# What is the operator norm of the following?

This is indeed meant to be a homework question but I just figured out the answer I prepared was false all this time. Kind of panicking, could anyone give me a lift?

$T:\ell^2 \rightarrow \ell^1: Tx = \big( (2^{-k})x_k \big)_{n=1}^{\infty}$

Thanks!

Edit: Here are my thoughts:

Observe that for all $x \in \ell^2$ in the unit ball we can attend the bounds $||Tx||_{2} = \frac{1}{2}, \frac{1}{4}, \frac{1}{8},...$ Another such bound can be achieved by the series $(\frac{1}{2^{k/2}})_{j \in \mathbb{N}}$ which goes to $1$. However looking for an upper bound we have

$$|| Tx ||_{1} = \sum_{k=1}^{\infty} | \frac{x_k}{2^k} |$$ which I think naturally should be applied Holder's inequality. I have tried various but non of them would coincide with the lower bounds I gave above.The best I could do was $1/\sqrt{3}$ for which I cannot get a lower bound for.

• No. show your thoughts – mathworker21 Mar 28 '18 at 8:42
• Attached are my thoughts. @mathworker21 I made a small mistake for which I thought the bounds would coincide some days ago and I just noticed its wrong in the very last seconds before it is due lol – Meagain Mar 28 '18 at 8:55
• see my answer below – Guy Fsone Mar 29 '18 at 14:14

About Hölder and your upper bound, OK: $$\|Tx\|_1 = \sum_{k=1}^\infty\left|\frac{x_k}{2^k}\right|\le \left(\sum_{k=1}^\infty 2^{-2k}\right)^{1/2} \left(\sum_{k=1}^\infty x_k^2\right)^{1/2} = \frac1{\sqrt 3}\|x\|_2.$$ About finding $x$ with $\|Tx\|_1\approx\frac1{\sqrt 3}\|x\|_2$ (equality can be impossible), see the conditions for equality in the Hölder inequality.

• Never remembered the equality condition. Learned simething new, quiet happy, thanks! – Meagain Mar 28 '18 at 11:17
• @Meagain you check the answer below – Guy Fsone Mar 29 '18 at 14:14
• @GuyFsone, is OK. Equality iff linear dependence. – Martín-Blas Pérez Pinilla Mar 29 '18 at 14:27

Using Cauchy-Schwarz inequality we have

$$\|Tx\|_1=\sum_{k=1}^\infty\left|2^{-k}x_k\right|\leq\left(\sum_{k=1}^\infty 4^{-k}\right)^{1/2}\cdot \left(\sum_{k=1}^\infty|x_k|^2\right)^{1/2}\ = \frac1{\sqrt3}\|x\|_2.$$

Hence $$\|T\| = \sup_{\|x\|_2=1}\|Tx\|_1\le \frac{1}{\sqrt 3}$$

Now let $x = (x_k=\frac{1}{2^k})_k$. then $$\|x\|_2= \left(\sum_{k=1}^\infty 4^{-k}\right)^{1/2} = \frac{1}{\sqrt 3}$$

and $Tx = (\frac{1}{4^k})_k$ $$\|Tx\|_1= \sum_{k=1}^\infty 4^{-k}=\left(\sum_{k=1}^\infty 4^{-k}\right)^{1/2} \left(\sum_{k=1}^\infty 4^{-k}\right)^{1/2} = \frac{1}{\sqrt 3}\|x\|_2$$

It readily follows that $$\|Tx\| = \frac{1}{\sqrt 3}$$