0
$\begingroup$

Let $U$ be a normed vector space with two norms: $ || . ||,|| . ||^{'} $

For every sequence $\{x_n \}$ that $||x_n-x||\rightarrow 0 $ & $||x_n-y||^{'}\rightarrow 0 $,we can conclude $x=y$.

Prove that these norms are equivalent. I know the definition of equivalency for norms,but i have no idea for this problem.

Help me with your hints please.

Thanks.

$\endgroup$
2
  • $\begingroup$ Are you sure you do not need $U$ to be complete w.r.t. both norms? $\endgroup$
    – user228113
    Dec 10, 2015 at 15:35
  • $\begingroup$ @G.Sassatelli it is not mentioned in the exercise $\endgroup$
    – user115608
    Dec 10, 2015 at 15:43

1 Answer 1

1
$\begingroup$

I think it's false. Let $U=L^2[0,1]\subseteq L^1[0,1]$.

$\|\bullet\|_2,\ \|\bullet\|_1$ are not equivalent on $L^2[0,1]$, since there exist sequences of functions $L^2[0,1]\ni f_n\stackrel{L^1}{\longrightarrow} f\in L^1[0,1]\setminus L^2[0,1]$.

More concisely, $(L^2[0,1],\|\bullet\|_2)$ is a Banach space, while $(L^2[0,1],\|\bullet\|_1)$ is not.

Nevertheless, let's show that $$(\|f_n-f\|_2\to 0\wedge \|f_n-g\|_1\to0)\implies f=g$$

Indeed $\|f_n-f\|_2\to 0$ implies the existence of $f_{n_k}-f\stackrel{a.e.}{\longrightarrow} 0$. Since $\|f_{n_k}-g\|_1\to 0$, there exists $f_{n_{k_h}}-g\stackrel{a.e.}{\longrightarrow} 0$. By construction, $f_{n_{k_h}}-f\stackrel{a.e.}{\longrightarrow} 0$ as well.

Therefore, since both $f$ and $g$ exist in $L^2[0,1]$, we can conclude that $f=g\ $ a.e., hence $f=g$.

$\endgroup$
2

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .