# Significance of equivalent norms?

So two norms $|| \cdot ||_1$ and $|| \cdot ||_2$ are equivalent if $$\exists \ c \in \mathbb{R}$$ such that $\forall x \in X$ we have that $$\frac{1}{c}|| \cdot ||_1 \le || \cdot ||_2 \le c|| \cdot ||_1$$

What is the intuition I should be taking from this definition? Why is $|| \cdot ||_2$ equivalent to $|| \cdot ||_1$ just because it is within a certain range that is dependent on the value of $|| \cdot ||_1$?

• The two norms then induce the same topology. – Daniel Fischer Sep 21 '15 at 10:59

## 1 Answer

As Daniel Fischer said, they induce the same topology, but this might not be so intuitive.

What kind of understanding is directly linked to the concept of a norm?

Right, it's the principle of convergence. So now just observe, that equivalent norms - I know this definition which is slightly different to yours, but they are equivalent - $\left\|x\right\|_a,\left\|x\right\|_b$, i.e. $$\exists c,C>0 \text{ such that }\forall x\in V:c \left\|x\right\|_a\leq\left\|x\right\|_b\leq C\left\|x\right\|_a \tag 1$$ have the same behavior when it comes to convergence, so the property stays invariant under the change of equivalent norms. So if a sequence $(x_n)$ converges with respect to $\left\|x\right\|_a$ so it does also converge with respect to the equivalent norm $\left\|x\right\|_b$.

EDIT/REMARK(doesn't concern the actual question)

Why are both definitions equivalent? Clearly $$1/c \left\|x\right\|_a\leq\left\|x\right\|_b\leq c\left\|x\right\|_a\Rightarrow (1)$$ and the same with the other direction $$(1)\Rightarrow 1/c \left\|x\right\|_a\leq\left\|x\right\|_b\leq c\left\|x\right\|_a$$ since if we found a $C$ for which $\left\|x\right\|_b\leq C\left\|x\right\|_a$ holds, we can choose $C$ big enough until we found a $c:=1/C$ which satisfies $c \left\|x\right\|_a\leq\left\|x\right\|_b$. This is possible since $c:=1/C\to0$ for $C\to \infty$.