# Find the norm of an $l_2 \to l_2$ operator

I am struggling to find the norm of the following linear operator: $$A:l_2 \rightarrow l_2, \ A(x_1,x_2,...,x_n,...) = (x_1, \frac{1}{2}x_2,...,\frac{1}{2^{n-1}}x_n,...)$$

I considered $$||Ax||^2_{l_2} = \sum_{k=1}^{\infty} |\frac{1}{2^{k-1}} x_k|^2 = \sum_{k=1}^{\infty} (\frac{1}{2^{k-1}})^2| x_k|^2, \tag{1}$$ but I don't know what to do next. Bounding $$(\frac{1}{2^{k-1}})^2$$ above with 1 we get $$||Ax||^2_{l_2} \leq \sum_{k=1}^{\infty} |x_k|^2 = ||x||^2_{l_2},$$ which means that $$||A|| \leq 1$$, but this is too crude because it's easy to see that if we consider $$x_0 = (\frac{1}{2}, \frac{1}{4}, ..., \frac{1}{2^n})$$ we get $$||Ax_0||_{l_2}=\frac{2}{3}|x_0|_{l_2},$$ suggesting that $$||A|| \leq \frac{2}{3}$$. But I'm missing the way to derive that from (1). Any tips?

$$\|Ax_0\|=\frac 2 3 \|x_0\|$$ does not tell you that $$\|A\| \leq \frac 2 3$$. (We may even have $$\|Ax_0\|=0(\|x_0\|)$$ for some non-zero $$x_0$$ and we cannot conclude that $$\|A\| \leq 0$$ right?)
In general if $$A(x_n)=(a_nx_n)$$ then $$\|Ax\| \leq \sup_n |a_n|\|x\|$$ so $$\|A\|\leq \sup_n |a_n|$$. Also $$Ae_n=a_ne_n$$ so $$\|A\| \geq |a_n|$$ for each $$n$$. Hence $$\|A\| \geq \sup_n |a_n|$$ proving that $$\|A\|=\sup_n |a_n|$$. In the present case the norm is $$1$$.
• I managed to show that $||A|| \leq \frac{2}{\sqrt{3}}$, but plugging in $x=a$ yields $||Ax|| = \sqrt\frac{16}{15} = \frac{4}{\sqrt15} = \frac{2}{\sqrt3} * \frac{2}{\sqrt5} \neq \frac{2}{\sqrt3}||x||.$ Am I missing something?
Take $$e_1=(1,0,0,0,...)$$. Clearly, $$\lVert e_1\rVert = 1$$, and $$Ae_1=e_1$$, so...?