Show that if the norms ∥⋅∥1 and ∥⋅∥2 are equivalent, then if space (X,∥⋅∥1) is complete, then the space (X,∥⋅∥2) is not necessary complete.

Question

X is complete if Cauchy sequence converges to it's limit in X ∀ε>0 ∃N: ∀n>N∶ ∥xn-x∥1<ε So let (X,∥⋅∥1) – separable and xn - Cauchy sequence in (X,∥⋅∥1) and xn→x ∀ε>0 ∃N: ∀n>N∶ ∥xn-x∥1<ε We have equivalent norms, which means: ∃a,b: a*∥⋅∥2≤∥⋅∥1≤b*∥⋅∥2 a*∥xn-x∥2≤∥xn-x∥1≤ε ∥xn-x∥2≤ε/a That means (X,∥⋅∥2) is complete, isn’t it? How can I prove it is not necessary complete?

Answer 1

Let $a,b>0$ such that for all $x\in X$ it is $\|x\|_1\leq a\|x\|_2$ and $\|x\|_2\leq b\|x\|_1$. Suppose that $(X,\|\cdot\|_1)$ is complete. We will show that $(X,\|\cdot\|_2)$ is also complete.

If $(x_n)\subset X$ is a Cauchy sequence for $(X,\|\cdot\|_2)$, then for any $\varepsilon>0$ there exists $n_0$ such that for all $m,n\geq n_0$ it is $\|x_n-x_m\|_2\leq\varepsilon$. So, if $\varepsilon>0$, consider $\varepsilon/a>0$ and find $n_0$ such that for all $n,m\geq n_0$ it is $\|x_n-x_m\|_2<\varepsilon/a$. Then for such $n,m$ it is $\|x_n-x_m\|_1<\varepsilon$, so $(x_n)$ is Cauchy for $(X,\|\cdot\|_1)$. This space is complete, so $x_n\to x$ with respect to $\|\cdot\|_1$. But then $0\leq\|x_n-x\|_2\leq b\|x_n-x\|_1\to0$, so $x_n\to x$ with respect to $\|\cdot\|_2$, so $(x_n)$ is convergent in $(X,\|\cdot\|_2)$ and we are done.

A comment: keeping things neat helps out!

Answer 2

Suppose $(X,\|\cdot\|_1)$ is complete. Let $(x_n)$ be a Cauchy sequence in $(X,\|\cdot\|_2)$. In order to prove that $(X,\|\cdot\|_2)$ is complete, you want to find an $x\in X$ such that $\|x_n-x\|_2\to 0$. To do so, consider the following:

• Prove that $(x_n)$ is also Cauchy in $\|\cdot\|_1$, by using that the norms are equivalent.
• Since $(X,\|\cdot\|_1)$ is complete, there exists an $x\in X$ such that $\|x_n-x\|_1\to 0$. This is our candidate limit.
• Prove that $\|x_n-x\|_2\to 0$ using that the norms are equivalent.