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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.

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

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  • $\begingroup$ Wow, sometimes when you look at a problem for so long continuously, you don't even realise the solution is right in front of you. Thank you so much! $\endgroup$ – Jack Jan 10 at 7:51

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