For each pair of functions $x,y\in L_2[0,1]$ let us denote by $x\cdot y$ their pointwise product $$ (x\cdot y)(t)=x(t)\cdot y(t),\quad t\in [0,1]. $$ It belongs to $L_1[0,1]$ due to the Cauchy-Bunyakovsky inequality: $$ x,y\in L_2[0,1]\quad\Longrightarrow\quad x\cdot y\in L_1[0,1]. $$

And for each pair of sets $A,B\subseteq L_2[0,1]$ by $A\cdot B$ we denote the corresponding "element-wise product", $$ A\cdot B=\{x\cdot y; \ x\in A, \ y\in B\}, $$ which is contained in $L_1[0,1]$: $$ A,B\subseteq L_2[0,1]\quad\Longrightarrow\quad A\cdot B\subseteq L_1[0,1]. $$ Let us denote by $\overline{\operatorname{absconv}}(A\cdot B)$ the closed absolutely convex hull of $A\cdot B$ in $L_1[0,1]$.

And for each $p>1$ let $B_p$ be the unit ball in $L_p[0,1]$.

I wonder in which case $B_p$ is contained in a set of the form $\overline{\operatorname{absconv}}(K\cdot B_2)$ where $K$ is a compact set in $L_2[0,1]$: $$ B_p\subseteq \overline{\operatorname{absconv}}(K\cdot B_2). $$ For $p\ge 2$ this is trivially true, since in this case we can take $K=\{1\}$, the set consisting of just one function, the constant identity ($1(t)=1$, $t\in[0,1]$): $$ B_p\subseteq B_2\subseteq \overline{\operatorname{absconv}}B_2= \overline{\operatorname{absconv}}(\{1\}\cdot B_2) $$

But for $1<p<2$ this seems to be not true:

If $1<p<2$ then there is no a compact set $K\subseteq L_2[0,1]$ such that $$ B_p\subseteq \overline{\operatorname{absconv}}(K\cdot B_2). $$

Am I right?

For each pair of functions $x,y\in L_2[0,1]$ let us denote by $x\cdot y$ their pointwise product $$ (x\cdot y)(t)=x(t)\cdot y(t),\quad t\in [0,1]. $$ It belongs to $L_1[0,1]$ due to the Cauchy-Bunyakovsky inequality: $$ x,y\in L_2[0,1]\quad\Longrightarrow\quad x\cdot y\in L_1[0,1]. $$

And for each pair of sets $A,B\subseteq L_2[0,1]$ by $A\cdot B$ we denote the corresponding "element-wise product", $$ A\cdot B=\{x\cdot y; \ x\in A, \ y\in B\}, $$ which is contained in $L_1[0,1]$: $$ A,B\subseteq L_2[0,1]\quad\Longrightarrow\quad A\cdot B\subseteq L_1[0,1]. $$ Let us denote by $\overline{\operatorname{absconv}}(A\cdot B)$ the closed absolutely convex hull of $A\cdot B$ in $L_1[0,1]$.

And for each $p>1$ let $B_p$ be the unit ball in $L_p[0,1]$.

I wonder in which case $B_p$ is contained in a set of the form $\overline{\operatorname{absconv}}(K\cdot B_2)$ where $K$ is a compact set in $L_2[0,1]$: $$ B_p\subseteq \underbrace{\overline{\operatorname{absconv}}(K\cdot B_2)}_{\scriptsize\begin{matrix}\text{closed absolutely convex hull in $L_1[0,1]$}\end{matrix}}. $$ For $p\ge 2$ this is trivially true, since in this case we can take $K=\{1\}$, the set consisting of just one function, the constant identity ($1(t)=1$, $t\in[0,1]$): $$ B_p\subseteq B_2\subseteq \overline{\operatorname{absconv}}B_2= \overline{\operatorname{absconv}}(\{1\}\cdot B_2) $$

But for $1<p<2$ this seems to be not true:

If $1<p<2$ then there is no a compact set $K\subseteq L_2[0,1]$ such that $$ B_p\subseteq \underbrace{\overline{\operatorname{absconv}}(K\cdot B_2)}_{\scriptsize\begin{matrix}\text{closed absolutely convex hull in $L_1[0,1]$}\end{matrix}}. $$

Am I right?

For each pair of functions $x,y\in L_2[0,1]$ let us denote by $x\cdot y$ their pointwise product $$ (x\cdot y)(t)=x(t)\cdot y(t),\quad t\in [0,1]. $$ It belongs to $L_1[0,1]$ due to the Cauchy-Bunyakovsky inequality: $$ x,y\in L_2[0,1]\quad\Longrightarrow\quad x\cdot y\in L_1[0,1]. $$

And for each pair of sets $A,B\subseteq L_2[0,1]$ by $A\cdot B$ we denote the corresponding "element-wise product", $$ A\cdot B=\{x\cdot y; \ x\in A, \ y\in B\}, $$ which is contained in $L_1[0,1]$: $$ A,B\subseteq L_2[0,1]\quad\Longrightarrow\quad A\cdot B\subseteq L_1[0,1]. $$ Let us denote by $\overline{\operatorname{absconv}}(A\cdot B)$ the closed absolutely convex hull of $A\cdot B$ in $L_1[0,1]$.

And for each $p>1$ let $B_p$ be the unit ball in $L_p[0,1]$.

I wonder in which case $B_p$ is contained in a set of the form $\overline{\operatorname{absconv}}(K\cdot B_2)$ where $K$ is a compact set in $L_2[0,1]$: $$ B_p\subseteq \overline{\operatorname{absconv}}(K\cdot B_2). $$ For $p\ge 2$ this is true, since in this case we can take $K=\{1\}$, the set consisting of just one function, the constant identity ($1(t)=1$, $t\in[0,1]$): $$ B_p\subseteq B_2\subseteq \overline{\operatorname{absconv}}B_2= \overline{\operatorname{absconv}}(\{1\}\cdot B_2) $$

But for $1<p<2$ this seems to be not true:

If $1<p<2$ then there is no a compact set $K\subseteq L_2[0,1]$ such that $$ B_p\subseteq \overline{\operatorname{absconv}}(K\cdot B_2). $$

Am I right?

For each pair of functions $x,y\in L_2[0,1]$ let us denote by $x\cdot y$ their pointwise product $$ (x\cdot y)(t)=x(t)\cdot y(t),\quad t\in [0,1]. $$ It belongs to $L_1[0,1]$ due to the Cauchy-Bunyakovsky inequality: $$ x,y\in L_2[0,1]\quad\Longrightarrow\quad x\cdot y\in L_1[0,1]. $$

And for each pair of sets $A,B\subseteq L_2[0,1]$ by $A\cdot B$ we denote the corresponding "element-wise product", $$ A\cdot B=\{x\cdot y; \ x\in A, \ y\in B\}, $$ which is contained in $L_1[0,1]$: $$ A,B\subseteq L_2[0,1]\quad\Longrightarrow\quad A\cdot B\subseteq L_1[0,1]. $$ Let us denote by $\overline{\operatorname{absconv}}(A\cdot B)$ the closed absolutely convex hull of $A\cdot B$ in $L_1[0,1]$.

And for each $p>1$ let $B_p$ be the unit ball in $L_p[0,1]$.

I wonder in which case $B_p$ is contained in a set of the form $\overline{\operatorname{absconv}}(K\cdot B_2)$ where $K$ is a compact set in $L_2[0,1]$: $$ B_p\subseteq \overline{\operatorname{absconv}}(K\cdot B_2). $$ For $p\ge 2$ this is trivially true, since in this case we can take $K=\{1\}$, the set consisting of just one function, the constant identity ($1(t)=1$, $t\in[0,1]$): $$ B_p\subseteq B_2\subseteq \overline{\operatorname{absconv}}B_2= \overline{\operatorname{absconv}}(\{1\}\cdot B_2) $$

But for $1<p<2$ this seems to be not true:

If $1<p<2$ then there is no a compact set $K\subseteq L_2[0,1]$ such that $$ B_p\subseteq \overline{\operatorname{absconv}}(K\cdot B_2). $$

Am I right?