Let $A$ be a $n \times n$ square matrix, and
$$\|A\|_1 := \max_{j=1}^{n} \sum_{i=1}^{n} |A_{i,j}|$$
and
$$\|A\|_{\max} := \max_{i,j} |A_{i,j}|$$
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Does $\|A\|_1 \le 1$ imply that $\|A\|_{\max} \le 1$?
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Is there any relation between $\|A\|_1$ and $\|A\|_{\max}$?
Best Answer
Since $$|A_{i,j}|\leq \max_{1\leq i\leq n}|A_{i,j}|,$$ for each $1\leq j \leq n$, then $$\sum_{i=1}^{n}|A_{i,j}|\leq \sum_{i=1}^{n}\max_{1\leq i\leq n}|A_{i,j}|= n \max_{1\leq i\leq n}|A_{i,j}|$$ Therefore
$$ \max_{1\leq j\leq n}\sum_{i=1}^{n} |A_{i,j}|\leq n\max_{1\leq i,j\leq n} |A_{i,j}|.$$ That is
$$ \|A\|_{1}\leq n \|A\|_{\infty}.\qquad (1)$$
The estimate (1) is sharp. Meaning the factor $n$ cannot be replaced by a smaller positive constant. To see this consider the matrix $A$ with $A_{i,j}=1$ for which (1) is actually an equality.
Now, does $\|A\|_{1}$ control $\|A\|_{\infty}$ ?
We have that $$\max_{1\leq i\leq n} |A_{i,j}|\leq \sum_{i=1}^{n} |A_{i,j}|\Longrightarrow \max_{1\leq j\leq n}\max_{1\leq i\leq n} |A_{i,j}|\leq \max_{1\leq j\leq n}\sum_{i=1}^{n} |A_{i,j}|$$ So, it is true that
$$ \|A\|_{\infty}\leq \|A\|_{1}\qquad (2)$$
Again the constant one in (2) is sharp. When $A=I_{n}$, the unit matrix, we have an equality in (2).