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?



Is the unit ball in $L_p$, $1<p<2$, contained in a "compact perturbation" of the unit ball in $L_2$?
You are right, the answer to the question is no: if $1<p<2$, then there is no compact set $K \subset L\_2$ such that $$B\_p \subset \overline{\textrm{absconv}} (K \cdot B\_2).$$


Assume by contradiction that such $K$ exists, and define $C=\max\_K \|f\|\_2$. By duality (this is an equivalence) we have 
$$ \|g\|\_q \leq \sup\_{f \in K} \|fg\|\_2$$
for every $g \in L\_\infty$, where $q = \frac{p}{p-1}$ is the conjugate exponent of $p$.


The idea is that if $g$ is independant from $K$, this inequality becomes $\|g\|\_q \leq C \|g\|\_2$, which is not true because $q>2$.


More details : for convenience let me work with $L\_p(\{0,1\}^{\mathbf{N}})$ (dyadic coordinates). Let $\varepsilon>0$ be arbitrary. By compactness, there are $f\_1,\dots,f\_k$ of $L\_2$-norm less than $C$ such that $K$ is contained in the $\varepsilon$-neighborhood of $\{f\_1,\dots,f\_k\}$, and so the previous inequality implies
$$ \|g\|\_q \leq \max\_i \|f\_ig\|\_2+\varepsilon \|g\|\_\infty.$$


By density we can even assume that each $f\_i \colon \{0,1\}^{\mathbf N}\to \mathbf C$ depends on finitely many coordinates, say the first $N$. If $g$ does not depend on the first $N$ coordinates, it is then independant from each $f\_i$, $\|f\_i g\|\_2 = \|f\_i\|\_2 \|g\|\_2 \leq C \|g\|\_2$, and
$$ \|g\|\_q \leq C \|g\|\_2+\varepsilon \|g\|\_\infty.$$
If $g \in L\_\infty$ is arbitrary, by applying this to the function $(\omega\_n)\_{n \in \mathbf N} \mapsto g((\omega\_{n+N})\_{n \in \mathbf N})$ we obtain that this inequality holds for every $g$. Making $\varepsilon \to 0$ we obtain $\|g\|\_q \leq C \|g\|\_2$, a contradiction.


