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Let $V$ be the Banach space of all sequences $v=\{\eta_j\}_{j=1}^\infty$ such that $\lim_{j\rightarrow\infty}\eta_j$ exists. The norm on $V$ is given by $\|v\|=\sup_{j\in \mathbb{N}}\eta_j$.

Consider the operator $T:l^2\rightarrow V$ defined by

$Tu = (\xi_1,\xi_1+\frac{1}{2}\xi_2,\xi_1+\frac{1}{2}\xi_2+\frac{1}{3}\xi_3,...) $

$u=\{\xi_j\}\in l^2$

where $l^2$ is the sequence space "little $l^2$"

(a) Show that T is bounded and find its norm.

I am currently stuck showing that it is bounded.

So far I see that:

$\|Tu\|=\sup_{j\in \mathbb{N}}|\sum_{i=1}^j\xi_i|\leq\sup_{j\in \mathbb{N}}\sum_{i=1}^j|\xi_i|$

and I need to show that: $\|Tu\|\leq C\|u\|_{l^2}=C(\sum_{i=1}^j|\xi_i|^2)^\frac{1}{2}$

To acheive this bound I have attempted:

$\|Tu\|^2\leq(\sup_{j\in \mathbb{N}}\sum_{i=1}^j|\xi_i|)^2=(\sum_{i=1}^\infty|\xi_i|)^2$

$=\sum_{i=1}^\infty\frac{|\xi_i|^2}{i^2}+\sum_{i=1,i\neq j}^\infty(\frac{|\xi_i|}{i}\sum_{j=1}^\infty\frac{|\xi_j|}{j})\leq\sum_{i=1}^\infty |\xi_i|^2+\sum_{i=1,j= 1}^\infty\frac{|\xi_i|}{i}\frac{|\xi_j|}{j}$

$=\|u\|_{l^2}^2+\sum_{i=1,j= 1}^\infty\frac{|\xi_i|}{i}\frac{|\xi_j|}{j}$

But now I am stuck, and would appreciate any help.

Taking on board the comment I have produced what I believe to be a correct proof:

$\|Tu\|=\sup_{j\in \mathbb{N}}|\sum_{n=1}^j\frac{\xi_n}{n}|\leq \sup_{j\in \mathbb{N}}\{(\sum_{n=1}^j |\xi_n|^2)^\frac{1}{2}(\sum_{n=1}^j\frac{1}{n^2})^\frac{1}{2}\}=(\sum_{n=1}^\infty |\xi_n|^2)^\frac{1}{2}(\sum_{n=1}^\infty\frac{1}{n^2})^\frac{1}{2}$

$=\frac{\pi}{\sqrt{6}}\|u\|_{l^2}$, i.e. $\|T\|\leq\frac{\pi}{\sqrt{6}}$

Now if we consider $w=\{w_n\}_{n\in \mathbb{N}}$, where, $ w_n = \frac{\sqrt{6}}{n\pi}$, then $\|w\|_{l^2}=1$.

So: $\|T\|=\sup_{\|u\|_{l^2}=1}\|Tu\|\geq\|Tw\|=\frac{\sqrt{6}}{\pi}\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi}{\sqrt{6}}$

Thus $\|T\|\leq\frac{\pi}{\sqrt{6}}$, and $\|T\|\geq\frac{\pi}{\sqrt{6}} \Rightarrow \|T\|=\frac{\pi}{\sqrt{6}} $

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1 Answer 1

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At some places you seem to be forgetting the coefficients. You have, using Cauchy-Schwarz, $$ \left|\sum_{k=1}^n\frac{\xi_k}k\right|\leq\left(\sum_{k=1}^n|\xi_k|^2\right)^{1/2}\left(\sum_{k=1}^n\frac1{k^2}\right)^{1/2}\leq\left(\sum_{k=1}^\infty|\xi_k|^2\right)^{1/2}\left(\sum_{k=1}^\infty\frac1{k^2}\right)^{1/2}=\|u\|\,\frac\pi{\sqrt 6}, $$ so $\|Tu\|\leq \frac\pi{\sqrt6}\,\|u\|$, i.e. $\|T\|\leq\pi/\sqrt6$.

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  • $\begingroup$ Thank you! I knew there was something simple I was missing! also does the first inequality hold if you start with an infinite sum? $\endgroup$
    – Ellya
    Commented Mar 23, 2014 at 22:55
  • $\begingroup$ You are welcome. Now prove that the inequality is sharp, i.e. $\|T\|=\pi/\sqrt6$. $\endgroup$ Commented Mar 23, 2014 at 22:56

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