Suppose we have a banach space $X$ with two norms $||\cdot||_1, ||\cdot||_2$. Let $B_1 (r) \equiv \{ x \in X : ||x||_1 \leq r \}$ and similarly for $B_2$.
If one is given the inclusion $(\star): B_1(r) \subseteq B_2(s) $, then one can prove that $||x||_2 \leq s/r ||x||_1$:
- Let $x \in X$. Consider $\hat x \equiv r x / ||x||_1$. Since $\hat x \in B_1$ by construction, $\hat x \in B_2$ by $(\star)$. Thus $||\hat x||_2 \leq s$. Substituting $\hat x$, $|| r x / ||x||_1 ||_2 \leq s$, which on rearrangement gives $||x ||_2 \leq s/r ||x||_1$.
Now I am confused, because I would have expected that depending on the values of $s, r$, I should be able to either lower or upper bound $||x||_2$ by $||x||_1$. But it appears that regardless of the values of $s, r$, $||x||_2$ is always upper bounded by $||x||_1$. Isn't this strange?
Why is the fact that a unit ball $B_1(1)$, arbitrarily rescaled by $r$ getting trapped in the ball $B_2(s)$ allow us to upper bound $||x||_2$ by $||x||_1$?
Intuitively, I do understand what happens if both balls are unit radius. Then the containment of the balls tells us that one of the norms increases faster, and thus its unit ball is smaller. But in this case, we are comparing balls of different radii ($r$, $s$), so I don't see how this analysis applies any more.
Help with intuition on how to think about this scenario would be very appreciated.
You received an answer yesterday in the case of unit balls. Now, simply realize that the ball (centered at $0$) of radius $r$ for some norm $N$ is nothing but the unit ball for the norm $N/r.$