## § Bounding L2 norm by L1 norm and vice versa

- We can bound a function along the x-axis (in its domain) or along the y axis (in its range).

#### § Bounded domain

- If a function's norm is well defined in a bounded domain, then it has not increased too rapidly.
- Intuitively, with L2 norm, large numbers become larger, thus it is "harder" for a function to stay finite.
- Thus, in bounded domain, L2 is a subset of L1.

#### § Bounded range

- If a bounded function's norm is well defined on an unbounded domain, then it has vanished to zero sufficiently quickly.
- Intuitively, with L2 norm, smaller numbers becomes smaller, thus L2 allows functions to decay faster (eg. $|1/n|_1 = \sum_k 1/k = \infty$ versus $|1/n|_2 = \sqrt{\sum_k 1/k^2 < \infty}$.
- Thus, in bounded range, L1 is a subset of L2, because L2 allows more functions to decay to zero.

- intuitively, l2 error will fail to reduce small values.
- but l1 error will reduce all values equally.
- thus, l1 norm is larger than l2 norm.