I'm working on an exercise from functional analysis.
Let $E$ be a vector space and $\|\cdot\|_1$ and $\|\cdot\|_2$ be two complete norms on $E$. Now suppose that $E$ satisfies the following property:
$\bullet$ if $(x_n)$ is a sequence in $E$ and $x,y\in E$ such that $\|x_n-x\|_1\to 0$ and $\|x_n-y\|_2\to 0$, then $x=y$.
Now we want to show that the norms $\|\cdot\|_1$ and $\|\cdot\|_2$ are equivalent.
My idea is as follows: If for any $n>0$, there is an element $x_n\in E$ such that $\|x_n\|_1>n\|x_n\|_2$. Then consider $(\frac{x_n}{\|x_n\|_1})_{n\geq 1}$. Clearly, $(\frac{x_n}{\|x_n\|_1})_{n\geq 1}$ converges to $0$. However, I cann't get a contradicition from this. Maybe my idea is wrong.
In fact, I even don't konw how to show that a Cauchy sequence in norm $\|\cdot\|_1$ is also a Cauchy sequence in norm $\|\cdot\|_2$.
Anyone can give me some hints or a counter example? Thank you very much.