Computing the norm of the operator $T: (\Bbb R^n, \|\cdot\|_2)\to (\Bbb R^n, \|\cdot\|_1)$ with $Tx = (\delta^kx_k)_{k=1,...,n}$

Question

Let $0<\delta<1$ consider the linear operator

$$T: (\Bbb R^n, \|\cdot\|_2)\to (\Bbb R^n, \|\cdot\|_1)$$ such that for $x=(x_k)_{k=1,...,n}$ we have $$Tx = (\delta^kx_k)_{k=1,...,n}$$

recall that $$ \|x\|_2^2=\sum_{k=1}^n|x_k|^2$$

and $$ \|x\|_1=\sum_{k=1}^n|x_k|$$

what is the norm of the operator T.

I attempted this by using Cauchy-Schwarz inequality as follow: Let set $\delta=(\delta^k)_{k=1,...,n}$

$$ \|Tx\|_1=\sum_{k=1}^n|\delta^kx_k|\le \|x\|_2 \|\delta \|_2=\|x\|_2\sqrt{\sum_{k=1}^n\delta^{2k}}= \delta\|x\|_2\sqrt{\frac{1-\delta^{2n}}{1-\delta^{2}}}$$

Therefore, I can conclude that,

$$ \|T\|\le \delta\sqrt{\frac{1-\delta^{2n}}{1-\delta^{2}}}$$

I don't how to go further with the lower bound.

Is this is inequality sharp?

Does anyone has an idea how to prove wether this is inequality is sharp? namely how can find an element $x$ such that $ \|x\|_2=1$ and $$ \|Tx\|_1= \delta\sqrt{\frac{1-\delta^{2n}}{1-\delta^{2}}}$$

Answer 1

The Cauchy–Schwarz inequality is an equality when the two vectors are linearly dependent. So just try $x = (\delta, \delta^2, \dots, \delta^n)$. Let $C = \sqrt{\sum_{i=1}^n \delta^{2i}}$ be the number you found. You have: $$\|x\|_2 = \sqrt{\sum_{i=1}^n \delta^{2i}} = C.$$

Now $Tx = (\delta^2,\delta^4,\dots,\delta^{2k}$. Thus: $$\|Tx\|_1 = \sum_{i=1}^n \delta^{2i} = C \|x\|_2.$$

Since you already know that $\|T\| \le C$, it follows that $\|T\| = C$.