0
$\begingroup$

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 $|| \cdot ||_1$?

$\endgroup$
1
  • 3
    $\begingroup$ The two norms then induce the same topology. $\endgroup$ – Daniel Fischer Sep 21 '15 at 10:59
1
$\begingroup$

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 $C\to \infty$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.