induced by vector norms [2].
We then use Theorem 1 in [3], which states: "Assume $A$ is diagonally dominant by rows and set $\alpha = \min_k(\vert a_{kk} \vert - \sum_{j\neq k} \vert a_{kj} \vert)$. Then $\Vert A^{-1}\Vert_{\infty}<1/\alpha$" (note: they must have a typo because the inequality is actually not strict). We apply their theorem to our $A^\top$ and obtain that
$$
\Vert A^{-\top}\Vert_{\infty} \le 1/K.
$$
This gives
$$
\Vert\vec{x}\Vert_{\infty}
\le
\frac1K \Vert\vec{b}\Vert_{\infty}.
$$
Rearranging proves the (modified, i.e. for $A^\top$ rather than $A$) statement of the question.
A: I've been thinking about the problem and I got here:
Let's consider $$K||x||_{\infty}\leq ||b||_{\infty}=||Ax||_{\infty}\leq||A||_{\infty}·||x||_{\infty}$$
Simplifying, it results in:
$$K\leq||A||_{\infty}$$
or $$1\leq \dfrac{||A||_{\infty}}{K}$$
Given that $$||A||_{\infty}=\text{max}_{i\leq j \leq n}\sum_{j=1}^n|a_{ij}|=|d_j|+|a_j|+|c_j|$$ for some $j\in\mathbb{N}$
and that $$K = \min\limits_{1 \leq i \leq n} \left\{|d_i|-|a_i|-|c_i| \right\}=|d_i|-|a_i|-|c_i|$$ for some $i\in\mathbb{N}$
We can rewrite the third inequality as
$$1\leq\dfrac{|d_j|+|a_j|+|c_j|}{|d_i|-|a_i|-|c_i|}$$
We know that because A is a tridiagonal matrix that
$$|d_i|\geq |a_i|+|c_i|$$ for all $i\in\mathbb{N}$
That is $$|d_i|-|a_i|-|c_i|\geq 0$$
Then we have two possibilities $i=j$ or $i\neq j$
If $i=j$ then $|d_j|+|a_j|+|c_j|\geq |d_j|-|a_j|-|c_j|$
Now, $K$ is defined as the minimum of the difference of $|d_i|-|a_i|-|c_i|$ so it follows that if $i\neq j$ then $|d_j|+|a_j|+|c_j|\geq |d_i|-|a_i|-|c_i|$.
So we have proven that $1\leq \dfrac{||A||_{\infty}}{K}$ and thus the first statement.
Is this solution correct?
