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.
Best Answer
Hint: Define $\Vert \dot\, \Vert_3 := \Vert \dot\, \Vert_1 + \Vert \dot\, \Vert_2$. Show that $\Vert \dot\, \Vert_3$ is a complete norm on $E$. Now use the fact, that the map $f: (E, \Vert \dot\, \Vert_3) \to (E,\Vert \dot\, \Vert_1)$ with $f(x) = x$ for all $x\in E$ is a continuous bijection between Banach spaces.