Continuous mapping

Question

It's given mapping
$T:l^1\rightarrow l^{\infty}$, $T(\langle x_1,x_2,x_3,...\rangle):=\langle \sum^\infty_{j=1}x_j,\sum^\infty_{j=2}x_j,\sum^\infty_{j=3}x_j,...\rangle)$
where
$l^{\infty}$ is space of bounded sequences $x=\langle x_n\rangle $ with norm $||x||_{\infty}=||\langle x_n\rangle||_{\infty}=\sup |x_n|$
$l^1$ is space of absolute summable sequences $x=\langle x_n\rangle$ with norm $||x||=||\langle x_n\rangle||=\sum^\infty_{n=1} |x_n|$.

How to show that this mapping is continuous?

Note $T$ is linear. Can you show that $T$ is bounded? It only requires the definitions... — David Mitra, Jun 14 '13 at 10:22
$T$ is bounded if $\exists M>0$ such that $|\sum^\infty_{j=k} x_j|<M, k=1,2,...$ ? — user23709, Jun 14 '13 at 10:37
Yes, if that holds for all $x\in\ell_1$ with $\Vert x\Vert_{\ell_1}\le 1$. (You don't need strict inequality.) — David Mitra, Jun 14 '13 at 10:45

score 1 · Answer 1 · answered Jun 14 '13 at 11:08

Hint (This is what David Mitra meant)

Given Banach spaces $X$ and $Y$, a linear operator $T:X\to Y$ is bounded provided there is a constant $M$ such that $$\|Tx\|_Y\leq M\|x\|_X$$ for all $x\in X$.

Thm A linear operator $T:X\to Y$ is continuous if and only if it is bounded.

Proof of the part we need

-Try to prove the converse by using that $T$ is continuous at $x=0$.

Now back to the operator, can you find a number $M$ that bounds $T$?

Continuous mapping

1 Answers1