Notion of convergence of equivalent norms is the same

I would like to make clear the proof for the following theorem which states that two norms over a vector space are equivalent iff their notion of convergence is the same.

I have an hint for the proof from the ($\Leftarrow$):

1)Define $F:X\rightarrow \mathbb{R},f\mapsto \lVert f\rVert_b$

2)Prove that the function $F$ Is $\lVert .\rVert _a$-continuous

3)note that $\exists r>0: B_r^b(0)\subseteq F^{-1}((-1,1))=B_1^a(0)$

And the result follows. The continuity of this function is a bit unclear. I tried the epsilon-delta criterion but I'm unsure.

Take $\{f_n\}_n \subset X$ such that $f_n\rightarrow f\in X$ So $\forall n>N\lVert f_n -f\rVert_a<\delta$ We have $\lvert \lVert f_n\rVert_b -\lVert f\rVert_b \rvert\leq \lVert f_n - f\rVert_b$ But then i don't know how to continue.

3) I really don't see why these two set are equal.

Thank you for any help

For (1), You are on the right way. To prove continuity of $F$, for $f_n \to f$, we only need to prove that $$\lim_{n\to \infty}F(f_n)=F(f),$$ i.e., for any $\epsilon >0$, there exists $N>0$ such that $n>N$, $|F(f_n)-F(f)|=\big|||f_n||_b-||f||_b\big|<\epsilon$.

I'm sure you know how to proceed futher from here.

• Ok. If X is finite dimensional then I can write the elements as a finite linear combination of the basis elements and so I would have continuity. Do you have some advice for point 3)? – sky90 Nov 3 '14 at 14:51

I present my proof that would be of no use to you, but is trivial for anyone who knows topological vector spaces (TVS). ;)

Normed space is a sequential TVS. Because two norms are sequentialy equivalent, and topologies induced by them are sequential, we have that this in fact is the same topology, name it $\tau$

Open unit balls $B_1, B_2$ in respective norms are open, bounded, absorbing neighbourhoods of $0$ in topology $\tau$. Hence, there exists $m,n>0$ such that $B_1 \subset m B_2$ and $B_2 \subset n B_1$. So we get:

$$\frac 1n B_2 \subset B_1 \subset m B_2$$

But interpreting what this means for norms (and maybe adding some justification in terms of Minkowski functionals) we get:

$$\frac 1n \| x \|_2 \leq \|x\|_1 \leq m \|x\|_2$$