# Equivalency of two norms

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.

Thanks.

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

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 $f=g$.

• ok thanx a lot but if u assume the completeness , how do u prove the statement? Dec 10, 2015 at 15:58
• would u pls have a look at math.stackexchange.com/questions/1569194/…? Dec 10, 2015 at 16:17