Define $F : \mathbb{R}^ω → \mathbb{R}^ω$ by $F(x)_n = \sum^n_{k=1} x_k$. Determine whether $F$ restricts to give a welldefined map $F : (\ell_p, d_p) → (\ell_q, d_q)$
Define F : $\mathbb{R}^? \rightarrow \mathbb{R}^?$ by $F(x)_n = \sum^n_{k=1} x_k$, that is by $F(x1, x2, x3, · · ·) = ( x1, x1+x2, x1+ x2+ x3, · · ·)$ . For each $p, q \in \{1, 2, ?\}$, determine whether $F$ restricts to give a welldefined map $F : (\ell_p, d_p) \rightarrow (\ell_q, d_q)$ and, if so, whether that map is continuous
Answer
For convenience one shows the following statements:
(i) If $F$ restricts to a map $\ell^p\to\ell^q$ then $q=\infty$.
(ii) If $F$ restricts to a map $\ell^p\to\ell^\infty$ then $p=1$.
(iii) Restricting $F$ to $\ell^1\to\ell^\infty$ one ges a continuous map.
(i)
Note that the sequence $2^{n}$ is in $\ell^p$ for all $p\in[1,\infty]$. But
$$F((2^{k})_k)_n = \frac{12^{n+1}}{12}$$
and the only $\ell^q$ containing this image is $\ell^\infty$. This means that if $q\neq\infty$ the map $F$ does not restrict to a map $\ell^p\to\ell^q$ for any $p$.
(ii)
Suppose $p>1$, then look at
$$x_n:= \dfrac1{n^{1/p + \epsilon(p)}}$$
here $\epsilon(p)$ is any number $>0$ so that $1/p + \epsilon(p) <1$ (eg $\epsilon(p) = 1\frac2{p}$ is one example). You have that:
$$x_n^p < \frac1{n^{1+\epsilon(p)}}$$
where the righthand side is summable, so $(x_n)_{n\in\Bbb N}\in \ell^p$. But:
$$x_n > \frac1n$$
which is not summable, so
$$F(x)_n = \sum_{k=1}^n \frac1{n^{1/p+\epsilon(p)}}$$
is unbounded and $F(x)$ does not lie in $\ell^\infty$.
(iii)
For all $x\in \ell^1$ one has that $x$ is an absolutely summable sequence. So one gets:
$$F(x)_n = \left\sum_{k=1}^n x_k\right ?\sum_{k=1}^n x_k ? \sum_{k=1}^\infty x_k = \x\_{\ell^1} <\infty$$
So $F(x)_n$ is bounded by $\x\_{\ell^1}$ and the map is well defined as a map $\ell^1\to\ell^\infty$. The same calculation shows that it is continuous:
$$\sup_{x\in \ell^1, \x\_{\ell^1}?1}\F(x)\_{\ell^\infty}? \sup_{x\in \ell^1, \x\_{\ell^1}?1} \x\_{\ell^1} ? 1$$

For (ii) I forgot to say why $F$ does not restrict to a map $\ell^\infty\to\ell^\infty$. Note here that $\ell^p\subseteq \ell^\infty$ to get this.
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