# Operator norm of $T:l^{2}\rightarrow l^{1}$ where $Tx=(x_{1},x_{2}/2,x_{3}/3,x_{4}/4,…)$

As the title states, I need to compute the operator norm of a linear operator $$T:l^{2}\rightarrow l^{1}$$, where

$$Tx=\left(x_{1},\frac{x_{2}}{2},\frac{x_{3}}{3},\frac{x_{4}}{4},... \right)$$

Using Holder's inequality for any sequence $$(x_{i})_{i\geq 1}\in l^{2}$$, we can show

\begin{align} |Tx|_{1}&=\sum_{i=1}^{\infty}=|x_{1}|+\left|\frac{x_{2}}{2}\right|+\left|\frac{x_{3}}{3}\right|+\cdots\\ &=|x_{1}||1|+|x_{2}|\left|\frac{1}{2}\right|+|x_{3}|\left|\frac{1}{3}\right|+\cdots \\ &\leq \left|x_{i}\right|_{2}\left|\frac{1}{i}\right|_{2} \\ &=\frac{\pi}{\sqrt{6}}|(x_{i})|_{2} \end{align}

Hence

$$\displaystyle |Tx|_{1}\leq\frac{\pi}{\sqrt{6}}|(x_{i})|_{2}\implies||T||\leq\frac{\pi}{\sqrt{6}}$$

However, I am unable to find a sequence in $$l_{2}$$ which has norm $$|(x_{i})|\leq 1$$ so that I may use the property $$||T||=\text{sup}_{|(x_{i})|_{2}=1}|Tx|_{1}$$.

Any help is appreciated. Thank you.

• You can use \|T\| to get $\|T\|$. It's claimed that this is better than ||T||, which gives $||T||$. – DanielWainfleet Jan 11 at 6:16

Let $$x_i=\frac c i, i=1,2\cdots$$ where $$c$$ is such that $$c\sum\limits_{i=1}^{\infty} x_i^{2}=1$$. In other words, $$c=\frac {\sqrt 6} {\pi}$$. Then $$\|T(x_i)\|=\sum\limits_{i=1}^{\infty} \frac c {i^{2}}$$ which is exactly $$\frac {\pi} {\sqrt 6}$$.