Let $F\subset \mathbb{R}^n$ be a finite set and $\sigma$ be uniformly distributed over $\{-1,1\}^n$. The usual Rademacher average of $F$ (modulo normalizing factors) is
$$ R\_n(F)=\mathbb{E}\_\sigma \max\_{f\in F}\sum\_{i=1}^n \sigma\_if\_i.
$$
Now let us define two operations on $F$: $\mathrm{conv}(F)$ and $[F]\_\vee$. The former is just the convex hull of the vector-set $F$ in $\mathbb{R}^n$. The latter is defined by
$$ [F]\_\vee=\{ f\vee g :f,g \in F\},
$$
where $(f\vee g)\_i=\max\{f\_i,g\_i\}$ is the coordinate-wise maximum.
It is well-known (and easy to show) that
$
R\_n( \mathrm{conv}(F))
=
R\_n( F)
$.


Question: is it true that
$$
R\_n( [\mathrm{conv}(F)]\_\vee)
=
R\_n( [F]\_\vee)
$$
?


The inequality 
$
R\_n( [F]\_\vee)
\le
R\_n( [\mathrm{conv}(F)]\_\vee)
$
holds due to set containment. Also, $\mathrm{conv}(F)$ is an infinite set, so the $\max$ in the definition should be replaced by a $\sup$.


**Update**. Fedor Petrov has constructed a counterexample, which I've accepted. The more general conjecture I had was as follows. Define the $k$-fold max operator $[F]\_k$ by
$$ [F]\_k=\{ f\_1\vee f\_2\ldots\vee f\_k :f\_i \in F\}.
$$
Is there a universal constant $c$ (independent of $n$ and $k$) such that
$$
R\_n( [\mathrm{conv}(F)]\_k)
\le
c R\_n( [F]\_k)
$$?



Rademacher, maxima, convex hulls
It looks that no. Take $n=4$ and $F$ containing four vectors: $f=(2,-2,1,-5); g=(-2,2,1,-5); h=(0,0,-5,1)$. We have $f\vee g=(2,2,1,-5)$, $f\vee h=(2,0,1,1)$, $g\vee h=(0,2,1,1)$. Thus $\max\_{w\in [F]\_\vee} (w\_3+w\_4-w\_1-w\_2)=0$. On the other hand $\frac{f+g}2\vee h=(0,0,1,1)$, therefore $\max\_{w\in [\mathrm{conv}(F)]\_\vee} (w\_3+w\_4-w\_1-w\_2)\geqslant 2$.


