Stronger Norm

Last updated Nov 1, 2022

Let $(X, ||\cdot||)$ be a Normed Vector Subspace and let $Y \subset X$ be a Vector Subspace equipped with Norm $|||\cdot|||$. We say $||\cdot||$ is a Stronger Norm than $|||\cdot|||$ if there exists a $C > 0$ so that $$|||y||| \leq C||y|| \text{ } \forall y \in Y$$

1. Stronger Norm defines a finer Topology