1 Introduction and main results
In many problems arising in probability theory and its applications one needs to study variables of the form
$S=\sup _{t\in T}|\!\sum _{i=1}^{n}t_{i}X_{i}|$
, where
$X_{1},\ldots ,X_{n}$
are independent random variables and
$T$
is a non-empty subset of
$\mathbb{R}^{n}$
. In particular it is of interest to estimate tails of
$S$
. Such estimates are strictly related to bounds for
$L_{p}$
-norms of
$S$
(i.e.
$\Vert S\Vert _{p}:=(\mathbb{E}|S|^{p})^{1/p}$
) for
$p\geqslant 1$
(see Corollary 1.3 and its proof in §4 below). There is a trivial lower estimate:
$$\begin{eqnarray}\biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\geqslant \max \biggl\{\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|,\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\biggr\}.\end{eqnarray}$$
It turns out that in some situations this obvious lower bound may be reversed, i.e. there exist numerical constants
$C_{1}$
and
$C_{2}$
such that
$$\begin{eqnarray}\biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\leqslant C_{1}\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|+C_{2}\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}.\end{eqnarray}$$
This is for example the case (with
$C_{1}=1$
), when
$X_{i}$
are normally distributed. This is an easy consequence of the Gaussian concentration (cf. [Reference Ledoux and Talagrand11, Ch. 3]). Dilworth and Montgomery-Smith [Reference Dilworth and Montgomery-Smith3] established the inequality (1.2) for
$X_{i}$
being symmetric Bernoulli random variables. This result was generalized in [Reference Latała6] to symmetric variables with logarithmically concave tails and in [Reference Latała and Tkocz9, Theorem 2.3] to symmetric random variables such that
$\Vert X_{i}\Vert _{q}\leqslant (q/p)\unicode[STIX]{x1D6FC}\Vert X_{i}\Vert _{p}$
for all
$q\geqslant p\geqslant 2$
.
The main result of this paper is the following.
Theorem 1.1. Let
$X_{1},\ldots ,X_{n}$
be independent mean zero random variables with finite moments such that
$$\begin{eqnarray}\Vert X_{i}\Vert _{2p}\leqslant \unicode[STIX]{x1D6FC}\Vert X_{i}\Vert _{p}\quad \text{for every }p\geqslant 2\text{ and }i=1,\ldots ,n,\end{eqnarray}$$
where
$\unicode[STIX]{x1D6FC}$
is a finite positive constant. Then for every
$p\geqslant 1$
and every non-empty set
$T\subset \mathbb{R}^{n}$
we have
$$\begin{eqnarray}\biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\leqslant C(\unicode[STIX]{x1D6FC})\biggl[\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|+\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\biggr],\end{eqnarray}$$
where
$C(\unicode[STIX]{x1D6FC})$
is a constant which depends only on
$\unicode[STIX]{x1D6FC}$
.
It turns out that Theorem 1.1 may be reversed in the independent and identically distributed (i.i.d.) case.
Theorem 1.2. Let
$X_{1},X_{2},\ldots$
be i.i.d. random variables. Assume that there exists a constant
$L$
such that for every
$p\geqslant 1$
, every
$n$
and every non-empty set
$T\subset \mathbb{R}^{n}$
we have
$$\begin{eqnarray}\biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\leqslant L\biggl[\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|+\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\biggr].\end{eqnarray}$$
Then
$$\begin{eqnarray}\Vert X_{1}\Vert _{2p}\leqslant \unicode[STIX]{x1D6FC}(L)\Vert X_{1}\Vert _{p}\quad \text{for }p\geqslant 2,\end{eqnarray}$$
where
$\unicode[STIX]{x1D6FC}(L)$
is a constant which depends only on
$L\geqslant 1$
.
It will be clear from the proof of Theorem 1.2 that it suffices to assume (1.5) for
$T=\{\pm e_{j}:j\in \{1,\ldots ,n\}\}$
only, where
$\{e_{1},\ldots ,e_{n}\}$
is the canonical basis of
$\mathbb{R}^{n}$
.
The comparison of weak and strong moments (1.4) also yields a deviation inequality for
$\sup _{t\in T}|\!\sum _{i=1}^{n}t_{i}X_{i}|$
.
Corollary 1.3.
$\!$
Assume
$\!X_{1},X_{2},\ldots$
satisfy the assumptions of Theorem 1.1. Then for any
$u\geqslant 0$
and any non-empty set
$T$
in
$\mathbb{R}^{n}$
,
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathbb{P}\biggl(\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|\geqslant C_{1}(\unicode[STIX]{x1D6FC})\biggl[u+\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|\biggr]\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant C_{2}(\unicode[STIX]{x1D6FC})\sup _{t\in T}\mathbb{P}\biggl(\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|\geqslant u\biggr),\end{eqnarray}$$
where constants
$C_{1}(\unicode[STIX]{x1D6FC})$
and
$C_{2}(\unicode[STIX]{x1D6FC})$
depend only on the constant
$\unicode[STIX]{x1D6FC}$
in (1.3).
Another consequence of the main theorem is the following Khintchine–Kahane type inequality.
Corollary 1.4. Assume
$X_{i}$
,
$1\leqslant i\leqslant n$
satisfy the assumptions of Theorem 1.1. Then for any
$p\geqslant q\geqslant 2$
and any non-empty set
$T$
in
$\mathbb{R}^{n}$
we have,
$$\begin{eqnarray}\biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\leqslant C_{3}(\unicode[STIX]{x1D6FC})\biggl(\frac{p}{q}\biggr)^{\max \{1/2,\log _{2}\unicode[STIX]{x1D6FC}\}}\biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{q}\biggr)^{1/q},\end{eqnarray}$$
where a constant
$C_{3}(\unicode[STIX]{x1D6FC})$
depends only on the constant
$\unicode[STIX]{x1D6FC}$
in (1.3).
We postpone proofs of the above results and first present a number of remarks and open questions.
Remark 1.5. Exponent
$\max \{1/2,\log _{2}\unicode[STIX]{x1D6FC}\}$
in Corollary 1.4 is optimal.
Indeed, since
$\Vert g\Vert _{p}\sim \sqrt{p/e}$
as
$p\rightarrow \infty$
one cannot go below
$1/2$
by the central limit theorem.
To see that
$\log _{2}\unicode[STIX]{x1D6FC}$
term cannot be improved it is enough to consider
$\unicode[STIX]{x1D6FC}>\sqrt{2}$
. Let
$r=1/\log _{2}\,\unicode[STIX]{x1D6FC}\in (0,2)$
and let
$X$
be a symmetric random variable given by
$\mathbb{P}(|X|\geqslant t)=e^{-t^{r}}$
(with
$2>r>0$
), i.e.
$X=|{\mathcal{E}}|^{1/r}\operatorname{sgn}{\mathcal{E}}$
, where
${\mathcal{E}}$
has the symmetric exponential distribution. By Stirling’s formula
$\unicode[STIX]{x1D6E4}(x+1)=(x/e)^{x}\sqrt{2\unicode[STIX]{x1D70B}x}e^{f(x)}$
with
$f(x)\in (0,1/12)$
for
$x\geqslant 1$
, so for
$p\geqslant 2$
,
$$\begin{eqnarray}\frac{\Vert X\Vert _{2p}}{\Vert X\Vert _{p}}=\frac{\unicode[STIX]{x1D6E4}(2p/r+1)^{1/(2p)}}{\unicode[STIX]{x1D6E4}(p/r+1)^{1/p}}\leqslant 2^{1/r}\biggl(\frac{r}{\unicode[STIX]{x1D70B}p}\biggr)^{1/(4p)}e^{1/(24p)}\leqslant 2^{1/r}=\unicode[STIX]{x1D6FC}.\end{eqnarray}$$
Moreover,
$\Vert X\Vert _{p}\sim (p/er)^{1/r}$
for
$p\rightarrow \infty$
, so the assertion of Corollary 1.4 cannot hold with any exponent better than
$\log _{2}\unicode[STIX]{x1D6FC}$
.
Remark 1.6. If the variables
$X_{i}$
are symmetric then one may replace the term
$\mathbb{E}\sup _{t\in T}|\!\sum _{i=1}^{n}t_{i}X_{i}|$
in (1.4) by
$\mathbb{E}\sup _{t\in T}\sum _{i=1}^{n}t_{i}X_{i}$
.
Proof. Let
$s$
be any point in
$T$
. Then
$T\subset T-T+s$
, so by the triangle inequality
$$\begin{eqnarray}\biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\leqslant \biggl(\mathbb{E}\sup _{t\in T-T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}+\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}s_{i}X_{i}\biggr|^{p}\biggr)^{1/p}.\end{eqnarray}$$
Estimate (1.4) applied to the set
$T-T$
yields
$$\begin{eqnarray}\displaystyle \biggl(\mathbb{E}\sup _{t\in T-T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p} & {\leqslant} & \displaystyle C(\unicode[STIX]{x1D6FC})[\mathbb{E}\sup _{t\in T-T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|\nonumber\\ \displaystyle & & \displaystyle +\,\sup _{t\in T-T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p} ].\nonumber\end{eqnarray}$$
The set
$T-T$
is symmetric, so
$$\begin{eqnarray}\mathbb{E}\sup _{t\in T-T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|=\mathbb{E}\sup _{t\in T-T}\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\leqslant 2\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}X_{i},\end{eqnarray}$$
where the last estimate follows, since
$(X_{i})_{i=1}^{n}$
and
$(-X_{i})_{i=1}^{n}$
are equally distributed. Moreover,
$$\begin{eqnarray}\sup _{t\in T-T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\leqslant 2\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p},\end{eqnarray}$$
which finishes the proof of the remark. ◻
Remark 1.7. If the variables
$X_{i}$
are not centered then (1.4) holds provided that the assumption (1.3) is replaced by
$$\begin{eqnarray}\Vert X_{i}-\mathbb{E}X_{i}\Vert _{2p}\leqslant \unicode[STIX]{x1D6FC}\Vert X_{i}-\mathbb{E}X_{i}\Vert _{p}\quad \text{for }p\geqslant 2\text{ and }i=1,\ldots ,n.\end{eqnarray}$$
Proof. We have
$$\begin{eqnarray}\biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\leqslant \biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}(X_{i}-\mathbb{E}X_{i})\biggr|^{p}\biggr)^{1/p}+\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}\mathbb{E}X_{i}\biggr|.\end{eqnarray}$$
Theorem 1.1 applied to centered variables
$X_{i}-\mathbb{E}X_{i}$
,
$i=1,\ldots ,n$
, yields
$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}(X_{i}-\mathbb{E}X_{i})\biggr|^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant C(\unicode[STIX]{x1D6FC})\biggl[\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}(X_{i}-\mathbb{E}X_{i})\biggr|+\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}(X_{i}-\mathbb{E}X_{i})\biggr|^{p}\biggr)^{1/p}\biggr].\nonumber\end{eqnarray}$$
To conclude it is enough to observe that
$$\begin{eqnarray}\displaystyle \mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}(X_{i}-\mathbb{E}X_{i})\biggr| & {\leqslant} & \displaystyle \mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|+\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}\mathbb{E}X_{i}\biggr|,\nonumber\\ \displaystyle \sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}(X_{i}-\mathbb{E}X_{i})\biggr|^{p}\biggr)^{1/p} & {\leqslant} & \displaystyle \sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}+\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}\mathbb{E}X_{i}\biggr|,\nonumber\end{eqnarray}$$
and
$$\begin{eqnarray}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}\mathbb{E}X_{i}\biggr|\leqslant \mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|.\end{eqnarray}$$
Open questions. For Gaussian random vectors (1.2) holds with
$C_{1}=1$
. This is also the case for
$X_{i}$
symmetric, independent with log-concave distributions [Reference Latała and Wojtaszczyk10, Remark 3.16 and Corollary 2.19]. However, we do not know the general conditions for the distributions of
$X_{i}$
which are sufficient for (1.2) to hold with
$C_{1}=1$
.
It is of interest to study the comparison of weak and strong moments for random vectors
$X=(X_{1},\ldots ,X_{n})$
with dependent coordinates. A natural and important class to investigate in this context are vectors with log-concave distributions (cf. [Reference Brazitikos, Giannopoulos, Valettas and Vritsiou2] for an up to date survey of the properties of such vectors). Paouris [Reference Paouris12] showed that (1.2) holds for log-concave vectors and sets
$T$
being balls in Euclidean spaces (see also [Reference Adamczak, Latała, Litvak, Oleszkiewicz, Pajor and Tomczak-Jaegermann1]). This was generalized in [Reference Latała and Strzelecka8] to balls in
$L_{r}$
-spaces with
$1\leqslant r<\infty$
. Unfortunately there are very few classes of log-concave vectors such that (1.2) is known to be satisfied for all sets
$T$
– this includes vectors uniformly distributed on
$l_{r}^{n}$
-balls (with
$1\leqslant r\leqslant \infty$
) [Reference Latała and Wojtaszczyk10, Remark 3.16 and Theorem 5.27], or more generally vectors with densities of the form
$\exp (-\unicode[STIX]{x1D711}(\Vert x\Vert _{r}))$
, where
$\unicode[STIX]{x1D711}:[0,\infty )\rightarrow (-\infty ,\infty ]$
is non-decreasing and convex, and
$1\leqslant r\leqslant \infty$
[Reference Latała7, Proposition 6.5].
The organization of this paper is as follows. In §2 we prove Theorem 1.1 for unconditional sets
$T$
only. Using this result we generalize it to the case of an arbitrary
$T$
in §3. In §4 we prove Corollaries 1.3 and 1.4. Finally, in §5 we present the proof of Theorem 1.2.
Throughout this paper we denote universal constants by the letter
$C$
and by
$C(\unicode[STIX]{x1D6FC})$
constants depending only on the parameter
$\unicode[STIX]{x1D6FC}$
. The values of the constants
$C$
,
$C(\unicode[STIX]{x1D6FC})$
may differ at each occurrence. We will also frequently work with a Bernoulli sequence
$\unicode[STIX]{x1D700}_{i}$
of i.i.d. symmetric random variables taking values
$\pm 1$
. We assume that variables
$\unicode[STIX]{x1D700}_{i}$
are independent of other random variables.
2 The case of unconditional sets
In this section we show that Theorem 1.1 holds under additional assumptions that the set
$T$
is unconditional and the variables
$X_{i}$
are symmetric. Recall that a set
$T$
in
$\mathbb{R}^{n}$
is called unconditional if it is symmetric with respect to the coordinate axes, i.e.
$(\unicode[STIX]{x1D702}_{i}t_{i})_{i=1}^{n}\in T$
for any
$t=(t_{i})_{i=1}^{n}\in T$
and any choice of signs
$\unicode[STIX]{x1D702}_{1},\ldots ,\unicode[STIX]{x1D702}_{n}\in \{-1,1\}$
.
Proposition 2.1. Let
$r\in (0,1)$
and
$L\geqslant 1$
. Assume that variables
$Y_{1},\ldots ,Y_{n}$
are independent and symmetric and
$$\begin{eqnarray}\biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}Y_{i}\biggr|^{p}\biggr)^{1/p}\leqslant L\biggl[\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}Y_{i}+\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}Y_{i}\biggr|^{p}\biggr)^{1/p}\biggr]\end{eqnarray}$$
for all
$p\geqslant 1$
and all unconditional sets
$T$
. Then variables
$X_{i}:=|Y_{i}|^{1/r}\operatorname{sgn}Y_{i}$
satisfy
$$\begin{eqnarray}\biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\leqslant (2L)^{1/r}\biggl[\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}X_{i}+\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\biggr]\end{eqnarray}$$
for all
$p\geqslant 1$
and all unconditional sets
$T\subset \mathbb{R}^{n}$
.
Proof. Definition of
$X_{i}$
and unconditionality of
$T$
yield
$$\begin{eqnarray}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|=\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}|Y_{i}|^{1/r}\operatorname{sgn}Y_{i}\biggr|=\sup _{t\in T}\mathop{\sum }_{i=1}^{n}|t_{i}||Y_{i}|^{1/r}.\end{eqnarray}$$
Let
$s=(1-r)^{-1}$
and let
$B_{s}^{n}$
denote the unit ball of
$\ell _{s}^{n}$
. Then
$1/s+r=1$
and by Hölder’s duality we have
$$\begin{eqnarray}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}|t_{i}||Y_{i}|^{1/r}\biggr|^{r}=\sup _{t\in T}\sup _{u\in B_{s}^{n}}\mathop{\sum }_{i=1}^{n}u_{i}|t_{i}|^{r}Y_{i}=\sup _{t\in T_{r}}\mathop{\sum }_{i=1}^{n}t_{i}Y_{i},\end{eqnarray}$$
where
$$\begin{eqnarray}T_{r}:=\{(u_{i}|t_{i}|^{r})_{i=1}^{n}:~t\in T,u\in B_{s}^{n}\}\end{eqnarray}$$
is unconditional in
$\mathbb{R}^{n}$
. Therefore (2.1) applied with
$p/r$
and
$T_{r}$
instead of
$p$
and
$T$
yields
$$\begin{eqnarray}\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\leqslant L^{p/r}\biggl[\mathbb{E}\sup _{t\in T_{r}}t_{i}Y_{i}+\sup _{t\in T_{r}}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}Y_{i}\biggr|^{p/r}\biggr)^{r/p}\biggr]^{p/r}.\end{eqnarray}$$
We have
$$\begin{eqnarray}\mathbb{E}\sup _{t\in T_{r}}\mathop{\sum }_{i=1}^{n}t_{i}Y_{i}=\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{r}\leqslant \biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|\biggr)^{r}.\end{eqnarray}$$
Moreover,
$$\begin{eqnarray}\displaystyle \sup _{t\in T_{r}}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}Y_{i}\biggr|^{p/r}\biggr)^{r/p} & {\leqslant} & \displaystyle \sup _{t\in T}\biggl(\mathbb{E}\sup _{u\in B_{s}^{n}}\biggl|\mathop{\sum }_{i=1}^{n}u_{i}|t_{i}|^{r}Y_{i}\biggr|^{p/r}\biggr)^{r/p}\nonumber\\ \displaystyle & = & \displaystyle \sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}|t_{i}||X_{i}|\biggr|^{p}\biggr)^{r/p}\nonumber\\ \displaystyle & = & \displaystyle \sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}|X_{i}|\biggr|^{p}\biggr)^{r/p}.\nonumber\end{eqnarray}$$
Estimates above together with the inequality
$(a+b)^{1/r}\leqslant 2^{1/r-1}(a^{1/r}+b^{1/r})$
yield
$$\begin{eqnarray}\biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{\!1/p}\!\leqslant \!\frac{1}{2}(2L)^{1/r}\biggl[\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|+\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}|X_{i}|\biggr|^{p}\biggr)^{\!1/p}\biggr].\end{eqnarray}$$
Hence, in order to prove (2.2) it suffices to show that
$$\begin{eqnarray}\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}|X_{i}|\biggr|^{p}\biggr)^{1/p}\leqslant \mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}X_{i}+2\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}.\end{eqnarray}$$
Let
$(X_{1}^{\prime },\ldots ,X_{n}^{\prime })$
be an independent copy of
$(X_{1},\ldots ,X_{n})$
. By the triangle inequality for the
$p$
th integral norm and Jensen’s inequality we get
$$\begin{eqnarray}\displaystyle \sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}|X_{i}|\biggr|^{p}\biggr)^{1/p} & {\leqslant} & \displaystyle \sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}(|X_{i}|-\mathbb{E}|X_{i}^{\prime }|)\biggr|^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & & \displaystyle +\,\sup _{t\in T}\biggl|\mathbb{E}\mathop{\sum }_{i=1}^{n}t_{i}|X_{i}|\biggr|\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}(|X_{i}|-|X_{i}^{\prime }|)\biggr|^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & & \displaystyle +\,\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}|X_{i}|\biggr|\nonumber\\ \displaystyle & = & \displaystyle \sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}(|X_{i}|-|X_{i}^{\prime }|)\biggr|^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & & \displaystyle +\,\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}X_{i},\end{eqnarray}$$
where the equation follows by the unconditionality of
$T$
.
Let
$(\unicode[STIX]{x1D700}_{i})_{i=1}^{n}$
be the Bernoulli sequence, independent of all
$X_{i}$
and
$X_{i}^{\prime }$
. Then the sequence
$(|X_{i}|-|X_{i}^{\prime }|)_{i=1}^{n}$
has the same distribution as
$(\unicode[STIX]{x1D700}_{i}(|X_{i}|-|X_{i}^{\prime }|))_{i=1}^{n}$
and for every
$t\in \mathbb{R}^{n}$
,
$$\begin{eqnarray}\displaystyle \biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}(|X_{i}|-|X_{i}^{\prime }|)\biggr|^{p}\biggr)^{1/p} & = & \displaystyle \biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}(|X_{i}|-|X_{i}^{\prime }|)\biggr|^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}|X_{i}|\biggr|^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & & \displaystyle +\,\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}|X_{i}^{\prime }|\biggr|^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & = & \displaystyle 2\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}.\end{eqnarray}$$
Putting (2.4) and (2.5) together we get (2.3), which completes the proof of (2.2). ◻
Corollary 2.2. Let
$X_{1},\ldots ,X_{n}$
be independent symmetric random variables with finite moments such that
$$\begin{eqnarray}\Vert X_{i}\Vert _{2p}\leqslant \unicode[STIX]{x1D6FC}\Vert X_{i}\Vert _{p}\quad \text{for }p\geqslant 2\text{ and }i=1,\ldots ,n,\end{eqnarray}$$
where
$\unicode[STIX]{x1D6FC}$
is a finite positive constant. Then for every
$p\geqslant 1$
and every unconditional set
$T\subset \mathbb{R}^{n}$
we have
$$\begin{eqnarray}\biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\leqslant C(\unicode[STIX]{x1D6FC})\biggl[\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}X_{i}+\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\biggr],\end{eqnarray}$$
where
$C(\unicode[STIX]{x1D6FC})$
is a constant, which depends only on
$\unicode[STIX]{x1D6FC}$
.
Proof. Let us first note, that the assumption (2.6) applied
$k$
times yields that
$$\begin{eqnarray}\Vert X_{i}\Vert _{2^{k}p}\leqslant \unicode[STIX]{x1D6FC}^{k}\Vert X_{i}\Vert _{p}\quad \text{for }p\geqslant 2.\end{eqnarray}$$
Therefore
$$\begin{eqnarray}\Vert X_{i}\Vert _{q}\leqslant \unicode[STIX]{x1D6FC}^{\lceil \log _{2}(q/p)\rceil }\Vert X_{i}\Vert _{p}\leqslant \unicode[STIX]{x1D6FC}\biggl(\frac{q}{p}\biggr)^{\log _{2}\unicode[STIX]{x1D6FC}}\Vert X_{i}\Vert _{p}\quad \text{for }q\geqslant p\geqslant 2.\end{eqnarray}$$
Let
$Y_{i}:=|X_{i}|^{1/\log _{2}\,\unicode[STIX]{x1D6FC}}\operatorname{sgn}X_{i}$
. Then
$X_{i}=|Y_{i}|^{1/r}\operatorname{sgn}Y_{i}$
with
$r:=1/\log _{2}\,\unicode[STIX]{x1D6FC}$
and
$$\begin{eqnarray}\Vert Y_{i}\Vert _{q}\leqslant 2\frac{q}{p}\Vert Y_{i}\Vert _{p}\quad \text{for }q\geqslant p\geqslant 2\log _{2}\unicode[STIX]{x1D6FC}.\end{eqnarray}$$
If
$\unicode[STIX]{x1D6FC}\leqslant 2$
we have
$$\begin{eqnarray}\Vert Y_{i}\Vert _{q}\leqslant 2\frac{q}{p}\Vert Y_{i}\Vert _{p}\quad \text{for }q\geqslant p\geqslant 2.\end{eqnarray}$$
Otherwise, take
$2\log _{2}\unicode[STIX]{x1D6FC}=q\geqslant p\geqslant 2$
. Then by Hölder’s inequality and (2.8) with exponents
$p(q-1)/(p-1)$
and
$q$
we get
$$\begin{eqnarray}\displaystyle \Vert Y_{i}\Vert _{q}^{q}=\mathbb{E}|Y_{i}||Y_{i}|^{q-1} & {\leqslant} & \displaystyle (\mathbb{E}|Y_{i}|^{p})^{1/p}(\mathbb{E}|Y_{i}|^{p(q-1)/(p-1)})^{(p-1)/p}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \Vert Y_{i}\Vert _{p}\Vert Y_{i}\Vert _{q}^{q-1}\biggl(2\frac{p(q-1)}{q(p-1)}\biggr)^{q-1}.\nonumber\end{eqnarray}$$
Observe that
$$\begin{eqnarray}\biggl(2\frac{p(q-1)}{q(p-1)}\biggr)^{q-1}\leqslant 4^{q-1}\leqslant \frac{1}{4}\unicode[STIX]{x1D6FC}^{4},\end{eqnarray}$$
so
$$\begin{eqnarray}\Vert Y_{i}\Vert _{q}\leqslant {\textstyle \frac{1}{4}}\unicode[STIX]{x1D6FC}^{4}\Vert Y_{i}\Vert _{p}\quad \text{for }2\log _{2}\unicode[STIX]{x1D6FC}=q\geqslant p\geqslant 2.\end{eqnarray}$$
Thus for any value of
$\unicode[STIX]{x1D6FC}$
we get
$$\begin{eqnarray}\Vert Y_{i}\Vert _{q}\leqslant \max \biggl\{2,\frac{1}{2}\unicode[STIX]{x1D6FC}^{4}\biggr\}\frac{q}{p}\Vert Y_{i}\Vert _{p}\quad \text{for }q\geqslant p\geqslant 2.\end{eqnarray}$$
Hence, by [Reference Latała and Tkocz9, Theorem 2.3] the variables
$Y_{1},\ldots ,Y_{n}$
satisfy (2.1) (in fact for arbitrary, not only unconditional sets
$T$
) and the assertion follows by Proposition 2.1.◻
3 Symmetrization argument
We will use the following proposition to prove that we may skip the unconditionality assumption in Corollary 2.2.
Proposition 3.1. Let
$(X_{i})_{i=1}^{n}$
be a sequence of independent random variables with finite second moments and let
$(\unicode[STIX]{x1D700}_{i})_{i=1}^{n}$
be a Bernoulli sequence independent of
$(X_{i})_{i=1}^{n}$
. Then for any
$T\subset \mathbb{R}^{n}$
and
$p\geqslant 1$
,
$$\begin{eqnarray}\mathbb{E}_{X}\sup _{t\in T}\biggl(\mathbb{E}_{\unicode[STIX]{x1D700}}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}X_{i}\biggr|^{p}\biggr)^{\!1/p}\leqslant C\biggl[\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}X_{i}+\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}X_{i}\biggr|^{p}\biggr)^{\!1/p}\biggr].\end{eqnarray}$$
Proof. Since this is only a matter of normalization we may and will assume that
$\mathbb{E}X_{i}^{2}=1$
for all
$i$
.
Let
$m$
be such an integer that
$2m\leqslant p<2(m+1)$
. Then, by the symmetry of
$X_{i},\unicode[STIX]{x1D700}_{i}$
, and the independence of
$X_{1},\ldots ,X_{n},\unicode[STIX]{x1D700}_{1},\ldots ,\unicode[STIX]{x1D700}_{n}$
we have
$$\begin{eqnarray}\displaystyle \mathop{\biggl\|\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}X_{i}\biggr\|}\nolimits_{p} & {\geqslant} & \displaystyle \mathop{\biggl\|\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}X_{i}\biggr\|}\nolimits_{2m}\nonumber\\ \displaystyle & = & \displaystyle \biggl(\mathop{\sum }_{i_{1}+\cdots +i_{n}=m}c_{i_{1},\ldots ,i_{n}}t_{1}^{2i_{1}}\cdots t_{n}^{2i_{n}}\mathbb{E}X_{1}^{2i_{1}}\cdots \mathbb{E}X_{n}^{2i_{n}}\biggr)^{1/2m}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \biggl(\mathop{\sum }_{i_{1}+\cdots +i_{n}=m}c_{i_{1},\ldots ,i_{n}}t_{1}^{2i_{1}}\cdots t_{n}^{2i_{n}}\biggr)^{1/2m}\nonumber\\ \displaystyle & = & \displaystyle \biggl(\mathop{\sum }_{i_{1}+\cdots +i_{n}=m}c_{i_{1},\ldots ,i_{n}}t_{1}^{2i_{1}}\cdots t_{n}^{2i_{n}}\mathbb{E}\unicode[STIX]{x1D700}_{1}^{2i_{1}}\cdots \mathbb{E}\unicode[STIX]{x1D700}_{n}^{2i_{n}}\biggr)^{1/2m}\nonumber\\ \displaystyle & = & \displaystyle \mathop{\biggl\|\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}\biggr\|}\nolimits_{2m},\nonumber\end{eqnarray}$$
where
$$\begin{eqnarray}c_{i_{1},\ldots ,i_{n}}=\frac{(2i_{1}+\cdots +2i_{n})!}{(2i_{1})!\cdots (2i_{n})!}.\end{eqnarray}$$
Moreover by the result of Hitczenko [Reference Hitczenko4],
$$\begin{eqnarray}\mathop{\biggl\|\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}\biggr\|}\nolimits_{2m}\geqslant \frac{1}{C}\biggl[\mathop{\sum }_{i\leqslant 2m}t_{i}^{\ast }+\sqrt{2m}\sqrt{\mathop{\sum }_{i>2m}|t_{i}^{\ast }|^{2}}\biggr],\end{eqnarray}$$
where
$(t_{i}^{\ast })_{i=1}^{n}$
denotes the non-increasing rearrangement of
$(|t_{i}|)_{i=1}^{n}$
.
Therefore to establish (3.1) it is enough to show that
$$\begin{eqnarray}\mathbb{E}\sup _{t\in T}\biggl(\mathbb{E}_{\unicode[STIX]{x1D700}}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\leqslant C\biggl(\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}X_{i}+pa\biggr),\end{eqnarray}$$
where
$$\begin{eqnarray}a:=\frac{1}{p}\sup _{t\in T}\biggl(\mathop{\sum }_{i\leqslant p}t_{i}^{\ast }+\sqrt{p}\biggl(\mathop{\sum }_{i>p}|t_{i}^{\ast }|^{2}\biggr)^{1/2}\biggr).\end{eqnarray}$$
To this end, observe that since
$$\begin{eqnarray}\displaystyle \mathop{\biggl\|\mathop{\sum }_{i=1}^{n}u_{i}\unicode[STIX]{x1D700}_{i}\biggr\|}\nolimits_{p}\leqslant C\sqrt{p}\Vert u\Vert _{2},\qquad \mathop{\biggl\|\mathop{\sum }_{i=1}^{n}u_{i}\unicode[STIX]{x1D700}_{i}\biggr\|}\nolimits_{p}\leqslant \Vert u\Vert _{1}, & & \displaystyle \nonumber\\ \displaystyle & & \displaystyle \nonumber\end{eqnarray}$$
and
$$\begin{eqnarray}\displaystyle \mathop{\biggl\|\mathop{\sum }_{i=1}^{n}u_{i}\unicode[STIX]{x1D700}_{i}\biggr\|}\nolimits_{p}=\mathop{\biggl\|\mathop{\sum }_{i=1}^{n}|u_{i}|\unicode[STIX]{x1D700}_{i}\biggr\|}\nolimits_{p}, & & \displaystyle \nonumber\end{eqnarray}$$
we have
$$\begin{eqnarray}\mathop{\biggl\|\mathop{\sum }_{i=1}^{n}u_{i}\unicode[STIX]{x1D700}_{i}\biggr\|}\nolimits_{p}\leqslant \mathop{\sum }_{i=1}^{n}(|u_{i}|-a)_{+}+C\sqrt{p}\biggl(\mathop{\sum }_{i=1}^{n}\min \{u_{i}^{2},a^{2}\}\biggr)^{1/2}.\end{eqnarray}$$
Thus
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathbb{E}_{X}\sup _{t\in T}\biggl(\mathbb{E}_{\unicode[STIX]{x1D700}}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}(|t_{i}X_{i}|-a)_{+}+C\sqrt{p}\biggl(\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}\min \{(t_{i}X_{i})^{2},a^{2}\}\biggr)^{1/2}.\quad\end{eqnarray}$$
To estimate the first term above observe that
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}(|t_{i}X_{i}|-a)_{+}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \sup _{t\in T}\mathbb{E}\mathop{\sum }_{i=1}^{n}(|t_{i}X_{i}|-a)_{+}+\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}((|t_{i}X_{i}|-a)_{+}-\mathbb{E}(|t_{i}X_{i}^{\prime }|-a)_{+}),\nonumber\end{eqnarray}$$
where
$(X_{i}^{\prime })_{i}$
is a copy of
$(X_{i})$
, independent of
$(\unicode[STIX]{x1D700}_{i})$
and
$(X_{i})$
.
Observe that for any
$u$
and
$i$
$$\begin{eqnarray}\mathbb{E}(|uX_{i}|-a)_{+}\leqslant |u|\mathbb{E}|X_{i}|\leqslant |u|\Vert X_{i}\Vert _{2}=|u|\end{eqnarray}$$
and, by the Cauchy–Schwarz inequality and the Markov inequality
$$\begin{eqnarray}\displaystyle \mathbb{E}(|uX_{i}|-a)_{+} & {\leqslant} & \displaystyle |u|\mathbb{E}|X_{i}|I_{\{|X_{i}|\geqslant a/|u|\}}\leqslant |u|\Vert X_{i}\Vert _{2}(\mathbb{P}(|X_{i}|\geqslant a/|u|))^{1/2}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle |u|\Vert X_{i}\Vert _{2}^{2}\frac{|u|}{a}=\frac{u^{2}}{a}.\nonumber\end{eqnarray}$$
Hence for any
$t\in T$
$$\begin{eqnarray}\mathop{\sum }_{i=1}^{n}\mathbb{E}(|t_{i}X_{i}|-a)_{+}\leqslant \mathop{\sum }_{i\leqslant p}t_{i}^{\ast }+\frac{1}{a}\mathop{\sum }_{i>p}(t_{i}^{\ast })^{2}\leqslant 2pa.\end{eqnarray}$$
Moreover, by the Jensen inequality
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}((|t_{i}X_{i}|-a)_{+}-\mathbb{E}(|t_{i}X_{i}^{\prime }|-a)_{+})\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}((|t_{i}X_{i}|-a)_{+}-(|t_{i}X_{i}^{\prime }|-a)_{+})\nonumber\\ \displaystyle & & \displaystyle \quad =\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}\unicode[STIX]{x1D700}_{i}((|t_{i}X_{i}|-a)_{+}-(|t_{i}X_{i}^{\prime }|-a)_{+})\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}\unicode[STIX]{x1D700}_{i}(|t_{i}X_{i}|-a)_{+}+\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}-\unicode[STIX]{x1D700}_{i}(|t_{i}X_{i}^{\prime }|-a)_{+}\nonumber\\ \displaystyle & & \displaystyle \quad =2\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}\unicode[STIX]{x1D700}_{i}(|t_{i}X_{i}|-a)_{+}.\nonumber\end{eqnarray}$$
Function
$x\mapsto (|x|-a)_{+}$
is
$1$
-Lipschitz, so Talagrand’s comparison theorem for Bernoulli processes [Reference Talagrand13, Theorem 2.1] yields
$$\begin{eqnarray}\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}\unicode[STIX]{x1D700}_{i}(|t_{i}X_{i}|-a)_{+}\leqslant \mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}X_{i}.\end{eqnarray}$$
Therefore
$$\begin{eqnarray}\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}(|t_{i}X_{i}|-a)_{+}\leqslant 2pa+2\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}X_{i}.\end{eqnarray}$$
Now we turn our attention to the other term in (3.3). We have
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}\min \{(t_{i}X_{i})^{2},a^{2}\}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \sup _{t\in T}\mathbb{E}\mathop{\sum }_{i=1}^{n}\min \{(t_{i}X_{i})^{2},a^{2}\}\nonumber\\ \displaystyle & & \displaystyle \qquad +\,\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}(\min \{(t_{i}X_{i})^{2},a^{2}\}-\mathbb{E}\min \{(t_{i}X_{i})^{2},a^{2}\}).\nonumber\end{eqnarray}$$
We have
$$\begin{eqnarray}\mathop{\sum }_{i=1}^{n}\mathbb{E}\min \{(t_{i}X_{i})^{2},a^{2}\}\leqslant \mathop{\sum }_{i=1}^{n}\min \{a^{2},t_{i}^{2}\mathbb{E}X_{i}^{2}\}\leqslant pa^{2}+\mathop{\sum }_{i>p}(t_{i}^{\ast })^{2}\leqslant 2pa^{2}.\end{eqnarray}$$
Moreover, by the Jensen inequality
$$\begin{eqnarray}\displaystyle & & \displaystyle \mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}(\min \{(t_{i}X_{i})^{2},a^{2}\}-\mathbb{E}\min \{(t_{i}X_{i}^{\prime })^{2},a^{2}\})\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant \mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}(\min \{(t_{i}X_{i})^{2},a^{2}\}-\min \{(t_{i}X_{i}^{\prime })^{2},a^{2}\})\nonumber\\ \displaystyle & & \displaystyle \quad =\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}\unicode[STIX]{x1D700}_{i}(\min \{(t_{i}X_{i})^{2},a^{2}\}-\min \{(t_{i}X_{i}^{\prime })^{2},a^{2}\})\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant 2\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}\unicode[STIX]{x1D700}_{i}\min \{(t_{i}X_{i})^{2},a^{2}\}.\nonumber\end{eqnarray}$$
Function
$x\mapsto \min \{x^{2},a^{2}\}$
is
$2a$
-Lipschitz, so using the comparison theorem for Bernoulli processes again we get
$$\begin{eqnarray}\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}\unicode[STIX]{x1D700}_{i}\min \{(t_{i}X_{i})^{2},a^{2}\}\leqslant 2a\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}X_{i}.\end{eqnarray}$$
Thus
$$\begin{eqnarray}\displaystyle p\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}\min \{(t_{i}X_{i})^{2},a^{2}\} & {\leqslant} & \displaystyle 2p^{2}a^{2}+4pa\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}X_{i}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \biggl(2pa+\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}\unicode[STIX]{x1D700}_{i}X_{i}\biggr)^{2}.\end{eqnarray}$$
Proof of Theorem 1.1.
Since it is enough to consider
$T\cup (-T)$
instead of
$T$
, we may and will assume that the set
$T$
is symmetric, i.e.
$T=-T$
.
Assume first that the variables
$X_{i}$
are also symmetric. Let
$\unicode[STIX]{x1D700}=(\unicode[STIX]{x1D700}_{i})_{i=1}^{n}$
be a Bernoulli sequence independent of
$(X_{i})_{i=1}^{n}$
. Weak and strong moments of
$(\unicode[STIX]{x1D700}_{i})_{i=1}^{n}$
are comparable
$$\begin{eqnarray}\biggl(\mathbb{E}\sup _{s\in S}\biggl|\mathop{\sum }_{i=1}^{n}s_{i}\unicode[STIX]{x1D700}_{i}\biggr|^{p}\biggr)^{1/p}\leqslant C\biggl[\mathbb{E}\sup _{s\in S}\biggl|\mathop{\sum }_{i=1}^{n}s_{i}\unicode[STIX]{x1D700}_{i}\biggr|+\sup _{s\in S}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}s_{i}\unicode[STIX]{x1D700}_{i}\biggr|^{p}\biggr)^{1/p}\biggr].\end{eqnarray}$$
Hence the symmetry of
$X_{i}$
yields
$$\begin{eqnarray}\displaystyle \biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p} & = & \displaystyle \biggl(\mathbb{E}_{X}\mathbb{E}_{\unicode[STIX]{x1D700}}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\unicode[STIX]{x1D700}_{i}\biggr|^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 2C[\biggl(\mathbb{E}_{X}\biggl(\mathbb{E}_{\unicode[STIX]{x1D700}}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\unicode[STIX]{x1D700}_{i}\biggr|\biggr)^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & & \displaystyle +\,\biggl(\mathbb{E}_{X}\sup _{t\in T}\mathbb{E}_{\unicode[STIX]{x1D700}}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\unicode[STIX]{x1D700}_{i}\biggr|^{p}\biggr)^{1/p} ],\end{eqnarray}$$
since
$(a+b)^{p}\leqslant 2^{p}(a^{p}+b^{p})$
.
Since
$T$
is symmetric, we have for
$x\in \mathbb{R}^{n}$
,
$$\begin{eqnarray}\mathbb{E}_{\unicode[STIX]{x1D700}}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}x_{i}\unicode[STIX]{x1D700}_{i}\biggr|=\sup _{t\in T_{1}}\mathop{\sum }_{i=1}^{n}t_{i}x_{i},\end{eqnarray}$$
where
$$\begin{eqnarray}T_{1}:=\{(\mathbb{E}_{\unicode[STIX]{x1D700}}s_{i}(\unicode[STIX]{x1D700})\unicode[STIX]{x1D700}_{i})_{i=1}^{n}:~s:\{-1,1\}^{n}\rightarrow T\}\end{eqnarray}$$
is an unconditional subset of
$\mathbb{R}^{n}$
. Estimate (2.7) applied for
$T_{1}$
instead of
$T$
yields
$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl(\mathbb{E}_{X}\biggl(\mathbb{E}_{\unicode[STIX]{x1D700}}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\unicode[STIX]{x1D700}_{i}\biggr|\biggr)^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant C(\unicode[STIX]{x1D6FC})\biggl[\mathbb{E}_{X}\mathbb{E}_{\unicode[STIX]{x1D700}}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\unicode[STIX]{x1D700}_{i}\biggr|+\sup _{t\in T_{1}}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\biggr].\nonumber\end{eqnarray}$$
By the symmetry of
$X_{i}$
we have
$$\begin{eqnarray}\mathbb{E}_{X}\mathbb{E}_{\unicode[STIX]{x1D700}}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\unicode[STIX]{x1D700}_{i}\biggr|=\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}X_{i}.\end{eqnarray}$$
Moreover,
$$\begin{eqnarray}T_{1}\subset S(T):=\text{conv}\{(\unicode[STIX]{x1D702}_{i}t_{i})_{i=1}^{n}:\unicode[STIX]{x1D702}\in \{-1,1\}^{n},t\in T\},\end{eqnarray}$$
hence
$$\begin{eqnarray}\sup _{t\in T_{1}}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\leqslant \sup _{t\in S(T)}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}=\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}.\end{eqnarray}$$
Thus
$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl(\mathbb{E}_{X}\biggl(\mathbb{E}_{\unicode[STIX]{x1D700}}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\unicode[STIX]{x1D700}_{i}\biggr|\biggr)^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant C(\unicode[STIX]{x1D6FC})\biggl[\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|+\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\biggr].\end{eqnarray}$$
Let
$q=p/(p-1)$
be the Hölder’s dual of
$p$
. For
$x\in \mathbb{R}^{n}$
we have
$$\begin{eqnarray}\biggl(\sup _{t\in T}\mathbb{E}_{\unicode[STIX]{x1D700}}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}x_{i}\unicode[STIX]{x1D700}_{i}\biggr|^{p}\biggr)^{1/p}=\sup _{t\in T_{2}}\mathop{\sum }_{i=1}^{n}t_{i}x_{i},\end{eqnarray}$$
where
$$\begin{eqnarray}T_{2}=\{(\mathbb{E}_{\unicode[STIX]{x1D700}}t_{i}\unicode[STIX]{x1D700}_{i}h(\unicode[STIX]{x1D700}))_{i=1}^{n}:~t\in T,h:\{-1,1\}^{n}\rightarrow \mathbb{R},\mathbb{E}_{\unicode[STIX]{x1D700}}|h(\unicode[STIX]{x1D700})|^{q}\leqslant 1\}\end{eqnarray}$$
is an unconditional subset of
$\mathbb{R}^{n}$
. Estimate (2.7) applied for
$T_{2}$
instead of
$T$
yields
$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl(\mathbb{E}_{X}\sup _{t\in T}\mathbb{E}_{\unicode[STIX]{x1D700}}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\unicode[STIX]{x1D700}_{i}\biggr|^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant C(\unicode[STIX]{x1D6FC})\biggl[\mathbb{E}_{X}\biggl(\sup _{t\in T}\mathbb{E}_{\unicode[STIX]{x1D700}}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\unicode[STIX]{x1D700}_{i}\biggr|^{p}\biggr)^{1/p}+\sup _{t\in T_{2}}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\biggr].\nonumber\end{eqnarray}$$
Proposition 3.1 and the symmetry of
$X_{i}$
gives
$$\begin{eqnarray}\mathbb{E}_{X}\biggl(\sup _{t\in T}\mathbb{E}_{\unicode[STIX]{x1D700}}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\unicode[STIX]{x1D700}_{i}\biggr|^{p}\biggr)^{1/p}\leqslant C\biggl[\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}X_{i}+\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\biggr].\end{eqnarray}$$
Since
$T_{2}\subset S(T)$
we have
$$\begin{eqnarray}\sup _{t\in T_{2}}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\leqslant \sup _{t\in S(T)}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}=\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}.\end{eqnarray}$$
Thus
$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl(\mathbb{E}_{X}\sup _{t\in T}\mathbb{E}_{\unicode[STIX]{x1D700}}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\unicode[STIX]{x1D700}_{i}\biggr|^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant C(\unicode[STIX]{x1D6FC})\biggl[\mathbb{E}\sup _{t\in T}\mathop{\sum }_{i=1}^{n}t_{i}X_{i}+\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\biggr].\end{eqnarray}$$
Estimate (1.4) follows (for symmetric
$X_{i}$
) by (3.6)–(3.8)
In the case when the variables
$X_{i}$
are centered, but not necessarily symmetric let
$(X_{1}^{\prime },\ldots ,X_{n}^{\prime })$
be an independent copy of
$(X_{1},\ldots ,X_{n})$
. Then
$X_{i}-X_{i}^{\prime }$
are symmetric. The Jensen inequality and the assumption on
$X_{i}$
imply that for any
$p\geqslant 2$
we have
$$\begin{eqnarray}\Vert X_{i}-X_{i}^{\prime }\Vert _{2p}\leqslant 2\Vert X_{i}\Vert _{2p}\leqslant 2\unicode[STIX]{x1D6FC}\Vert X_{i}-\mathbb{E}X_{i}\Vert _{p}\leqslant 2\unicode[STIX]{x1D6FC}\Vert X_{i}-X_{i}^{\prime }\Vert _{p}.\end{eqnarray}$$
Therefore, Theorem 1.1 applied to
$(X_{1}-X_{1}^{\prime },\ldots ,X_{n}-X_{n}^{\prime })$
implies
$$\begin{eqnarray}\displaystyle & & \displaystyle \biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & & \displaystyle \quad =\biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}(X_{i}-\mathbb{E}X_{i}^{\prime })\biggr|^{p}\biggr)^{1/p}\leqslant \biggl(\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}(X_{i}-X_{i}^{\prime })\biggr|^{p}\biggr)^{1/p}\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant C(2\unicode[STIX]{x1D6FC})\biggl[\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}(X_{i}-X_{i}^{\prime })\biggr|+\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}(X_{i}-X_{i}^{\prime })\biggr|^{p}\biggr)^{1/p}\biggr]\nonumber\\ \displaystyle & & \displaystyle \quad \leqslant 2C(2\unicode[STIX]{x1D6FC})\biggl[\mathbb{E}\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|+\sup _{t\in T}\biggl(\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)^{1/p}\biggr],\nonumber\end{eqnarray}$$
which completes the proof in the general case. ◻
Remark 3.2. It follows by the proof of [Reference Latała and Tkocz9, Theorem 2.3] that if
$(X_{i})_{i=1}^{n}$
are symmetric, independent and for any
$i$
moments of
$X_{i}$
grow
$\unicode[STIX]{x1D6FD}$
-regularly (i.e. (2.9) holds with
$\unicode[STIX]{x1D6FD}$
instead of
$2$
), then the comparison of weak and strong moments of suprema of linear combinations of variables
$X_{i}$
holds with a constant
$C(\unicode[STIX]{x1D6FD})=C\unicode[STIX]{x1D6FD}^{11}$
. Therefore, we may follow the constants in the proofs above to obtain that Theorem 1.1 holds with
$C(\unicode[STIX]{x1D6FC})=C^{\log _{2}^{2}\unicode[STIX]{x1D6FC}}$
.
4 From comparison of weak and strong moments to comparison of weak and strong tails
In this section we prove Corollaries 1.3 and 1.4. To this end we need the following lemma.
Lemma 4.1. Assume
$X_{1},X_{2},\ldots$
satisfy the assumptions of Theorem 1.1. Then for any
$t\in \mathbb{R}^{n}$
,
$$\begin{eqnarray}\mathop{\biggl\|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr\|}\nolimits_{p}\leqslant C(\unicode[STIX]{x1D6FC})\biggl(\frac{p}{q}\biggr)^{\max \{1/2,\log _{2}\unicode[STIX]{x1D6FC}\}}\mathop{\biggl\|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr\|}\nolimits_{q}\quad \text{ for }p\geqslant q\geqslant 2.\end{eqnarray}$$
Proof. Let
$\unicode[STIX]{x1D6FD}:=\max \{1/2,\log _{2}\unicode[STIX]{x1D6FC}\}$
. It is enough to show that for positive integers
$k\leqslant l$
we have
$$\begin{eqnarray}\mathop{\biggl\|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr\|}\nolimits_{2k}\leqslant C\unicode[STIX]{x1D6FC}\biggl(\frac{k}{l}\biggr)^{\unicode[STIX]{x1D6FD}}\mathop{\biggl\|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr\|}\nolimits_{2l}.\end{eqnarray}$$
A standard symmetrization argument shows that we may assume that the random variables
$X_{i}$
are symmetric (see the proof of Theorem 1.1 in the non-symmetric case).
Using the hypercontractivity method [Reference Kwapień and Woyczyński5, §3.3], it is enough to show that for
$1\leqslant i\leqslant n$
,
$$\begin{eqnarray}\biggl\|s+\frac{t}{2\sqrt{2}e\unicode[STIX]{x1D6FC}}\biggl(\frac{l}{k}\biggr)^{\unicode[STIX]{x1D6FD}}X_{i}\biggr\|_{2k}\leqslant \Vert s+tX_{i}\Vert _{2l}\quad \text{for all }s,t\in \mathbb{R}.\end{eqnarray}$$
This reduces to the following claim.
Claim. Suppose that
$Y$
is a symmetric random variable such that
$\Vert Y\Vert _{2p}\leqslant \unicode[STIX]{x1D6FC}\Vert Y\Vert _{p}$
for some
$\unicode[STIX]{x1D6FC}\geqslant 1$
and every
$p\geqslant 2$
. Let
$k\geqslant l$
be positive integers. Then
$$\begin{eqnarray}\Vert 1+\unicode[STIX]{x1D70E}Y\Vert _{2k}\leqslant \Vert 1+Y\Vert _{2l},\quad \text{where }\unicode[STIX]{x1D70E}:=\frac{1}{2\sqrt{2}e\unicode[STIX]{x1D6FC}}\biggl(\frac{l}{k}\biggr)^{\unicode[STIX]{x1D6FD}}.\end{eqnarray}$$
To show the claim observe first that
$$\begin{eqnarray}\Vert Y\Vert _{q}\leqslant \unicode[STIX]{x1D6FC}\biggl(\frac{q}{p}\biggr)^{\log _{2}\unicode[STIX]{x1D6FC}}\Vert Y\Vert _{p}\leqslant \unicode[STIX]{x1D6FC}\biggl(\frac{q}{p}\biggr)^{\unicode[STIX]{x1D6FD}}\Vert Y\Vert _{p}\quad \text{for }q\geqslant p\geqslant 2.\end{eqnarray}$$
Moreover we have
$$\begin{eqnarray}\displaystyle \mathbb{E}|1+\unicode[STIX]{x1D70E}Y|^{2k} & = & \displaystyle 1+\mathop{\sum }_{j=1}^{k}\binom{2k}{2j}\mathbb{E}|\unicode[STIX]{x1D70E}Y|^{2j}\leqslant 1+\mathop{\sum }_{j=1}^{k}\biggl(\frac{ek}{j}\unicode[STIX]{x1D70E}\Vert Y\Vert _{2j}\biggr)^{2j}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle 1+\sup _{1\leqslant j\leqslant k}\biggl(\frac{\sqrt{2}ek}{j}\unicode[STIX]{x1D70E}\Vert Y\Vert _{2j}\biggr)^{2j},\nonumber\end{eqnarray}$$
so it is enough to show that
$$\begin{eqnarray}1+\biggl(\frac{k^{1-\unicode[STIX]{x1D6FD}}l^{\unicode[STIX]{x1D6FD}}}{2j\unicode[STIX]{x1D6FC}}\Vert Y\Vert _{2j}\biggr)^{2j}\leqslant \Vert 1+Y\Vert _{2l}^{2k}\quad \text{ for }j=1,2,\ldots ,k.\end{eqnarray}$$
To this end we will use the following deterministic inequality:
$$\begin{eqnarray}(1+u)^{p}\geqslant \biggl(1+\frac{p}{q}u\biggr)^{q}\geqslant 1+\biggl(\frac{p}{q}u\biggr)^{q}\quad \text{for }p\geqslant q\geqslant 1\text{ and }u\geqslant 0,\end{eqnarray}$$
and a simple lower bound for
$\Vert 1+Y\Vert _{2l}^{2l}$
:
$$\begin{eqnarray}\mathbb{E}|1+Y|^{2l}=1+\mathop{\sum }_{r=1}^{l}\binom{2l}{2r}\mathbb{E}|Y|^{2r}\geqslant 1+\mathop{\sum }_{r=1}^{l}\biggl(\frac{l}{r}\Vert Y\Vert _{2r}\biggr)^{2r}.\end{eqnarray}$$
Assume first that
$1\leqslant j\leqslant k/l$
. Estimate (4.2) applied with
$p=2j$
and
$q=2$
yields
$$\begin{eqnarray}\frac{k^{1-\unicode[STIX]{x1D6FD}}l^{\unicode[STIX]{x1D6FD}}}{2j\unicode[STIX]{x1D6FC}}\Vert Y\Vert _{2j}\leqslant \frac{k^{1-\unicode[STIX]{x1D6FD}}l^{\unicode[STIX]{x1D6FD}}}{j^{1-\unicode[STIX]{x1D6FD}}}\Vert Y\Vert _{2}\leqslant \sqrt{\frac{kl}{j}}\Vert Y\Vert _{2},\end{eqnarray}$$
where the last inequality holds since
$\unicode[STIX]{x1D6FD}\geqslant 1/2$
and
$k\geqslant jl$
. Inequalities (4.5) and (4.4) (applied with
$p=k/l$
and
$q=j$
) yield
$$\begin{eqnarray}\Vert 1+Y\Vert _{2l}^{2k}\geqslant (1+(l\Vert Y\Vert _{2})^{2})^{k/l}\geqslant 1+\biggl(\sqrt{\frac{kl}{j}}\Vert Y\Vert _{2}\biggr)^{2j}\end{eqnarray}$$
so (4.3) holds for
$j\leqslant k/l$
.
If
$j\geqslant k/l$
we choose
$r=\lceil jl/k\rceil$
, then
$jl\leqslant kr\leqslant 2jl$
. Since
$1\leqslant r\leqslant l$
, the estimate (4.5) gives
$$\begin{eqnarray}\displaystyle \Vert 1+Y\Vert _{2l}^{2k} & {\geqslant} & \displaystyle \biggl(1+\biggl(\frac{l}{r}\Vert Y\Vert _{2r}\biggr)^{2r}\biggr)^{k/l}\geqslant \biggl(1+\biggl(\frac{l}{r}\Vert Y\Vert _{2r}\biggr)^{2r}\biggr)^{j/r}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle 1+\biggl(\frac{l}{r}\Vert Y\Vert _{2r}\biggr)^{2j},\nonumber\end{eqnarray}$$
where to get the last two inequalities we used
$k/l\geqslant j/r$
and
$j/r\geqslant 1$
. Applying estimate (4.2) with
$2j$
and
$2r$
instead of
$p$
and
$q$
we get
$$\begin{eqnarray}\frac{k^{1-\unicode[STIX]{x1D6FD}}l^{\unicode[STIX]{x1D6FD}}}{2j\unicode[STIX]{x1D6FC}}\Vert Y\Vert _{2j}\leqslant \frac{k^{1-\unicode[STIX]{x1D6FD}}l^{\unicode[STIX]{x1D6FD}}}{2j}\biggl(\frac{j}{r}\biggr)^{\unicode[STIX]{x1D6FD}}\Vert Y\Vert _{2r}\leqslant \frac{k}{2j}\Vert Y\Vert _{2r}\leqslant \frac{l}{r}\Vert Y\Vert _{2r},\end{eqnarray}$$
which completes the proof of the claim in the remaining case. ◻
Proof of Corollary 1.3.
Let
$$\begin{eqnarray}S:=\sup _{t\in T}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|.\end{eqnarray}$$
By the Paley–Zygmund inequality and (4.1) we have for
$t\in T$
and
$p\geqslant 2$
,
$$\begin{eqnarray}\displaystyle \mathbb{P}\biggl(\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|\geqslant \frac{1}{2}\mathop{\biggl\|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr\|}\nolimits_{p}\biggr) & = & \displaystyle \mathbb{P}\biggl(\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\geqslant 2^{-p}\mathbb{E}\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|^{p}\biggr)\nonumber\\ \displaystyle & {\geqslant} & \displaystyle (1-2^{-p})^{2}\biggl({\displaystyle \frac{\Vert \!\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\Vert _{p}}{\Vert \!\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\Vert _{2p}}}\biggr)^{2p}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle e^{-C_{4}(\unicode[STIX]{x1D6FC})p}.\end{eqnarray}$$
In order to show (1.7) we consider 3 cases.
Case 1.
$2u<\sup _{t\in T}\Vert \!\sum _{i=1}^{n}t_{i}X_{i}\Vert _{2}$
. Then by (4.6)
$$\begin{eqnarray}\sup _{t\in T}\mathbb{P}\biggl(\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|\geqslant u\biggr)\geqslant e^{-2C_{4}(\unicode[STIX]{x1D6FC})}\end{eqnarray}$$
and (1.7) obviously holds if
$C_{2}(\unicode[STIX]{x1D6FC})\geqslant \exp (2C_{4}(\unicode[STIX]{x1D6FC}))$
.
Case 2.
$\sup _{t\in T}\Vert \!\sum _{i=1}^{n}t_{i}X_{i}\Vert _{2}\leqslant 2u<\sup _{t\in T}\Vert \sum _{i=1}^{n}t_{i}X_{i}\Vert _{\infty }$
. Let us then define
$$\begin{eqnarray}p:=\sup \biggl\{q\geqslant 2C_{4}(\unicode[STIX]{x1D6FC}):\sup _{t\in T}\mathop{\biggl\|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr\|}\nolimits_{q/C_{4}(\unicode[STIX]{x1D6FC})}\leqslant 2u\biggr\}.\end{eqnarray}$$
By (4.6) we have
$$\begin{eqnarray}\sup _{t\in T}\mathbb{P}\biggl(\biggl|\mathop{\sum }_{i=1}^{n}t_{i}X_{i}\biggr|\geqslant u\biggr)\geqslant e^{-p}.\end{eqnarray}$$
By (4.1) we have
$\sup _{t\in T}\Vert \!\sum _{i=1}^{n}t_{i}X_{i}\Vert _{p}\leqslant C(\unicode[STIX]{x1D6FC})u$
, so by Theorem 1.1 and Chebyshev’s inequality we have
$$\begin{eqnarray}\mathbb{P}(S\geqslant C_{1}(\unicode[STIX]{x1D6FC})(\mathbb{E}S+u))\leqslant \mathbb{P}(S\geqslant e\Vert S\Vert _{p})\leqslant e^{-p}\end{eqnarray}$$
for
$C_{1}(\unicode[STIX]{x1D6FC})$
large enough. Thus (1.7) holds in this case.
Case 3.
$2u>\sup _{t\in T}\Vert \!\sum _{i=1}^{n}t_{i}X_{i}\Vert _{\infty }=\Vert S\Vert _{\infty }$
. Then
$\mathbb{P}(S\geqslant 2u)=0$
and (1.7) holds for any
$C_{1}(\unicode[STIX]{x1D6FC})\geqslant 2$
.◻
5 Comparison of weak and strong moments of suprema implies comparison of moments
$p$
and
$2p$
Proof of Theorem 1.2.
We will use the assumption (1.5) for
$T$
containing all vectors of the standard base of
$\mathbb{R}^{n}$
and their negatives, i.e. we will use only the inequality
$$\begin{eqnarray}\Bigl(\mathbb{E}\sup _{1\leqslant i\leqslant n}|X_{i}|^{p}\Bigr)^{1/p}\leqslant L\Bigl[\mathbb{E}\sup _{1\leqslant i\leqslant n}|X_{i}|+\Vert X_{1}\Vert _{p}\Bigr].\end{eqnarray}$$
Fix
$p\geqslant 2$
and let
$n:=\lfloor (4L)^{2p}\rfloor +1$
,
$A:=n^{1/p}\Vert X_{1}\Vert _{p}$
. If
$A\geqslant \Vert X_{1}\Vert _{2p}$
, then (1.6) holds with
$\unicode[STIX]{x1D6FC}=(4L)^{2}+1$
. Hence we may and will assume
$A\leqslant \Vert X_{1}\Vert _{2p}$
.
Obviously
$$\begin{eqnarray}\mathbb{P}\Bigl(\sup _{1\leqslant i\leqslant n}|X_{i}|\geqslant t\Bigr)\leqslant \min \{1,n\mathbb{P}(|X_{1}|\geqslant t)\}.\end{eqnarray}$$
Moreover, if
$\mathbb{P}(|X_{1}|\geqslant t)\leqslant 1/n$
,
$$\begin{eqnarray}\displaystyle \mathbb{P}\biggl(\sup _{1\leqslant i\leqslant n}|X_{i}|\geqslant t\biggr) & = & \displaystyle 1-\mathbb{P}(|X_{1}|<t)^{n}=\mathbb{P}(|X_{1}|\geqslant t)\mathop{\sum }_{k=0}^{n-1}\mathbb{P}(|X_{1}|<t)^{k}\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \mathbb{P}(|X_{1}|\geqslant t)\cdot n\biggl(1-\frac{1}{n}\biggr)^{n-1}\geqslant \frac{n}{3}\mathbb{P}(|X_{1}|\geqslant t).\nonumber\end{eqnarray}$$
Since
$\mathbb{P}(|X_{1}|\geqslant A)\leqslant 1/n$
(which follows by the Markov inequality) and
$A\leqslant \Vert X_{1}\Vert _{2p}$
, we have
$$\begin{eqnarray}\displaystyle \mathbb{E}\sup _{1\leqslant i\leqslant n}|X_{i}|^{2p} & {\geqslant} & \displaystyle 2p\int _{A}^{\infty }t^{2p-1}\mathbb{P}\Bigl(\sup _{1\leqslant i\leqslant n}|X_{i}|\geqslant t\Bigr)\,dt\nonumber\\ \displaystyle & {\geqslant} & \displaystyle 2p\int _{A}^{\infty }t^{2p-1}\frac{n}{3}\mathbb{P}(|X_{1}|\geqslant t)\,dt\nonumber\\ \displaystyle & = & \displaystyle \frac{n}{3}\mathbb{E}(|X_{1}|^{2p}-A^{2p})_{+}\geqslant \frac{n}{3}(\Vert X_{1}\Vert _{2p}^{2p}-A^{2p})\nonumber\\ \displaystyle & {\geqslant} & \displaystyle \frac{n}{3}(\Vert X_{1}\Vert _{2p}-A)^{2p}\nonumber\end{eqnarray}$$
and
$$\begin{eqnarray}\displaystyle \mathbb{E}\sup _{1\leqslant i\leqslant n}|X_{i}| & {\leqslant} & \displaystyle A+\int _{A}^{\infty }\mathbb{P}\Bigl(\sup _{1\leqslant i\leqslant n}|X_{i}|\geqslant t\Bigr)\,dt\leqslant A+n\int _{A}^{\infty }\mathbb{P}(|X_{1}|\geqslant t)\,dt\nonumber\\ \displaystyle & {\leqslant} & \displaystyle A+n\mathbb{E}(|X_{1}|\mathbf{1}_{\{|X_{1}|\geqslant A\}})\leqslant A+n\Vert X_{1}\Vert _{p}\mathbb{P}(|X_{1}|\geqslant A)^{1-1/p}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle A+n^{1/p}\Vert X_{1}\Vert _{p},\nonumber\end{eqnarray}$$
where in the last inequality we used again the fact that
$\mathbb{P}(|X_{1}|\geqslant A)\leqslant 1/n$
.
Thus our choice of
$n$
and
$A$
, and (5.1) (applied to
$2p$
instead of
$p$
) imply that
$$\begin{eqnarray}\displaystyle 2L\Vert X_{1}\Vert _{2p} & {\leqslant} & \displaystyle {\textstyle \frac{1}{2}}n^{1/2p}\Vert X_{1}\Vert _{2p}\leqslant {\textstyle \frac{1}{2}}n^{1/2p}A+\Bigl(\mathbb{E}\sup _{1\leqslant i\leqslant n}|X_{i}|^{2p}\Bigr)^{1/2p}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle {\textstyle \frac{1}{2}}n^{1/2p}A+L\biggl[\mathbb{E}\sup _{1\leqslant i\leqslant n}|X_{i}|+\Vert X_{1}\Vert _{2p}\biggr]\nonumber\\ \displaystyle & {\leqslant} & \displaystyle {\textstyle \frac{1}{2}}n^{1/2p}A+LA+Ln^{1/p}\Vert X_{1}\Vert _{p}+L\Vert X_{1}\Vert _{2p}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle \Vert X_{1}\Vert _{p}({\textstyle \frac{1}{2}}(4L+1)n^{1/p}+2Ln^{1/p})+L\Vert X_{1}\Vert _{2p}\nonumber\\ \displaystyle & {\leqslant} & \displaystyle (4L+{\textstyle \frac{1}{2}})((4L)^{2}+1)\Vert X_{1}\Vert _{p}+L\Vert X_{1}\Vert _{2p}.\nonumber\end{eqnarray}$$
Thus
$$\begin{eqnarray}\Vert X_{1}\Vert _{2p}\leqslant \biggl(4+\frac{1}{2L}\biggr)(16L^{2}+1)\Vert X_{1}\Vert _{p}.\square\end{eqnarray}$$
Remark 5.1. It is clear from the proof above that we may take
$\unicode[STIX]{x1D6FC}(L)=CL^{2}$
in Theorem 1.2.
