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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.

molan
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1 Answers1

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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.

Martin Sleziak
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Alexander Thumm
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    HINT : and you should use the open mapping theorem to conclude for $f$. – JBC Jun 18 '12 at 08:25
  • Nice. BTW, I use the Open Mapping Theorem to show that $f^{-1}$ is also continuous. Then there exists $C>0$ such that $|x|_3\leq C|x|_1$ for all $x\in E$. Then $|x|_2\leq (C-1)|x|_1$ for all $x\in E$. Right? Thank you so much. – molan Jun 18 '12 at 08:29
  • Yes. (some more characters) – Alexander Thumm Jun 18 '12 at 09:11