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Let $c$ be the vector space of all real convergent sequences, with the norm

$$\|(a_n)_{n\in \mathbb N}\|_\infty = \sup\{|a_n|:n\in \mathbb N\}$$

Let $c_0$ be the subspace of $c$ of sequences with limit $0$, with the standard $\sup$-norm.

Define $T:c \to c_0$ by

$$T((a_n))=(a,a_1-a,a_2-a,\dots)$$

where $a=\lim_{n \to \infty}a_n$.

Question: How do I prove that $T$ is a bijection (1-1 and onto), and also continuous?

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    Either $T ((a_n)) = (a, a_0-a, a_1-a, \dots) $ or your sequences are indexed on $\mathbb N ^*$, otherwise T is not a bijection. – 2017-01-21
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    For continuity, if $a_n\to a$ does there exist any relation between $\|a_n\|_\infty$ and $|a|$? $T(a_n)$ has norm $≤\|a_n\|_\infty +|a|$, can you use the previous result to get a Lipschitz constant? For injectivity, if you suppose two sequences are sent to the same vector, what follows for the limit of the sequences? Can you see how this implies all terms of the sequences must be the same (start with the first term)? – 2017-01-21

1 Answers 1

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Presumably you are using $\mathbb N = \{1,2,3,\dots\}$.

The first thing is to prove that the function $(l,a_1-l,a_2-l,\dots)$ actually converges to $0$. This is easy.

Now prove it is injective. suppose $f((a_n))=f((b_n))$. Then we must have that both limits $l_a$ and $l_b$ are equal, we also have $l_a-a_k=l_b-b_k$ for all $j\in \mathbb N$. Since $l_a=l_b$ we have $a_k=b_k$ for all $k$. So the two sequences are equal.

To prove it is surjective. suppose you have a sequence $(a_1,a_2,\dots )$ that converges to $0$. Then the sequence $(a_1+a_2,a_1+a_3,a_1+a_4,\dots)$ converges to $a_1$ (because $a_n$ converges to $0$).

It follows that $f(a_1+a_2,a_1+a_3,\dots)=(a_1,a_2,a_3,\dots)$. So it is surjective.