Show that the sum is the norm of $l_1$

Question

I have the space $\mathbb E = l_1$ and $Tx = (x_1, 0, x_3, 0, ...).$ I need to show that $T$ is continuous.

The norm of $l_1$ is $$\Vert x \Vert = \sum_{k=1}^∞ \vert x_k \vert.$$

My solution:

$\Vert Tx \Vert = \sum_{k=1}^∞ \vert (Tx)_k \vert = \vert 0 \vert + \sum_{k=0}^∞ \vert x_{2k+1} \vert $.

Question:

How to continue this equality to get $\sum_{k=1}^\infty \vert x_k \vert$ (the norm)?

Answer 1

It is :

$$\|Tx\| = \|(x_1,0,x_3,0,\dots)\| = \sum_{k=1}^\infty|x_{2k+1}| \leq \sum_{k=1}^\infty|x_k| = \|x\|.$$

since the absolute sum of the odd entries of the sequence is definitely smaller or equal to the total sum of absolute values of all entries.

Thus :

$$\|Tx\| \leq \|x\| \implies \|T\| \leq 1$$

Now, if you take the sequence $x = (x_n) \in \ell^1$ with $x = (1,0,\dots,0,\dots)$, then :

$$\|Tx\| = \|(1,0,\dots,0,\dots)\| = 1 \implies \|T\| = 1$$

Thus the operator $T$ is continuous.