Bijective Linear Map Construction for continuous and bounded maps: Linear Operator $A$ and $A^{-1}$

Question

Can any one construct a bijective linear map, with $C(E)\simeq C_0(E)$?

That is, consider a linear operator $A$, such that $A\colon C(E)\rightarrow C_0(E)$, which can make sense for both $A$ and its inverse $A^{-1}$ are bounded.

Note:

• E be Banach space
• $C(E)$ be the space of convergent sequences in $E$
• $C_0(E)$ be space of null sequences in $E$

I can not find any specific example about that. It's about the subject of functional analysis. Please provide with more specific explanation. Thanks!

Answer 1

Define $T: C(E)\to C_0(E)$ by $T(a_n)=(\lim a_n,a_1-\lim a_n, a_2-\lim a_n,\cdots)$. It is quite easy to see that this map is linear and injective. Given $c_n \to 0$ let $a_1=c_1+c_2, a_2=c_1+c_3,a_3=c_1+c_4,\cdots$. Then $T(a_n)=(c_n)$. This defines $T^{-1}$ explicitly. Now you can verify that $T$ and $T^{-1}$ are continuous.