- #1

- 613

- 3

##||A||_{\infty} = max_i \sum_{j} |a_{ij}|##

which reads as the ##\infty## norm is the max row sum of matrix A.

##i## is the row index and ##j## is the column index.

Here is what I thought of:

##||A||_{\infty} = sup_{x\neq 0} \frac{||Ax||_{\infty}}{||x||_{\infty}}##

The numerator can be written as:

##||Ax||_{\infty} = max_i \sum_{j} |a_{ij} x_j| \leq max_i \sum_{j} |a_{ij}|\ |x_j| \leq max_j |x_j| * max_i \sum_{j} |a_{ij}| = ||x||_{\infty} * max_i \sum_{j} |a_{ij}|##

Therefore,

##||Ax||_{\infty} \leq ||x||_{\infty} * max_i \sum_{j} |a_{ij}|##

Next,

##\frac{||Ax||_{\infty}}{||x||_{\infty}} \leq max_i \sum_{j} |a_{ij}|##

Next,

the supremum is defined as the "least upper bound." Thus, the supremum of the above expression is simply the equal sign. Thus,

##||A||_{\infty} = max_i \sum_{j} |a_{ij}|##

Do you guys see any flaws in this proof? I saw various other proofs that went on to prove that the "<" condition cannot exist, which is still correct, but I don't understand what the point is when you know that the supremum is the "LEAST" upper bound.