Let $M_{n}(\mathbb{C})$ denote the space of complex $n \times n$ matrices and, for $a>0$, $a \in \mathbb{R}$ fixed, let $A: [0,a) \to M_{n}(\mathbb{C})$ be a given function. I will write $A(t) = (A_{i,j}(t))$ the matrix. Suppose further that each entry $A_{ij}(t)$ is $L^{1}$ in $[0,a)$. Define: $$\|A\|_{\infty} = \sup_{i}\int_{0}^{a}dt \sum_{j=1}^{n}|A_{ij}(t)|.$$
We can discretize the interval $[0,a)$ by fixing $N > 1$ a natural number and setting: $$I_{N} = \{t_{m} = \frac{ma}{N}: m=1,...,N\}.$$ The spacing is $a/N$. In this case, I define: $$\|A\|_{N} = \sup_{i}\frac{a}{N}\sum_{m=1}^{N}\sum_{j=1}^{n}|A_{ij}(t_{m})|.$$ In other words, I replaced the integral over $t$ by its Riemann sum.
My question is: does it always hold that: $$\|A\|_{N} \le \|A\|_{\infty}$$ for every $N$? And if not, what conditions should $A$ have for this inequality to be true?
