Is it true that
$$\max_{1\le i\le n} \sum_{j=1}^n |a_{ij}| \to \sup_{i \in \mathbb{N} } \sum_{j=1}^\infty |a_{ij}|, \quad n \to \infty$$
Where the $a_{ij}$ are the coefficients of an infinite matrix.
Tried to find a counterexample but didnt succeed, so I tried proving which didn't work that well. So I mainly would like to know if it's true and get a hint to prove it.
Yes, it's true. First we observe that $$\max_{1\le i\le n} \sum_{j=1}^n |a_{ij}|\le\max_{1\le i\le n} \sum_{j=1}^\infty |a_{ij}|\le\sup_{i \in \mathbb{N} } \sum_{j=1}^\infty |a_{ij}|$$ If for some $\xi$ the sum $\sum_{j=1}^\infty|a_{\xi j}|$ diverges, then the proposition is obviously true since for $n\ge \xi$ we have $$\max_{1\le i\le n} \sum_{j=1}^n |a_{ij}|\ge\sum_{j=1}^n|a_{\xi j}|$$ which tends to infinity as $n\to\infty$. Otherwise all $\sum_{j=1}^\infty |a_{ij}|<\infty$, and for a fixed $\zeta$ and sufficiently large $N$ we have $\sum_{j=N}^\infty|a_{\zeta j}|<\varepsilon$, which leads to $$\sum_{j=1}^\infty|a_{\zeta j}|\le\sum_{j=1}^N |a_{\zeta j}|+\varepsilon\le\max_{1\le i\le N} \sum_{j=1}^N |a_{ij}|+\varepsilon\le\lim_{n\to\infty}\max_{1\le i\le n} \sum_{j=1}^n |a_{ij}|+\varepsilon$$ Since $\varepsilon$ is arbitrary, we see $$\sum_{j=1}^\infty|a_{\zeta j}|\le\lim_{n\to\infty}\max_{1\le i\le n} \sum_{j=1}^n |a_{ij}|,~\text{ for any }\zeta$$ Which concludes the proof along with the first inequality.