Let $T$ be a map from $\ell^3 \to \ell^1$ and $S$ be a map from $\ell^1 \to \ell ^3$. Are these maps continuous?

Question

Let $T:(x_n)_{n=1}^\infty \mapsto (n^{-1}x_n)_{n=1}^\infty$ be a map from $\ell^3 \to \ell^1$ and $S:(x_n)_{n=1}^\infty \mapsto (\log(n+2)x_n)_{n=1}^\infty$ be a map from $\ell^1 \to \ell ^3$. Are these maps continuous?

One needs to show boundedness of the norms $\|T\|_1$ and $\|S\|_3$ to conclude continuity. So starging with $T$ one has $$\|T(x_n)\|_1 = \|(n^{-1}x_n)\|_1 = \sum_{n=1}^\infty |n^{-1}x_n| = \sum_{n=1}^\infty \left| \frac{1}{n} \right| |x_n|$$ but the harmonic series diverges so $$\|T(x_n)\|_1 = \infty$$ which means that $T$ cannot be continuous.

For $S$ one has $$\|S(x_n)\|_3 = \|\log(n+2)x_n\|_3 = \left(\sum_{n=1}^\infty |\log(n+2)x_n|^3 \right)^{1/3}$$ and by Hölder's inequality $$\left(\sum_{n=1}^\infty |\log(n+2)x_n|^3 \right)^{1/3} \le \left( \left(\sum_{n=1}^\infty |\log(n+2)|^3 \right) \cdot \left(\sum_{n=1}^\infty |x_n|^3 \right) \right)^{1/3}$$ from which one gets $$\left( \left(\sum_{n=1}^\infty |\log(n+2)|^3 \right) \cdot \left(\sum_{n=1}^\infty |x_n|^3 \right) \right)^{1/3} = \left(\sum_{n=1}^\infty |\log(n+2)|^3 \right)^{1/3} \cdot \|x\|_3$$

but here also I checked that $\left(\sum_{n=1}^\infty |\log(n+2)|^3 \right)^{1/3}$ will diverge. Is there some known upper bound for $\log(n+2)$ which could be useful here?

Answer 1

$T$ is continuous, namely using Hölder's inequality for conjugated exponents $\frac32$ and $3$ we have $$\|Tx\|_1 = \sum_{n=1}^\infty \left|\frac1n\right||x_n| \le \sqrt[3/2]{\sum_{n=1}^\infty \frac1{n^{3/2}}}\sqrt[3]{\sum_{n=1}^\infty |x_n|^3} = \underbrace{\sqrt[3/2]{\sum_{n=1}^\infty \frac1{n^{3/2}}}}_{< +\infty} \|x\|_3.$$

On the other hand, is $S$ even well-defined? Since $\log(n+2)$ is increasing and unbounded, we can pick an increasing sequence $(p(n))_n$ in $\Bbb{N}$ such that $$\log(p(n)+2) \ge 2^n, \qquad \text{ for all }n \in \Bbb{N}.$$ Define $x$ as a sequence with $\frac1{2^n}$ at the position $p(n)$ for every $n \in \Bbb{N}$ and zeroes elsewhere. Then $$\sum_{n=1}^\infty |x_n| = \sum_{n=1}^\infty |x_{p(n)}| = \sum_{n=1}^\infty \frac1{2^n} = 1 \implies x \in \ell^1$$ but \begin{align} \sum_{n=1}^\infty |(Sx)_n|^3 &= \sum_{n=1}^\infty |\log(n+3)x_n|^3 = \sum_{n=1}^\infty |\log(p(n)+3)x_{p(n)}|^3\\ &= \sum_{n=1}^\infty \frac{\log^3(p(n)+3)}{6^n} \ge \sum_{n=1}^\infty \frac{(2^n)^3}{6^n} = \sum_{n=1}^\infty 1 = +\infty \end{align} so $Sx \notin \ell^3$.

Answer 2

The map $S:f\mapsto \log(\cdot+2) f$ does not map $\ell_3$ to $\ell_1$ since $g(n)=\log(n+2)$ is not bounded. In fact, it is not difficult to show that

For numeric maps on $\mathbb{N}$ and $1\leq p$. If the linear map $S_\phi: f\mapsto \phi f$ maps $\ell_p$ to $\ell_1$, then $\phi$ is bounded.

Proof: For each $k\in\mathbb{N}$, define $E_k=\{m\in\mathbb{N}: k\leq |\phi(m)|<k+1\}$. If $\phi$ is not bounded, then there are infinitely many $E_k$'s that are not empty. Suppose $E_{k_n}$, $k_n<{k_n+1}$, is such a sequence of non empty sets $E_k$, and choose $m_n\in E_{k_n}$. Define the sequence $$f(m)=\sum_n \frac{1}{k_n}\mathbb{1}_{\{m_n\}}(m)$$ Then $f\in \ell_p$: $$\sum_m|f(m)|^p=\sum_m\sum_n\frac{1}{k^p_n}\mathbb{1}_{m_n}(m)=\sum_n\frac{1}{k^p_n}<\infty$$ However $$\sum_m\phi(m)f(m)=\sum_m\sum_n\frac{1}{k_n} \phi(m)\mathbb{1}_{\{m_n\}}(m)\geq\sum_n\mathbb{1}=\infty$$

We also have the follwing result:

For numeric functions on $\mathbb{N}$, $p\geq1$ and $\frac1r\geq (1-\frac1p)$. If $\phi\in \ell_r$, then the linear map $S:f\mapsto \phi f$ maps $\ell_p$ into $\ell_1$.

Proof: With $\frac{1}{q}=1-\frac1p$, the hypothesis means that $1\leq r\leq q$ and so, $\ell_r\subset \ell_q$. Since $\phi\in \ell_r$, $\phi\in\ell_q$, and application of Hölder's inequality yields $$\sum_m|f(m)\phi(m)|\leq\Big(\sum_m|\phi(m)|^q\Big)^{1/q}\Big(\sum_m|f(m)|^p\Big)^{1/p}$$

Incidentaly, this shows that $S$ restricted to $\ell_p$ is a bounded operator: $\|S_\phi f\|_1\leq\|\phi\|_q\|f\|_p$ for all $f\in \ell_p$.