Is it true that the infinity norm of the matrix exponential $\|e^{At}\|_{l^\infty} \leq 1$ if $A$ is a negative diagonally dominant matrix?

Question

Assume $A = (a_{ij}) \in \mathbb{R}^{N\times N}$ is negative diagonally dominant matrix, i.e. $|a_{ii}| \geq \sum_{j = 1, j\neq i}^{N} |a_{ij}|$ with $ a_{ii} < 0, 1 \leq i \leq N$.

For example: \begin{equation} A = \left[ \begin{array}{ccccc} -2 & 1 & & & 1 \\ 1 & -2 & 1 & & \\ & \cdots & \cdots & \cdots & \\ & & 1 & -2 & 1\\ 1& & & 1 & -2 \\ \end{array} \right]_{N\times N} \end{equation} Is it true that the infinity norm of the matrix exponential $\|\mathrm{e}^{A t}\|_{l^\infty} \leq 1, \forall t \geq 0$?

EDIT 1 The $l^\infty$ norm of a matrix $B = (b_{ij}) \in \mathbb{R}^{N\times N}$ is given by \begin{align*} \|B\|_{l^\infty} = \mathop{max}_{i = 1, \cdots, N}\{\sum_{j = 1}^{N} |b_{ij}| \}. \end{align*} How to prove it?

Lemma 3.1 of (Du Qiang, et al., 2019, MAXIMUM PRINCIPLE PRESERVING EXPONENTIAL TIME DIFFERENCING SCHEMES FOR THE NONLOCAL ALLEN-CAHN EQUATION)[https://arxiv.org/pdf/1902.04998.pdf] shows that (I think there is a typo in the stricit diagonally dominant condition, $j \neq i$ is missing)

When $\kappa = 0$ is it still true that $\|\mathrm{e}^{A t}\|_{l^\infty} \leq 1$?

EDIT 2 I wrote a short matlab code to verify the inequality

N = 10;
for i = 1:100000
    A0 = 2*rand(N, N) - 1; % random value in [-1, 1]
    A = A0 + A0'; % construct symmetric matrix;
    v = -(sum(abs(A), 2) - abs(diag(A)));
    for i = 1:N
        A(i,i) = v(i); % Assign v to the diagonal elements
    end
    tmp = norm(expm(A), inf);
    if  tmp > 1
        tmp
    end
end

Thank you very much!

Answer 1

By a continuity argument, we may assume that $A$ is a strictly diagonally dominant matrix whose diagonal entries are negative. Let $t>0$ and $h=\frac{t}{m}$. When $m\ge1$ is sufficiently large, we have $$ \|e^{hA}\|_\infty =\|I+hA+o(h)\|_\infty =\max_i\left[1+h\left(a_{ii}+\sum_{j\ne i}|a_{ij}|\right)+o(h)\right]<1. $$ Therefore $\|e^{tA}\|_\infty=\|(e^{hA})^m\|_\infty\le\|e^{hA}\|_\infty^m\le1$.

Remark. I don't think the proof of lemma 3.1 in the cited paper is valid. The authors (Qiang et al.) tried to treat their lemma as a special case of theorem 2(b) of Lazer, Characteristic Exponents and Diagonally Dominant Linear Differential Systems, J. Math. Ana. and Appl. 35, 215-229 (1971). However, the matrix norm that Lazer used is $\|M\|=\max_j\|M_{\ast j}\|_2$ (the maximum Euclidean norm among the columns of $M$), not the induced $\infty$-norm.