$\newcommand{\vertiii}[1]{{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert #1 \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}}$ Let $X=\left(\ell_1, \|\cdot\|_2\right)$ and $(\phi_n)_n\subset\mathcal{B}(\ell_1\to \mathbb{R})$ be a sequence of bounded maps, where $\phi_n(x)=\sum\limits_{k=1}^n x_k$. Show that $(\phi_n)_n$ converges pointwise but $\sup_n \vertiii{\phi_n}:=\sup\limits_{n\in\mathbb{N}}\sup\limits_{x\in\ell_1, \|\cdot\|_2\le 1} |\phi_n(x)|=\infty$.

To prove that $(\phi_n)_n$ converges pointwise, let $x\in\ell_1$, then for every $\varepsilon > 0$, $\exists n_\varepsilon \in \mathbb{N}$ such that $n\ge n_\varepsilon\implies \sum\limits_{k=n}^\infty |x_k|<\varepsilon$. Let $\varepsilon>0$ and $n\ge n_\varepsilon$ and consider

$$|\phi_n(x)-\phi(x)|=\left|\sum\limits_{k=1}^n x_k - \sum\limits_{k=1}^\infty x_k \right| = \left| \sum\limits_{k=n+1}^\infty x_k\right|\le \sum\limits_{k+1}^\infty |x_k|<\varepsilon.$$

Hence, $(\phi_n)_n$ converges pointwise.

By the Uniform Boundedness Principle we also know that there exists $Y\subset \ell_1$ such that $\vertiii{\phi_n(x)}$ converges to $\vertiii{\phi(x)}$ for all $x\in Y$.

However, the problem I'm experiencing is that I do not understand how to calculate $\sup_n \vertiii{\phi_n}:=\sup\limits_{n\in\mathbb{N}}\sup\limits_{x\in\ell_1, \|\cdot\|_2\le 1} |\phi_n(x)|=\infty$. That is, I don't see the mechanics behind this definition. It is suggested to use the Hölder's inequality, but I don't even know how, in this case.

$\phi_n(x)$ appears to be a bounded operator, so its norm can probably be calculated as $$\sup\limits_{x\in Y}\frac{|\phi_n(x)|}{\|x\|_2},$$ but here the condition is that $\|x\|_2\le 1$.

Can someone please give me a clue of how to sort this out? And also how is the Uniform Boundedness principle related to the solution of this problem?

  • $\begingroup$ The definition of $\phi_n$ on the first line doesn't seem to be right - should it be a sum up to $n$ instead of $\infty$? $\endgroup$ – Nate Eldredge Nov 30 '17 at 3:17
  • $\begingroup$ @NateEldredge Sorry, yes. Editing. $\endgroup$ – sequence Nov 30 '17 at 6:04

$||\phi_n||=n^{1/2}$. It is clear (by Cauchy -Scwartz inequality) that $||\phi_n|| \leq n^{1/2}$. The maximum is attained when $x_i=n^{-1/2}$ for $i \leq n$ and 0 for $i>n$

  • $\begingroup$ Sorry, what does $E\|\phi_n\|$ mean? I haven't seen this notation. Can you please explain this in more detail - where exactly do you apply Cauchy-Schwartz inequality? $\endgroup$ – sequence Nov 30 '17 at 7:56
  • $\begingroup$ Very sorry! Simply delete E in both places. $\endgroup$ – Kavi Rama Murthy Nov 30 '17 at 9:29
  • $\begingroup$ @KaviRamaMurthy: Why not correct it yourself? Use the edit button. $\endgroup$ – Nate Eldredge Nov 30 '17 at 13:56

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