Operator norm of $T:l^2 \to l^1$ I need to find the norm of an operator from $l^2 \to l^1$, but I'm struggling because of the different norms on $l^2$ and $l^1$. 
The operator is defined by $T:l^2 \to l^1, x_i \mapsto 2^{-i}x_i$. 
Using the canonical basis, I have that $||T||\geq 1/2$, but I have a feeling this is not a very good lower bound. I also cant seem to find any upper bound, because I have that 
$$||Tx||_1 = \sum_{i=1}^{\infty}2^{-i}x_i$$ but I can't relate this to $||x||_2$ because $||x||_2= (\sum_i^\infty x_i^2)^{1/2}$. 
Thanks for any help!
 A: Using the Cauchy-Schwarz inequality you get
$$
\|Tx\|_1 = \sum_{i=1}^\infty 2^{-i} |x_i| \leq 
\left(\sum_{i=1}^\infty 4^{-i}\right)^{1/2}
\left(\sum_{i=1}^\infty|x_i|^2\right)^{1/2}
= \frac{1}{\sqrt{3}} \|x\|_2.
$$
On the other hand, if you choose $x=(x_i)$ with $x_i = 2^{-i}$, you check in a moment that you get equality in the above inequality, hence $\|T\| = 1/\sqrt{3}$.
Edit: same answer given in comments by acetone.
A: By Cauchy-Schwartz
$$\Big(\sum_{i=1}^{n}\frac{|x_i|}{2^i}\Big)^2\leqslant\Big(\sum_{i=1}^{n}\frac{1}{2^{2i}}\Big)\Big(\sum_{i=1}^nx^2_i\Big)$$ 
Taking limit $n\to\infty$
$$\lim_n\Big(\sum_{i=1}^{n}\frac{|x_i|}{2^i}\Big)^2\leqslant\lim_n\Big(\sum_{i=1}^{n}\frac{1}{2^{2i}}\Big)\lim_n\Big(\sum_{i=1}^nx^2_i\Big)$$
which yields
$$||Tx||_1^2=\Big(\sum_{i=1}^{\infty}\frac{|x_i|}{2^i}\Big)^2\leqslant\Big(\sum_{i=1}^{\infty}\frac{1}{2^{2i}}\Big)\Big(\sum_{i=1}^{\infty}x^2_i\Big)\Rightarrow ||Tx||_1^2\leqslant C||x||^2_2$$
with $$C:=\sum_{i=1}^{\infty}\frac{1}{2^{2i}}=\frac{1}{4}\cdot\frac{1}{1-\frac{1}{4}}=\frac{1}{3}$$
Therefore 
$$||T||:=\sup_{||x||_2\leqslant 1}||Tx||_1\leqslant \frac{\sqrt{3}}{3}$$
We need to show that $$||T||\geqslant\frac{\sqrt{3}}{3}$$
You can pick $x^*$ as given by @user545497 in the comments above with $x^*_k:=1/2^k$ then $x^*_k\in\mathcal{l}^2(\mathbb{N})$ and $||x^*||_2<1$. Also $||Tx^*||_1=C||x^*||_2$. Thus $||T||=\sqrt{3}/3$.
