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?

  • $\begingroup$ I removed a wrong answer. $\endgroup$ Jun 27, 2018 at 16:08
  • $\begingroup$ I see. If you'll understand how to correct it, I hope, you will share your ideas. $\endgroup$ Jun 27, 2018 at 16:12
  • $\begingroup$ New try, hoping it is more correct this time. $\endgroup$ Jun 28, 2018 at 12:54

1 Answer 1


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.

  • $\begingroup$ Mikael, this trick with replacing $[0,1]$ by $\{0,1\}^{\mathbb N}$, I did not see it before, and I have no intuition here. As far as I understand, you just treat real numbers $r\in[0,1]$ as binary sequences $r=0,r_1r_2...$. How is the Lebesgue measure represented in this picture? And the shift $\omega_n\mapsto\omega_{n+N}$ should not change this measure, or I don't understand something? $\endgroup$ Jun 28, 2018 at 20:26
  • $\begingroup$ @SergeiAkbarov : I am indeed using that $([0,1],Lebesgue)$ and $(\{0,1\}^{\mathbf N}, (\delta_0 /2 + \delta_1/2)^{\otimes \mathbf N})$ are isomorphic as measure spaces. For this, you can use the explicit isomorphism given by dyadic decomposition, or you might prefer to just invoke that they are both standard atomless probability measures, and hence are isomorphic en.wikipedia.org/wiki/Standard_probability_space $\endgroup$ Jun 29, 2018 at 8:18

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