1 IG of Poincaré inequalities …
9
The arguments used here differs from those used, for example, in [21], which are
based on the equation for the infinitesimal generator Eqs. (1.11) and (1.12). We will
come to that point later. Notice that we can take Φ(s) = s
2 and derive a Poincaré
inequality with a non-optimal constant > 1.
We can prove now a set of inequalities of the Poincaré type. The first example is
Φ(s) = e
s . In such a case, the equation for the moment generating function of the
Gaussian distribution gives
Φ(a) =
exp
π
2
az
γ (z) dz = exp
π
2 a
2
8
,
so that the inequality (1.14) becomes
exp
f (x) − f
γ (x) dx ≤
exp
π
2
8
|∇ f (x)|
2
γ (x) dx .
More clearly, we can change f to
2κ
π
f and write
exp
2κ
π
f (x) − f
γ (x) dx ≤
exp
κ
2
2
|∇ f (x)|
2
γ (x) dx =
(2π)
n/2
exp
−
1
2
|x|
2
− κ
2
|∇ f (x)|
2
dx .
(1.16)
The inequality above is non-trivial only if the RHS is bounded, that is
|∇ f (x)| < κ
−1
|x| , x ∈ R
n
,
that is, the function f is Lipschitz. We have found that f ∈ C
1
(R
n
) and globally
Lipschitz implies that f is sub-exponential in the Gaussian space.
The first case of bound for Orlicz norms we consider is the Lebesgue norm,
Φ(s) = s
2 p , p > 1/2. In such a case,
Φ(a) =
π
2
2 p
m(2 p) a
2 p
,
where m(2 p) is the 2 p-moment of the standard Gaussian distribution. It follows that
f −
f (y) γ (y) dy
L 2 p (γ )
≤
π
2
(m(2 p))
1/2 p
|∇ f || L 2 p (γ ) .
The cases Φ(a) = a
2 p are special in that we can use the in the proof the multiplicative property Φ(ab) = Φ(a)Φ(b). The argument generalizes to the case where
9
The arguments used here differs from those used, for example, in [21], which are
based on the equation for the infinitesimal generator Eqs. (1.11) and (1.12). We will
come to that point later. Notice that we can take Φ(s) = s
2 and derive a Poincaré
inequality with a non-optimal constant > 1.
We can prove now a set of inequalities of the Poincaré type. The first example is
Φ(s) = e
s . In such a case, the equation for the moment generating function of the
Gaussian distribution gives
Φ(a) =
exp
π
2
az
γ (z) dz = exp
π
2 a
2
8
,
so that the inequality (1.14) becomes
exp
f (x) − f
γ (x) dx ≤
exp
π
2
8
|∇ f (x)|
2
γ (x) dx .
More clearly, we can change f to
2κ
π
f and write
exp
2κ
π
f (x) − f
γ (x) dx ≤
exp
κ
2
2
|∇ f (x)|
2
γ (x) dx =
(2π)
n/2
exp
−
1
2
|x|
2
− κ
2
|∇ f (x)|
2
dx .
(1.16)
The inequality above is non-trivial only if the RHS is bounded, that is
|∇ f (x)| < κ
−1
|x| , x ∈ R
n
,
that is, the function f is Lipschitz. We have found that f ∈ C
1
(R
n
) and globally
Lipschitz implies that f is sub-exponential in the Gaussian space.
The first case of bound for Orlicz norms we consider is the Lebesgue norm,
Φ(s) = s
2 p , p > 1/2. In such a case,
Φ(a) =
π
2
2 p
m(2 p) a
2 p
,
where m(2 p) is the 2 p-moment of the standard Gaussian distribution. It follows that
f −
f (y) γ (y) dy
L 2 p (γ )
≤
π
2
(m(2 p))
1/2 p
|∇ f || L 2 p (γ ) .
The cases Φ(a) = a
2 p are special in that we can use the in the proof the multiplicative property Φ(ab) = Φ(a)Φ(b). The argument generalizes to the case where