0
$\begingroup$

Let's define $B_s$ as the of real valued sequences $(x_n)$, such that $sup_{N\in \mathbb N} |\sum_{k=0}^{N}{x_k}| $ is bounded, and make it a vector space considering the usual pointwise operations on sequences and scalars. Then define a norm by $||(x_n)||=sup_{N\in \mathbb N} |\sum_{k=0}^{N}{x_k}| $.

Clearly this space contains properly the set $C_s$ that consist of convergent series( for example $x_n = (-1)^n$).

I want to prove that:

i) $B_s$ is a banach space

$ii)$ $C_s$ is a closed subspace of $B_s$.

I tried by finding explicitly the limit of a Cauchy sequence, I think that if I have a Cauchy Sequence $(x_n)_n$ in $B_s$ given by $x_n=(x_{nk})_k$ , then the limit it's given by the pointwise limit ( I actually proved that the pointwise limit exist) But I can't prove convergence. Please I really need help with this problem )=

$\endgroup$
1
$\begingroup$

It's probably easier if you consider the space $\ell^\infty$ of all bounded (real) sequences and its subspace $c$ of convergent sequences. On $\ell^\infty$, you take the norm $\lVert y\rVert_\infty = \sup \lvert y_k\rvert$.

Then $I \colon \ell^\infty \to B_s$,

$$I(y) = (y_0, y_1 - y_0, y_2 - y_2, \dotsc)$$

is an isometric isomorphism.

$\endgroup$
  • $\begingroup$ Please could you write more explicitly the map $I$ I don't understand )= $\endgroup$ – Shanks Sep 12 '13 at 0:32
  • $\begingroup$ Let $x = I(y)$. Then $x_0 = y_0$, and for $k > 0$, we have $x_k = y_k - y_{k-1}$. The inverse is given by $S \colon B_s \to \ell^\infty$, with $S(x) = (x_0, x_0+x_1,\dotsc)$, or, if $y = S(x)$, then $y_k = \sum\limits_{j=0}^k x_j$. $\endgroup$ – Daniel Fischer Sep 12 '13 at 0:40
  • $\begingroup$ Your solution it's very elegant! $\endgroup$ – Shanks Sep 12 '13 at 0:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.