Generalized entropy, fractional logarithmic Sobolev inequality, and Beckner type inequality

Qing-Ping Liu^{∗} Nippon Institute of Technology

Ichiro Shigekawa^{†}^{‡} Kyoto University

Abstract

We discuss a generalization of entropy function in terms of Orlicz functional. Orlicz functional is defined by using Young function but we extend a scope of Young function adapted to entropy function. We introduce the ϕ-Sobolev inequality and discuss the relation between the spectral gap and the ϕ-Sobolev inequality. Taking function ϕ to be {log(1 + x)}^{γ}, we can define the fractional logarihtmic Sobolev inequality. The connection with the Latała-Oleszkiewicz inequality is also discussed.

1. Introduction

In this paper, we try to generalize the notion of entropy. To do this, we use Orlicz spaces. To define an Orlicz space, we need a Young function but our Young function belongs to a wider class than usual one. A Young function is defied as follows:

Φ(x) = ∫_{0}^{x} ϕ(t)dt.

Here ϕ is a non-negative right-continuous non-decreasing function with ϕ(0) = 0. In our definition, we do not assume the positivity of ϕ. By using this Young function, we can define an Orlicz functional and we see that the entropy function is a typical example of Orlicz functional. To be more precise, let (M, m) be a probability space. The entropy function is defied as follows:

Ent(f) = E[f log f] − E[f] log E[f].

∗e-mail: liu@math.kyoto-u.ac.jp

†e-mail: ichiro@math.kyoto-u.ac.jp, URL: http://www.math.kyoto-u.ac.jp/~ichiro/

‡This research was partially supported by the Ministry of Education, Culture, Sports, Science and Technology, Grant-in-Aid for Scientific Research (B), No. 17340036

Here E stands for the integration with respect to m. We will define an Orlicz functional ∥ ∥_{Φ} (for the definition, see Section 2) so that if we take ϕ(x) = log x, then

∥f∥_{Φ} = Ent(f).

Using this Orlicz functional, we define a ϕ-Sobolev inequality and discuss its properties. Under suitable conditions, we can show that the ϕ-Sobolev inequality is in between the Poincaré inequality and the logarithmic Sobolev inequality. Such kind of inequality was discussed by, e.g., F.-Y. Wang [9], D. Chafaï[3], and we give a different formulation. We also discuss the fractional logarithmic Sobolev inequality. In particular, we discuss these inequalities in L^{p} setting. A typical inequality is the following:

E [|f|^{p} log_{+}^{pβ}(|f|^{p}/∥f∥_{p}^{p})] ≤ K_{1}∥∇f∥_{p}^{p} + K_{2}∥f∥_{p}^{p}.

Further we consider the following Beckner type inequality:

\frac{E[|f|^{p}] − E[|f|^{q}]^{p/q}}{(p − q)^{δ}} ≤ L_{1}∥∇f∥_{p}^{p} + L_{2}∥f∥_{p}^{p}.

Latała-Oleszkiewicz [5] discussed this type of inequality in the case p = 2. We generalize it to the L^{p} case.

The organization of the paper is as follows. In Section 2, we discuss the Orlicz functional for a generalized Young function. We define ϕ-Sobolev inequality and defective ϕ-Sobolev inequality and show that ϕ-Sobolev inequality is stronger than the Poincaré inequality. In Section 3, we introduce the fractional logarithmic Sobolev inequality and discuss it in L^{p} setting. Last, in Section 4, we discuss the relation between the fractional logarithmic Sobolev inequality and the Beckner type inequality,

2. Generalized entropy and Poincaré inequality

Young functions play a fundamental role in the Orlicz space theory. In this section, we treat a little wider class of functions. Usually, we are given a function ϕ satisfying ϕ(0) = 0, ϕ(t) > 0 if t > 0 and ϕ(∞) = ∞. Φ is defined to be an integral of ϕ. Here, we generalize ϕ so that

1. ϕ: [0, ∞) → ℝ ∪ {−∞} is continuous and strictly increasing.

2. ϕ(∞) = ∞.

3. ϕ is integrable on any finite interval of [0, ∞).

4. ϕ is of class C^{1}.

In particular, ϕ can be negative and unbounded from below. Therefore it may happen that ϕ(0) = lim_{t→0} ϕ(t) = −∞. Typical example is ϕ(t) = log t. We will define a functional, which is an entropy when ϕ(t) = log t. Φ is defined as

Φ(x) = ∫_{0}^{x} ϕ(t) dt

and satisfies the following:

1. Φ is convex and Φ(0) = 0.

2. \frac{Φ(x)}{x} is strictly increasing, lim_{x→0} \frac{Φ(x)}{x} = ϕ(0), and lim_{x→∞} \frac{Φ(x)}{x} = ϕ(∞).

The inverse function ψ is defined by

ψ(u) = inf{t; ϕ(t) > u}.

Note that ψ(u) is defined for all u ∈ ℝ and ψ(u) = 0 for u ≤ ϕ(0). The complementary function Ψ is defined by

Ψ(y) = ∫_{−∞}^{y} ψ(u) du.

Then the following holds:

(2.1)  xy ≤ Φ(x) + Ψ(y),
(2.2)  xϕ(x) = Φ(x) + Ψ(ϕ(x)).

(2.1) is called the Young inequality. We also have

Ψ(y) = sup_{x≥0}{xy − Φ(x)}. (2.3)

Suppose we are given a measure space (M, m). We assume that a reference measure m is a probability measure. We denote the integration with respect to m by E[ ]. From the Young inequality, we have the following proposition.

Proposition 2.1. Let U, V ≥ 0 be functions on the measure space and let E denote a integration with respect to a finite measure. Suppose −∞ < E[Uϕ(U)] < ∞. If

E[Uϕ(U)] ≤ E[V ϕ(U)] + C, (2.4)

then we have

E[Φ(U)] ≤ E[Φ(V)] + C. (2.5)

Here we use the convention xϕ(x) = 0 for x = 0.

Proof. (2.4) means that P [V > 0, U = 0] = 0 when ϕ(0) = −∞. By the Young inequality, we have

E[V ϕ(U)] ≤ E[Φ(V)] + E[Ψ(ϕ(U))],
E[Uϕ(U)] = E[Φ(U)] + E[Ψ(ϕ(U))].

Since E[Uϕ(U)] < ∞, it follows that E[Ψ(ϕ(U))] < ∞ and

E[Φ(U)] = E[Uϕ(U)] − E[Ψ(ϕ(U))]
≤ E[V ϕ(U)] + C − E[Ψ(ϕ(U))]
≤ E[Φ(V)] + C,

which is the desired inequality.

We introduce some functionals in connection to Φ. Usually, these functions become norms but they do not in our case because ϕ is not positive.

First define N_{Φ} by

N_{Φ}(f) = inf{λ > 0; E[Φ(|f|/λ)] ≤ 1}. (2.6)

We denote by L^{Φ} the set of all functions with N_{Φ}(f) < ∞ and call it an Orlicz space. For f ∈ L^{Φ}, F (t) = E[Φ(t|f|)] is a convex function. It is easy to see that F (T) = 1 if and

only if T = \frac{1}{N_{Φ}(f)}. We denote the minimum of Φ by −m. N_{Φ} satisfies the following.

Proposition 2.2. We have

(1) N_{Φ}(f) ≥ 0.

(2) N_{Φ}(f) = 0 if and only if f = 0.

(3) N_{Φ}(cf) = |c|N_{Φ}(f).

(4) If f_{1}, f_{2} ≥ 0, then N_{Φ}(f_{1} + f_{2}) ≤ N_{Φ}(f_{1}) + N_{Φ}(f_{2}).

(5) For general f_{1}, f_{2}, N_{Φ}(f_{1} + f_{2}) ≤ (1 + m)(N_{Φ}(f_{1}) + N_{Φ}(f_{2})).

Proof. (1), (2), (3) are easy.

To show (4), set a_{i} = N_{Φ}(f_{i}), b = a_{1} + a_{2}. Then

E[Φ(\frac{f_{1} + f_{2}}{b} )] = E[Φ(\frac{a_{1}}{b} \frac{f_{1}}{a_{1}} + \frac{a_{2}}{b} \frac{f_{2}}{a_{2}} )]

≤ E[\frac{a_{1}}{b} Φ(\frac{f_{1}}{a_{1}} ) + \frac{a_{2}}{b} Φ(\frac{f_{2}}{a_{2}} )]

≤ \frac{a_{1}}{b} + \frac{a_{2}}{b} = 1.

Now N_{Φ}(f_{1} + f_{2}) ≤ b follows.

To show (5), set Φ_{+} = max{Φ, 0}. Φ_{+} is a non-negative, increasing and convex function satisfying Φ_{+}(0) = 0. In particular, Φ_{+}(θx) ≤ θΦ_{+}(x), 0 < θ < 1. We also note that Φ ≤ Φ_{+} ≤ Φ + m. Now, setting a_{i} = N_{Φ}(f_{i}), b = a_{1} + a_{2}, we have

E[Φ(\frac{|f_{1} + f_{2}|}{(1 + m)b})] ≤ E[Φ_{+}(\frac{|f_{1} + f_{2}|}{(1 + m)b})]

≤ E[Φ_{+}(\frac{|f_{1}| + |f_{2}|}{(1 + m)b})]

≤ E[Φ_{+}(\frac{a_{1}}{b} \frac{|f_{1}|}{(1 + m)a_{1}} + (\frac{a_{2}}{b} \frac{|f_{2}|}{(1 + m)a_{2}} )]

≤ \frac{a_{1}}{b} E[Φ_{+}(\frac{|f_{1}|}{(1 + m)a_{1}} )] + \frac{a_{2}}{b} E[Φ_{+}(\frac{|f_{2}|}{(1 + m)a_{2}} )]

≤ \frac{a_{1}}{b} \frac{1}{1 + m}E[Φ_{+}(\frac{|f_{1}|}{a_{1}})] + \frac{a_{2}}{b} \frac{1}{1 + m}E[Φ_{+}(\frac{|f_{2}|}{a_{2}})]

≤ \frac{a_{1}}{b} \frac{1}{1 + m}E[Φ(\frac{|f_{1}|}{a_{1}}) + m] + \frac{a_{2}}{b} \frac{1}{1 + m}E[Φ(\frac{|f_{2}|}{a_{2}}) + m]

= 1,

which yields (5).

The property (5) shows that N_{Φ} is a quasi-norm and so L^{Φ} is a topological linear space. This topology is stronger than L^{1} as will be shown below. Proposition 2.3. The following inequality holds: ∥f∥_{1} ≤ (1 + Ψ(1))N_{Φ}(f). (2.7) Proof. If N_{Φ}(f) = 1, then the Young inequality xy ≤ Φ(x) + Ψ(y) yields

E[|f|y] ≤ E[Φ(|f|) + Ψ(y)] = 1 + Ψ(y).

Setting y = 1,

∥f∥_{1} ≤ 1 + Ψ(1).

Now (2.7) follows easily.

We introduce another functional ∥ ∥_{Φ}. Define ∥f∥_{Φ} = sup{E[|f|g]; E[Ψ(g)] ≤ 1}. (2.8) We call ∥ ∥_{Φ} the Orlicz functional. If Φ is a nice Young function, then ∥ ∥_{Φ} becomes a norm. But it does not in our case as we remarked. Proposition 2.4. Let f ≥ 0 be a non-trivial function. Choose λ > 0 so that E[Ψ(ϕ(λf))] = 1. (2.9) Then we have

∥f∥_{Φ} = E[fϕ(λf)] = \frac{1}{λ}\{E[Φ(λf)] + 1\}. (2.10)

For general f, we have

∥f∥_{Φ} = inf_{κ>0} \frac{1}{κ}\{E[Φ(κ|f|)] + 1\}. (2.11)

Proof. First we assume E[Ψ(ϕ(f))] = 1, i.e., in the case of λ = 1. If g satisfies E[Ψ(g)] ≤ 1, then, by the Young inequality, we have

E[fg] ≤ E[Φ(f)] + E[Ψ(g)] ≤ E[Φ(f)] + 1.

If we take g = ϕ(f), then E[Ψ(g)] ≤ 1 and E[fg] = E[fϕ(f)] = E[Φ(f)] + E[Ψ(ϕ(f))] = E[Φ(f)] + 1.

This shows that ∥f∥_{Φ} = E[fϕ(f)] = E[Φ(f)] + 1.

Next, for non-negative f, we take λ satisfying (2.9). The above result implies

∥λf∥_{Φ} = E[λfϕ(λf)] = E[Φ(λf)] + 1

and (2.10) follows because of ∥λf∥_{Φ} = λ∥f∥_{Φ}.

Last we see (2.11). If E[Ψ(g)] ≤ 1, then, for κ > 0,

E[|f|g] = \frac{1}{κ}E[κ|f|g]

≤ \frac{1}{κ}\{E[Φ(κ|f|)] + E[Ψ(g)]\}

≤ \frac{1}{κ}\{E[Φ(κ|f|)] + 1\}.

Hence

E[|f|g] ≤ inf_{κ>0} \frac{1}{κ}\{E[Φ(κ|f|)] + 1\}.

Since g is arbitrary, we have

∥f∥_{Φ} ≤ inf_{κ>0} \frac{1}{κ}\{E[Φ(κ|f|)] + 1\}.

We have already seen that the equality holds if κ = λ. Now the proof is complete.

We see some properties of the Orlicz functional ∥f∥_{Φ}.

Proposition 2.5. The Orlicz functional ∥ ∥_{Φ} satisfies the following

(1) For f ≢ 0, Ψ^{−1}(1) ≤ \frac{∥f∥_{Φ}}{∥f∥_{1}} . The equality holds if and only if |f| is constant.

(2) ∥cf∥_{Φ} = |c|∥f∥_{Φ}.

(3) For f_{1}, f_{2} ≥ 0, ∥f_{1} + f_{2}∥_{Φ} ≤ ∥f_{1}∥_{Φ} + ∥f_{2}∥_{Φ}.

Proof. Take g so that g ≡ Ψ^{−1}(1). Then E[Ψ(g)] = 1 and so

E[|f|Ψ^{−1}(1)] ≤ ∥f∥_{Φ}.

This means that

Ψ^{−1}(1) ≤ \frac{∥f∥_{Φ}}{∥f∥_{1}} .

It is easy to see that the equality holds when |f| is constant.

We will show the converse: if the equality holds, then |f| is constant. We take λ so that E[Ψ(ϕ(λ|f|)] = 1. From Proposition 2.4, we have

∥f∥_{Φ} = \frac{1}{λ}\{E[Φ(λ|f|)] + 1\}.

Therefore

∥λf∥_{Φ} = E[Φ(λ|f|)] + 1
≥ Φ(E[λ|f|]) + 1 (∵ Φ is convex)
= Φ(E[λ|f|]) + Ψ(Ψ^{−1}(1))
≥ E[λ|f|]Ψ^{−1}(1) (∵ the Young inequality)
= ∥λf∥_{Φ}.

Hence, all inequalities above should be equalities. In particular, E[Φ(λf)] = Φ(E[λf]) and so f is constant since Φ is strictly convex.

(3) is easy.

Let us see the relation between N_{Φ} and ∥ ∥_{Φ}.

Proposition 2.6. The following inequality holds:

∥f∥_{Φ} ≤ 2N_{Φ}(f). (2.12)

Proof. For any λ, we have

∥f∥_{Φ} ≤ \frac{1}{λ}\{E[Φ(λf) + 1\}.

Taking λ = N_{Φ}(f)^{−1},

∥f∥_{Φ} ≤ N_{Φ}(f)\{E [Φ(\frac{f}{N_{Φ}(f)})] + 1\} ≤ 2N_{Φ}(f),

which is the desired result.

When we are given two Young functions Φ_{1}, Φ_{2}, we can have the following comparisons. If there exists a constant a so that Φ_{1}(x) ≤ Φ_{2}(ax), x ≥ 0, then

(1) N_{Φ_{1}}(f) ≤ aN_{Φ_{2}}(f).

(2) ∥f∥_{Φ_{1}} ≤ a∥f∥_{Φ_{2}}.

The proof is standard (e.g., see [8, II.2.2]) and will be omitted.

To get the comparison, we only need the inequality near ∞, i.e., Φ_{1}(x) ≤ Φ_{2}(ax), x ≥ x_{0} for some constant x_{0}. In this case, we can find a constant C so that N_{Φ_{1}}(f) ≤ CN_{Φ_{2}}(f).

From now on, we add the following assumption to ϕ:

sup_{x>0} xϕ′(x) = l < ∞. (2.13)

This means that the growth order of ϕ is less than that of log.

Proposition 2.7. Under the assumption (2.13), we have

N_{Φ}(f) ≤ max{∥f∥_{Φ}, l∥f∥_{1}}. (2.14)

Proof. Using xϕ′(x) ≤ l,

\frac{d}{dx}Ψ ∘ ϕ(x) = Ψ′(ϕ(x))ϕ′(x) = ϕ^{−1} ∘ ϕ(x)ϕ′(x) = xϕ′(x) ≤ l.

Since Ψ ∘ ϕ(0) = 0, we haveΨ ∘ ϕ(x) ≤ lx. Therefore

E [Ψ(ϕ(\frac{f}{l∥f∥_{1}} ))] ≤ E [l\frac{f}{l∥f∥_{1}} ] = 1.

Now we have

∥f∥_{Φ} = sup{E [fg] ; E [Ψ(g)] ≤ 1}

≥ E [fϕ(\frac{f}{max\{∥f∥_{Φ}, l∥f∥_{1}\}})]

= max\{∥f∥_{Φ}, l∥f∥_{1}\}E [\frac{f}{max\{∥f∥_{Φ}, l∥f∥_{1}\}}ϕ(\frac{f}{max\{∥f∥_{Φ}, l∥f∥_{1}\}})]

= max\{∥f∥_{Φ}, l∥f∥_{1}\}E [Φ(\frac{f}{max\{∥f∥_{Φ}, l∥f∥_{1}\}}) + Ψ(ϕ(\frac{f}{max\{∥f∥_{Φ}, l∥f∥_{1}\}}))] .

This brings

E [Φ(\frac{f}{max\{∥f∥_{Φ}, l∥f∥_{1}\}})] ≤ \frac{∥f∥_{Φ}}{max\{∥f∥_{Φ}, l∥f∥_{1}\}} − E [Ψ(ϕ(\frac{f}{max\{∥f∥_{Φ}, l∥f∥_{1}\}}))]
≤ 1.

This shows (2.14).

Under the assumption (2.14), let us further investigate ∥ ∥_{Φ}. Our aim is to generalize the inequality in Deuschel-Stroock [4, Chapter VI (6.1.26)]. Define a constant α as α = ϕ^{−1} ∘ Ψ^{−1}(1). (2.15) Proposition 2.8. Let f be a bounded function and set f̂ = f − E [f]. Then we have

(2.16)  ∥f^{2}∥_{Φ} ≤ ∥f̂^{2}∥_{Φ} + 2l∥f̂∥_{2}^{2} + Ψ^{−1}(1)E [f]^{2} ,
(2.17)  ∥f̂^{2}∥_{Φ} ≤ ∥f^{2}∥_{Φ} + 2l∥f∥_{2}^{2} − Ψ^{−1}(1)E [f]^{2}

and further

lim_{t↓0} \frac{∥(1 + tf)^{2}∥_{Φ} − Ψ^{−1}(1) − 2Ψ^{−1}(1)E [f] t}{t^{2}} = 2αϕ′(α)∥f̂∥_{2}^{2} + Ψ^{−1}(1)E [f^{2}] . (2.18)

Proof. We first show ∥(1 + tf)^{2}∥_{Φ} ≤ Ψ^{−1}(1) + Ψ^{−1}(1)E [f] t + t^{2}∥f^{2}∥_{Φ} + 2lt^{2}E [f^{2}] . (2.19) To do this, take any δ > 0 and set

g_{t} = ϕ(λ_{t}{(1 + tf)^{2} + δ}).

Here λ_{t} is taken so that E [Ψ(g_{t})] = 1. By Proposition 2.4, we have ∥(1 + tf)^{2} + δ∥_{Φ} = E [{(1 + tf)^{2} + δ}g_{t}]. Since E [Ψ(g_{t})] = 1, by differentiating in t, we have

0 = E [Ψ′(g_{t})g′_{t}] = E [ψ ∘ ϕ(λ_{t}{(1 + tf)^{2} + δ})g′_{t}] = λ_{t}E [{(1 + tf)^{2} + δ}g′_{t}] , which yields E [{(1 + tf)^{2} + δ}g′_{t}] = 0. (2.20) On the other hand, g′_{t} is computed as

g′_{t} = ϕ′λ′_{t}{(1 + tf)^{2} + δ} + ϕ′2λ_{t}(1 + tf)f.

Here we have omitted the variable λ_{t}{(1 + tf)^{2} + δ} of ϕ′. Substituting this into (2.20),

0 = E [{(1 + tf)^{2} + δ}ϕ′{λ′_{t}{(1 + tf)^{2} + δ} + 2λ_{t}(1 + tf)f}]
= λ′_{t}E [{(1 + tf)^{2} + δ}^{2}ϕ′] + 2λ_{t}E [ϕ′{(1 + tf)^{2} + δ}(1 + tf)f]

and therefore

λ′_{t} = −\frac{2λ_{t}E [ϕ′\{(1 + tf)^{2} + δ\}(1 + tf)f]}{E [\{(1 + tf)^{2} + δ\}^{2}ϕ′]}. (2.21)

Now we set

K_{t} = ∥(1 + tf)^{2} + δ∥_{Φ} − t^{2}∥f^{2}∥_{Φ} = E [{(1 + tf)^{2} + δ}g_{t}] − t^{2}∥f^{2}∥_{Φ}.

Differentiating this, we have

K′_{t} = E [{(1 + tf)^{2} + δ}g′_{t}] + E [2(1 + tf)fg_{t}] − 2t∥f^{2}∥_{Φ}
= E [2(1 + tf)fg_{t}] − 2t∥f^{2}∥_{Φ}. (∵ (2.20))

Further

K′′_{t} = 2E [f^{2}g_{t}] + E [2(1 + tf)fg′_{t}] − 2∥f^{2}∥_{Φ}

≤ E [2(1 + tf)fg′_{t}]

= E [2(1 + tf)f{ϕ′λ′_{t}{(1 + tf)^{2} + δ} + 2ϕ′λ_{t}(1 + tf)f}] = 2λ′_{t}E [ϕ′{(1 + tf)^{2} + δ}(1 + tf)f] + 4E [ϕ′λ_{t}(1 + tf)^{2}f^{2}]

= −\frac{4λ_{t}E [ϕ′\{(1 + tf)^{2} + δ\}(1 + tf)f]^{2}}{E [\{(1 + tf)^{2} + δ\}ϕ′]}

+ 4E [ϕ′λ_{t}{(1 + tf)^{2} + δ}f^{2}] − 4E [ϕ′λ_{t}δf^{2}] (∵ (2.21)) ≤ 4E [ϕ′λ_{t}{(1 + tf)^{2} + δ}f^{2}] .

The variable of ϕ′ above is omitted. Writing it explicitly, we have ϕ′λ_{t}{(1 + tf)^{2} + δ}f^{2} = ϕ′(λ_{t}{(1 + tf)^{2} + δ})λ_{t}{(1 + tf)^{2} + δ}f^{2} ≤ lf^{2}.

We now obtain

K′′_{t} ≤ 4lE [f^{2}] .

g_{0} satisfies Ψ(g_{0}) = 1 and so g_{0} = Ψ^{−1}(1). By noting that K_{0} = E[(1 + δ)g_{0}] = (1 + δ)Ψ^{−1}(1), K′_{0} = E[2fg_{0}] = 2Ψ^{−1}(1)E [f], we have

K_{t} ≤ (1 + δ)Ψ^{−1}(1) + 2Ψ^{−1}(1)E [f] t + 2lt^{2}E [f^{2}] .

Combining all of them,

∥(1 + tf)^{2} + δ∥_{Φ} ≤ (1 + δ)Ψ^{−1}(1) + 2Ψ^{−1}(1)E [f] t + t^{2}∥f^{2}∥_{Φ} + 2lt^{2}E [f^{2}] .

Letting δ ↓ 0, we get (2.19).

Next, we show (2.16), (2.17). If E [f] = 0, then (2.16), (2.17) are trivial since f = f̂. If E [f] ≠ 0, then we take f̂/E [f] instead of f. Then

∥(\frac{f}{E [f]})^{2}∥_{Φ} = ∥(1 + \frac{f̂}{E [f]})^{2}∥_{Φ} ≤ Ψ^{−1}(1) + ∥(\frac{f̂}{E [f]})^{2}∥_{Φ} + 2l∥\frac{f̂}{E [f]}∥_{2}^{2}

and (2.16) follows. If we take −f/E [f] instead of f, then

∥(\frac{f̂}{E [f]})^{2}∥_{Φ} = ∥(1 − \frac{f}{E [f]})^{2}∥_{Φ} ≤ Ψ^{−1}(1) − 2Ψ^{−1}(1) + ∥(\frac{f}{E [f]})^{2}∥_{Φ} + 2l∥\frac{f}{E [f]}∥_{2}^{2}.

which shows (2.17).

Lastly we show (2.18). Since t is small, we may assume that 1 + tf is positive. Hence we can put δ = 0 in the above argument. We set g_{t} = ϕ(λ_{t}(1 + tf)^{2}) and choose λ_{t} so that E [Ψ(g_{t})] = 1. Set

K_{t} = ∥(1 + tf)^{2}∥_{Φ} = E [(1 + tf)^{2}g_{t}] .

By differentiating in t,

K′_{t} = E [2(1 + tf)fg_{t}] + E [(1 + tf)^{2}g′_{t}] = E [2(1 + tf)fg_{t}]

and

K′′_{t} = E [2f^{2}g_{t}] + E [2(1 + tf)fg′_{t}]

Since g_{0} = Ψ^{−1}(1), K_{0} = Ψ^{−1}(1) and K′_{0} = 2Ψ^{−1}(1)E [f]. Let us compute K′′_{0}.

g′_{t} = ϕ′λ′_{t}(1 + tf)^{2} + ϕ′2λ_{t}(1 + tf)f

and

λ′_{t} = −\frac{2λ_{t}E [ϕ′(1 + tf)^{3}f]}{E [(1 + tf)^{4}ϕ′]}.

Noting that λ_{0} = ϕ^{−1} ∘ Ψ^{−1}(1) = α, we have

λ′_{0} = −\frac{2λ_{0}E [ϕ′(λ_{0})f]}{E [ϕ′(λ_{0})]} = −2αE [f] ,
g′_{0} = ϕ′(λ_{0})λ′_{0} + ϕ′(λ_{o})2λ_{0}f = −2αϕ′(α)E [f] + 2αϕ′(α)f = 2αϕ′(α)f̂.

Thus

K′′_{0} = E [2f^{2}g_{0}] + E [2fg′_{0}] = 2Ψ^{−1}(1)E [f^{2}] + 2E [f2αϕ′(α)f̂]

= 2Ψ^{−1}(1)E [f^{2}] + 4αϕ′(α)E \left[{ \atop f̂^{2}}\right] .

Now we have

lim_{t↓0} \frac{K_{t} − Ψ^{−1}(1) − 2Ψ^{−1}(1)E [f] t}{t^{2}} = \frac{K′′_{0}}{2} = Ψ^{−1}(1)E [f^{2}] + 2αϕ′(α)E \left[{ \atop f̂^{2}}\right] ,

which shows (2.18).

The entropy function is defined by

(2.22) Ent(f) = E[f log f] − E[f] log E[f].

Here f is a non-negative function. Usually f is assumed to satisfy E[f] = 1 but we do not assume it. If we take ϕ(t) = log t, then

Φ(x) = ∫_{0}^{x} log t dt = x log x − x,
Ψ(y) = ∫_{−∞}^{y} e^{u}du = e^{y}

Ent(f) = ∥f∥_{Φ}.

In fact, by Proposition 2.4

∥f∥_{Φ} = inf_{λ>0} E [f log f − f log λ − f + λ] .

The right hand side is nothing but the entropy. So our Orlicz functional is regarded as a generalization of the entropy.

We call the following inequality a ϕ-Sobolev inequality:

∥f^{2}∥_{Φ} ≤ 2λE(f, f) + Ψ^{−1}(1)∥f∥_{2}^{2}. (2.23)

Here E is a Dirichlet form on a probability space (M, m). λ is a non-negative constant and (2.23) holds for all f ∈ Dom(E). We assume that 1 ∈ Dom(E) and E(1, 1) = 0. So the associated generator has the maximal eigenvalue 0 and its eigenfunction 1. The term Ψ^{−1}(1)∥f∥_{2}^{2} is indispensable since the equality holds when f = 1. The following little weakened inequality is called a defective ϕ-Sobolev inequality:

∥f^{2}∥_{Φ} ≤ 2λE(f, f) + Ψ^{−1}(1)∥f∥_{2}^{2} + 2μ∥f∥_{2}^{2}. (2.24)

When ϕ(t) = log t, ∥ ∥_{Φ} is the entropy and (2.23) is a logarithmic Sobolev inequality. So our terminology is consistent with this.

On the other hand the following inequality is called the Poincaré inequality:

∥f̂∥_{2}^{2} ≤ νE(f, f). (2.25)

Here, f̂ = f − E [f]. Now we have the following theorem:

Theorem 2.9. We have the following:

(1) If the ϕ-Sobolev inequality (2.23) holds, then, setting ν = λ/αϕ′(α), the Poincaré inequality (2.25) holds.

(2) If we assume the defective ϕ-Sobolev inequality (2.24) and, in addition, the Poincaré inequality, the ϕ-Sobolev inequality (2.23) holds for λ + (μ + l)ν in place of λ.

Proof. To show (1), from (2.23), we note

∥(1 + tf)^{2}∥_{Φ} ≤ 2λt^{2}E(f, f) + Ψ^{−1}(1)∥1 + tf∥_{2}^{2}
= 2λt^{2}E(f, f) + Ψ^{−1}(1)(1 + 2tE [f] + t^{2}E [f^{2}]).

But, from Proposition 2.8 and (2.18), we have

lim_{t↓0} \frac{∥(1 + tf)^{2}∥_{Φ} − Ψ^{−1}(1)(1 + 2tE [f])}{t^{2}} = 2αϕ′(α)∥f̂∥_{2}^{2} + Ψ^{−1}(1)E [f^{2}]

and hence

2αϕ′(α)∥f̂∥_{2}^{2} ≤ 2λE(f, f).

This is what we wanted.

As for (2), by Proposition 2.8 and (2.16), we have

∥f^{2}∥_{Φ} ≤ Ψ^{−1}(1)E [f]^{2} + ∥f̂^{2}∥_{Φ} + 2l∥f̂∥_{2}^{2}

≤ Ψ^{−1}(1)E [f]^{2} + 2λE(f̂, f̂) + Ψ^{−1}(1)E \left[{ \atop f̂^{2}}\right] + 2μ∥f̂∥_{2}^{2} + 2l∥f̂∥_{2}^{2}

≤ 2λE(f̂, f̂) + 2(μ + l)νE(f, f) + Ψ^{−1}(1)E [f^{2}] = 2(λ + (μ + l)ν)E(f, f) + Ψ^{−1}(1)E [f^{2}] ,

which is the desired result.

3. Fractional Logarithmic Sobolev inequality

When we take ϕ = log^{γ}(1 + x), the inequality

∥f^{2}∥_{Φ} ≤ 2λE(f, f) + Ψ^{−1}(1)∥f∥_{2}^{2} (3.1)

is called a logarithmic Sobolev inequality of the fractional order γ. Here γ > 0 and when γ = 1, it is a usual logarithmic Sobolev inequality and it is natural to take ϕ(x) = log x in this case. Anyway, we call these inequalities fractional logarithmic Sobolev inequalities in general. We are interested in inequalities which is stronger than the Poincaré inequality and is weaker than the logarithmic Sobolev inequality. So, in the sequel, we always assume

0 < γ ≤ 1. (3.2)

We mention here some recent related results. Latała-Oleszkiewicz [5] proposed the intermediate inequality in between the Poincaré inequality and the logarithmic Sobolev inequality. F.-Y. Wang [9] proved the Latała-Oleszkiewicz inequality is equivalent to the strong Poincaré inequality and, at the same time, he proved that it is equivalent to the F -Sobolev inequality. Therefore, inequalities we have discussed are equivalent to the F - Sobolev inequality and we formulated them in the framework of Orlicz space. We have seen that the ϕ-Sobolev inequality is stronger than the Poincaré inequality. On the other hand, the defective logarithmic Sobolev inequality is not necessarily stronger than the Poincaré inequality. So, to be more precise, the F -Sobolev inequality corresponds to the defective logarithmic Sobolev inequality.

So far, our discussions are in L^{2} setting. We are now turning to L^{p} setting. To do so, we change the formulation. We assume that the Dirichlet form is of the gradient form:

E(f, g) = ∫_{M}(∇f, ∇g)dm. (3.3)

We consider the following inequality in L^{p} setting:

∥f∥_{Φ} ≤ K(∥∇f∥_{p} + ∥f∥_{p}). (3.4)

We are interested in this kind of L^{p} inequality. To investigate these inequalities, we prepare a general theory. So we consider a general norm ∥ ∥ for a while. We also need a class of functions Φ. For α > 0, we define Φ ∈ U(α) if and only if

xΦ′(x) ≤ αΦ(x), x ≥ 0.

Proposition 3.1. Assume Φ ∈ U(α). If the following inequality

∥f∥_{Φ} ≤ K∥f∥

holds, then

E [Φ(f)] ≤ K^{α}∥f∥^{α} + 1. (3.5)

Proof. Note

N_{Φ}(f) ≤ ∥f∥_{Φ} ≤ K∥f∥.

Using Φ ∈ U(α), if K∥f∥ ≥ 1, then we have

E [Φ(f)] = E [Φ(K∥f∥\frac{f}{K∥f∥})]

≤ E [K^{α}∥f∥^{α}Φ(\frac{f}{K∥f∥})]

≤ K^{α}∥f∥^{α}.

If K∥f∥ ≤ 1, then

E [Φ(f)] ≤ E [Φ(\frac{f}{K∥f∥})] ≤ 1.

In all cases, (3.5) holds.

We can also show the converse. Proposition 3.2. If

E [Φ(f)] ≤ K∥f∥^{α} + C

for α ≥ 1, then

∥f∥_{Φ} ≤ αK^{1/α}(\frac{C + 1}{α − 1})^{1−1/α}∥f∥. (3.6)

In case of α = 1, (3.6) should be read as

∥f∥_{Φ} ≤ K∥f∥.

Proof. By Proposition 2.4, we have

∥f∥_{Φ} ≤ \frac{1}{λ}\{E [Φ(λf)] + 1\}.

From the assumption,

\frac{1}{λ}\{E [Φ(λf)] + 1\} ≤ \frac{1}{λ}\{Kλ^{α}∥f∥^{α} + C + 1\}

≤ λ^{α−1}K∥f∥^{α} + \frac{1}{λ}(C + 1)

=: g(λ).

It is enough to minimize the g(λ). By an easy computation, g(λ) takes its minimum at λ = ∥f∥^{−1}(\frac{C+1}{K(α−1)})^{1/α} and the minimum is

g(∥f∥^{−1}(\frac{C + 1}{K(α − 1)})^{1/α}) = ∥f∥^{1−α}(\frac{C + 1}{K(α − 1)})^{(α−1)/α}K∥f∥^{α}

+ ∥f∥(\frac{K(α − 1)}{C + 1})^{1/α}(C + 1)

= ∥f∥(\frac{C + 1}{α − 1})^{1−1/α}(K^{−1+1/α} · K + K^{1/α}α)

= αK^{1/α}(\frac{C + 1}{α − 1})^{1−1/α}∥f∥,

which is (3.6).

When α = 1, we have

\frac{1}{λ}\{E [Φ(λf)] + 1\} ≤ \frac{1}{λ}\{Kλ∥f∥ + C + 1\} ≤ K∥f∥ + \frac{1}{λ}(C + 1).

Now letting λ → ∞,

∥f∥_{Φ} ≤ K∥f∥.

This is the desired result.

We now return to the inequality

∥f∥_{Φ} ≤ K(∥∇f∥_{p} + ∥f∥_{p}).

We assume that ϕ is of the following form:

ϕ(x) = ϕ_{p,β}(x) = x^{p−1} log^{pβ}(e + x).

Here p > 1 and β ∈ ℝ. We denote its integration by Φ = Φ_{p,β}:

Φ_{p,β}(x) = ∫_{0}^{x} ϕ_{p,β}(y)dy.

It is not difficult to see that Φ ∈ U(α) for sufficiently large α. From now on, to avoid the change of constant, we use the following convention: we denote A ≲ B if there exists a constant c so that A ≤ cB.

Proposition 3.3. Assume β ≥ 0. Then the inequality

∥f∥_{Φ_{p,β}} ≤ K(∥∇f∥_{p} + ∥f∥_{p}) (3.7)

is equivalent to

E [|f|^{p} log_{+}^{pβ}(|f|^{p}/∥f∥_{p}^{p})] ≤ K_{1}∥∇f∥_{p}^{p} + K_{2}∥f∥_{p}^{p}. (3.8)

Proof. Assume (3.7). From Proposition 3.1, there exists a constant α such that

E [Φ_{p,β}(f)] ≲ (∥∇f∥_{p} + ∥f∥_{p})^{α} + 1.

Using x^{p} log_{+}^{pβ} x ≤ KΦ_{p,β}(x), we have

E [|f|^{p} log_{+}^{pβ} |f|^{p}] ≲ (∥∇f∥_{p} + ∥f∥_{p})^{α} + 1.

We show that the left hand side dominates E [|f|^{p} log_{+}^{pβ}(|f|^{p}/∥f∥_{p}^{p})]. If ∥f∥_{p} ≥ 1, then log_{+}^{pβ}(|f|^{p}/∥f∥_{p}^{p}) ≤ log_{+}^{pβ} |f|^{p} and the result is clear. In the case of ∥f∥_{p} < 1, we have

E [|f|^{p} log_{+}^{pβ}(|f|^{p}/∥f∥_{p}^{p})] ≤ E [|f|^{p}\{log_{+} |f|^{p} + log_{+} \frac{1}{∥f∥_{p}^{p}} \}^{pβ}]

≤ 2^{pβ−1}\{E [|f|^{p} log_{+}^{pβ} |f|^{p}] + E [|f|^{p} log_{+}^{pβ} \frac{1}{∥f∥_{p}^{p}} ]\}

≤ 2^{pβ−1}\{E [|f|^{p} log_{+}^{pβ} |f|^{p}] + ∥f∥_{p}^{p} log_{+}^{pβ} \frac{1}{∥f∥_{p}^{p}} \}.

Since f(x) = x log^{pβ} \frac{1}{x} is bounded on (0, 1], we have

E [|f|^{p} log_{+}^{pβ}(|f|^{p}/∥f∥_{p}^{p})] ≲ E [|f|^{p} log_{+}^{pβ} |f|^{p}] + 1

which leads to

E [|f|^{p} log_{+}^{pβ}(|f|^{p}/∥f∥_{p}^{p})] ≲ (∥∇f∥_{p} + ∥f∥_{p})^{α} + 1.

Substitute λf in place of f,

E [|f|^{p} log_{+}^{pβ}(|f|^{p}/∥f∥_{p}^{p})] ≲ λ^{α−p}(∥∇f∥_{p} + ∥f∥_{p})^{α} + λ^{−p}.

Now we take λ = \frac{1}{∥∇f∥_{p}+∥f∥_{p}}. Then

E [|f|^{p} log_{+}^{pβ}(|f|^{p}/∥f∥_{p}^{p})] ≲ (∥∇f∥_{p} + ∥f∥_{p})^{p−α}(∥∇f∥_{p} + ∥f∥_{p})^{α} + (∥∇f∥_{p} + ∥f∥_{p})^{p}
≲ (∥∇f∥_{p} + ∥f∥_{p})^{p}
≲ ∥∇f∥_{p}^{p} + ∥f∥_{p}^{p}.

Next let us show the reversed inequality. So we assume (3.8). Then

E [|f|^{p} log_{+}^{pβ} |f|^{p}] = E [|f|^{p} log_{+}^{pβ}(|f|^{p}/∥f∥_{p}^{p}) · ∥f∥_{p}^{p}]
≲ E [|f|^{p}(log_{+}^{pβ} |f|^{p}/∥f∥_{p}^{p} + log_{+}^{pβ} ∥f∥_{p}^{p})]
≲ ∥∇f∥_{p}^{p} + ∥f∥_{p}^{p} + ∥f∥_{p}^{p} log_{+}^{pβ} ∥f∥_{p}^{p}.

Using Φ_{p,β}(x) ≲ x^{p} log_{+}^{pβ} x + 1, we have

E [Φ_{p,β}(f)] ≲ ∥∇f∥_{p}^{p} + ∥f∥_{p}^{p} + ∥f∥_{p}^{p} log_{+}^{pβ} ∥f∥_{p}^{p} + 1
≲ (∥∇f∥_{p} + ∥f∥_{p})^{q} + 1.

Here q is any number satisfying q > p. The rest is easy by Proposition 3.2.

Now we assume the fractional logarithmic Sobolev inequality of the following form: E [|f|^{2} log_{+}^{γ}(|f|^{2}/∥f∥_{2}^{2})] ≤ C_{1}∥∇f∥_{2}^{2} + C_{2}∥f∥_{2}^{2}. (3.9) This means that we only assume the defective inequality. From now on, until the end of this section, we always assume (3.9) and we will deduce L^{p} inequalities of the type (3.9).

We introduce a new Young function as follows. Setting (3.10) θ(x) = {x^{2} log^{γ}(e + x^{2})}^{(p−2)/4} log^{pβ/4}(k + x^{2} log^{γ}(e + x^{2})), we define Θ(x) = ∫_{0}^{x} θ(y)dy. (3.11) Then we have Proposition 3.4. There exist k ≥ 1 and K > 0 so that (3.12) x^{p} log^{pγ/2}(e + x^{2}) log^{pβ/2}(k + x^{2} log^{γ}(e + x^{2})) ≤ KΘ(x)^{2} log^{γ}(e + Θ(x)^{2}). Proof. We divide into two cases.

(i) In the case of β ≥ 0, k = 1.

We investigate the asymptotic behavior as x → 0.

LHS ∼ x^{p} · x^{(pβ/2)2} = x^{p(1+β)}.

Here A ∼ B means that lim \frac{A}{B} = 1. Further

θ(x) ∼ x^{(p−2)/2+(pβ)/2} = x^{p(β+1)/2−1}

and hence

Θ(x) ∼ \frac{2}{p(β + 1)}x^{p(β+1)/2},

Θ(x)^{2} ∼ \frac{4}{p^{2}(β + 1)^{2}} x^{p(β+1)}.

We can see that they have the same asymptotics up to constant.

Next, when x → ∞,

LHS ∼ x^{p}2^{pγ/2} log^{pγ/2} x2^{pβ/2} log^{pβ/2} x = 2^{p(β+γ)/2}x^{p} log^{p(β+γ)/2} x.

On the other hand

θ(x) ∼ x^{(p−2)/2}2^{(p−2)γ/4}(log^{(p−2)γ/4} x)2^{pβ/4} log^{pβ/4} x
= 2^{(pβ+pγ−2γ)/4}x^{(p−2)/2} log^{(pβ+pγ−2γ)/4} x,
Θ(x) ∼ (2/p)2^{(pγ+pγ−2γ)/4}x^{p/2} log^{(pβ+pγ−2γ)/4} x
= p^{−1}2^{(pβ+pγ−2γ+4)/4}x^{p/2} log^{(pβ+pγ−2γ)/4} x,

Θ(x)^{2} log^{γ}(e + Θ(x)^{2}) ∼ p^{−2}2^{(pβ+pγ−2γ+4)/2}x^{p}(log^{(pβ+pγ−2γ)/2} x)p^{γ} log^{γ} x
= p^{−2+γ}2^{(pβ+pγ−2γ+4)/2}x^{p} log^{(pβ+pγ)/2} x.

Again, they have the same asymptotics. (ii) In the case of β < 0.

This time, k is taken to be sufficiently large. The behavior near x = ∞ can be checked in a similar manner as in the case β ≥ 0. When x → 0, it is clear that LHS ∼ log^{pβ/2} kx^{p}. Further

θ(x) ∼ (log^{pβ/4} k)x^{(p−2)/2}

Θ(x) ∼ \frac{2 log^{pβ/4} k}{p} x^{p/2}

Θ(x)^{2} log^{γ}(e + Θ(x)^{2}) ∼ \frac{4 log^{pβ/2} k}{p^{2}} x^{p}

and we can see that they have the same asymptotics.

Using this proposition, we have the following

Theorem 3.5. For p > 2, β ∈ ℝ, we have

E[Φ_{p,(β+γ)/2}(|f|)] ≲ E[e + Φ_{p,(β+γ−(2γ/p))/2}(|f|)]{1 + log^{γ}(e + E[Φ_{p,(β+γ−(2γ/p))/2}(|f|)])} (3.13)
+ E[Φ_{p,β/2}(|∇f|)].

Proof. First we set g = \sqrt{e + Θ(|f|)^{2}}. Then, since ∇g = \frac{2Θ(|f|)Θ′(|f|)∇|f|}{2\sqrt{Θ(|f|)^{2}+e}}, we have |∇g| ≤

θ(|f|)|∇f|. Combining this with the fractional logarithmic Sobolev inequality

E[g^{2} log_{+}^{γ} g^{2}/∥g∥_{2}^{2}] ≤ K_{1}E[|∇g|^{2}] + K_{2}E[g^{2}],

we have

E[g^{2} log_{+}^{γ} g^{2}] ≲ E[g^{2} log_{+}^{γ} ∥g∥_{2}^{2}] + E[|∇g|^{2}] + E[g^{2}]

and hence

E[{e+Θ(|f|)^{2}} log^{γ}(e + Θ(|f|)^{2})]
≲ E[e + Θ(|f|)^{2}] log^{γ} E[e + Θ(|f|)^{2}] + E[|∇f|^{2}θ(|f|)^{2}] + E[e + Θ(|f|)^{2}].

Thus we have

E[Θ(|f|)^{2} log^{γ}(e + Θ(|f|)^{2})]
≲ E[e + Θ(|f|)^{2}](1 + log^{γ} E[e + Θ(|f|)^{2}]) + 2λE[|∇f|^{2}θ(|f|)^{2}].

We set ϕ(x) = x^{(p/2)−1} log^{pβ/2}(k + x) and U = |f|^{2} log^{γ}(e + |f|^{2}). Here k is taken to be large enough. Note that θ has been defined to be θ(|f|)^{2} = ϕ(U).

Uϕ(U) = |f|^{2} log^{γ}(e + |f|^{2}){|f|^{2} log^{γ}(e + |f|^{2})}^{(p/2)−1} log^{pβ/2}(k + |f|^{2} log^{γ}(e + |f|^{2}))

= |f|^{p} log^{pγ/2}(e + |f|^{2}) log^{pβ/2}(k + |f|^{2} log^{γ}(e + |f|^{2}))
≲ Θ(|f|)^{2} log^{γ}(e + Θ(|f|)^{2}). (∵ (3.12))

Hence

E[Uϕ(U)] ≲ E[e + Θ(|f|)^{2}](1 + log^{γ} E[e + Θ(|f|)^{2}]) + E[|∇f|^{2}ϕ(U)].

Now, by Proposition 2.1, we have

E[Φ(U)] ≲ E[e + Θ(|f|)^{2}](1 + log^{γ} E[e + Θ(|f|)^{2}]) + E[Φ(C|∇f|^{2})].

Since Φ is an integral of ϕ(x) = x^{(p/2)−1} log^{pβ/2}(k + x), we have

Φ(Cx^{2}) ≲ x^{2}ϕ(x^{2})
= x^{2}(x^{2})^{(p−2)/2} log^{pβ/2}(k + x^{2})
= x^{p} log^{pβ/2}(k + x^{2})
≲ Φ_{p,β/2}(x).

Further

Φ(x^{2} log^{γ}(e + x^{2})) ≳ x^{2} log^{γ}(e + x^{2})ϕ(x^{2} log^{γ}(e + x^{2}))
= x^{2} log^{γ}(e + x^{2}){x^{2} log^{γ}(e + x^{2})}^{(p−2)/2} log^{pβ/2}(k + x^{2} log^{γ}(e + x^{2}))
= x^{p} log^{pγ/2}(e + x^{2}) log^{pβ/2}(k + x^{2} log^{γ}(e + x^{2}))
≳ x^{p} log^{pγ/2}(e + x) log^{pβ/2}(k + x)
≳ x^{p} log^{p(β+γ)/2}(k + x)
≳ Φ_{p,(β+γ)/2}(x)

and

Θ(x)^{2} ≲ x^{2}θ(x)^{2}
= x^{2}{x^{2} log^{γ}(e + x^{2})}^{(p−2)/2} log^{pβ/2}(k + x^{2} log^{γ}(e + x^{2}))
≲ x^{p} log^{(p−2)γ/2}(e + x) log^{pβ/2}(k + x)
≲ x^{p} log^{(pβ+pγ−2γ)/2}(k + x)
= x^{p} log^{p(β+γ−(2γ/p))/2}(k + x)
≲ Φ_{p,(β+γ−(2γ/p))/2−(1/p)}(x).

Combining all of them, we eventually have

E[Φ_{p,(β+γ)/2}(|f|)] ≲ E[e + Φ_{p,(β+γ−(2γ/p))/2}(|f|)](1 + log^{γ} E[e + Φ_{p,(β+γ−(2γ/p))/2}(|f|)])
+ E[Φ_{p,β/2}(|∇f|)].

This is what we wanted.

By the above theorem, it follows that

∥f∥_{Φ_{p,(β+γ)/2}} ≲ ∥f∥_{Φ_{p,(β+γ−(2γ/p))/2}} + ∥∇f∥_{Φ_{p,β/2}}.

This implies that differentiability improves the integrability by the logarithmic order pγ/2.

When p = 2, the same result holds if we impose the additional condition β ≥ 0. We give a proof for completeness. Set

θ(x) = log^{β/2}(1 + x^{2} log(e + x^{2})) (3.14)

and

Θ(x) = ∫_{0}^{x} θ(y)dy. (3.15)

We need the monotonicity of θ, which forces β ≥ 0. Moreover, in the proof of Theorem 3.5, ϕ becomes ϕ = log^{β}(k + x), which also forces β ≥ 0. By being aware of this, we just repeat the same proof. So we have the following

Proposition 3.6. There exists a constant K > 0 so that

(3.16) x^{2} log^{γ}(e + x^{2}) log^{β}(1 + x^{2} log^{γ}(e + x^{2})) ≤ KΘ(x)^{2} log^{γ}(e + Θ(x)^{2}).

Proof. We investigate the asymptotic behavior.

Near x = 0, noting Θ(x) ∼ x^{β+1}, we have Θ(x)^{2} log(e + Θ(x)^{2}) ∼ x^{2β+2}. The left hand side is ∼ x^{2} · x^{2β} = x^{2β+2} and so they have same asymptotics.

Next we consider when x → ∞. This time,

lim_{x→∞} \frac{Θ(x)}{x log^{β/2} x^{2}} = lim_{x→∞} \frac{Θ′(x)}{log^{β/2} x^{2} + β log^{(β/2)−1} x^{2}}

= lim_{x→∞} \frac{log^{β/2}(k + x^{2} log(e + x))}{log^{β/2} x^{2}}

= lim_{x→∞} \frac{log^{β/2} x^{2}}{log^{β/2} x^{2}}

= 1,

which yields log^{γ}(e + Θ(x)^{2}) ∼ log^{γ} x^{2}. Combining them, we have

Θ(x)^{2} log^{γ}(e + Θ(x)^{2}) ∼ x^{2}(log^{β} x^{2}) log^{γ} x^{2} = x^{2} log^{β+γ} x^{2}.

It is easy to see that the left hand side has the same asymptotics. Thus we have obtained (3.16).

To proceed further, it is subtle to see xθ(x) ≲ Θ(x) ≲ xθ(x). In the case of p > 2, θ(x) ≍ xθ′(x) holds but in the case of p = 2, this does not hold. In fact

θ′(x) = \frac{βx\{(e + x^{2})^{γ} log(e + x^{2}) + γx^{2} log^{γ−1}(e + x^{2})\} log^{β/2}(1 + x^{2} log^{γ}(e + x^{2}))}{(e + x^{2})(1 + x^{2} log^{γ}(e + x^{2})) log(1 + x^{2} log^{γ}(e + x^{2}))}.

Hence

lim_{x→∞} \frac{xθ′(x)}{θ(x)} = 0,  lim_{x→0} \frac{xθ′(x)}{θ(x)} = β

and we can show xθ′(x) ≲ θ(x). But the reversed estimate does not hold. We should notice that the reversed estimate does hold for Θ. To see this,

lim_{x→∞} \frac{xθ(x)}{Θ(x)} = lim_{x→∞} \frac{θ(x) + xθ′(x)}{θ(x)} = 1, (∵ xθ′(x)/θ(x) → 0),
lim_{x→0} \frac{xθ(x)}{Θ(x)} = lim_{x→0} \frac{θ(x) + xθ′(x)}{θ(x)} = 1 + β,

which shows xθ(x) ≲ Θ(x) ≲ xθ(x). Noticing this, we have the following.

Theorem 3.7. For p = 2 and β ≥ 0, there exists a constant C > so that

(3.17)  E[Φ_{2,(β+γ)/2}(|f|)] ≤ CE[e + Φ_{p,β/2}(|f|)](2μ + log^{γ}(e + E[Φ_{p,β/2}(|f|)]))
+ CE[Φ_{p,β/2}(|∇f|)].

Proof. Set g = \sqrt{e + Θ(|f|)^{2}}. Then, by noting that |∇g| ≤ θ(|f|)|∇f| since ∇g = \frac{2Θ(|f|)Θ′(|f|)∇|f|}{2\sqrt{Θ(|f|)^{2}+e}}, and by using the defective logarithmic Sobolev inequality

E[g^{2} log^{γ}(g^{2}/∥g∥_{2}^{2})] ≤ 2λE[|∇g|^{2}] + 2μE[g^{2}],

we have

E[{e+Θ(|f|)^{2}} log^{γ}(e + Θ(|f|)^{2})]
≲ E[e + Θ(|f|)^{2}] log^{γ} E[e + Θ(|f|)^{2}] + E[|∇f|^{2}θ(|f|)^{2}] + E[e + Θ(|f|)^{2}]

and hence

E[Θ(|f|)^{2} log^{γ}(e + Θ(|f|)^{2})]
≤ E[e + Θ(|f|)^{2}](1 + log^{γ}(e + E[Θ(|f|)^{2}])) + E[|∇f|^{2}θ(|f|)^{2}].

Now we set ϕ(x) = log^{β}(1 + x) and U = |f|^{2} log^{γ}(e + |f|^{2}). Then, since θ satisfies θ(|f|)^{2} = ϕ(U), we have

Uϕ(U) = |f|^{2} log^{γ}(e + |f|^{2}) log^{β}(1 + |f|^{2} log^{γ}(e + |f|^{2}))
≤ KΘ(|f|)^{2} log^{γ}(e + Θ(|f|)^{2}). (∵ (3.16))

Combining this with the previous result, we obtain

E[Uϕ(U)] ≲ E[e + Θ(|f|)^{2}](1 + log^{γ}(e + E[Θ(|f|)^{2}])) + E[|∇f|^{2}ϕ(U)].

From Proposition 2.1,

E[Φ(U)] ≲ E[e + Θ(|f|)^{2}](1 + log^{γ}(e + E[Θ(|f|)^{2}])) + E[Φ(C|∇f|^{2})].

Since Φ is an integral of ϕ(x) = log^{β}(1 + x), we have

Φ(Cx^{2}) ≲ x^{2}ϕ(x^{2}) = x^{2} log^{β}(1 + x^{2}) ≲ Φ_{2,β/2}(x).

Further

Φ(x^{2} log^{γ}(e + x^{2})) ≳ x^{2} log^{γ}(e + x^{2})ϕ(x^{2} log^{γ}(e + x^{2}))
= x^{2} log^{γ}(e + x^{2}) log^{β}(1 + x^{2} log^{γ}(e + x^{2}))
≳ x^{2} log^{γ}(e + x) log^{β}(e + x) − 1
= x^{2} log^{β+γ}(e + x) − 1
≥ Φ_{2,(β+1)/2}(x) − 1

and

Θ(x)^{2} ≲ x^{2}θ(x)^{2}
= x^{2} log^{β}(1 + x^{2} log^{γ}(e + x^{2}))
≲ x^{2} log^{β}(1 + x^{2})
≲ Φ_{2,β/2}(x).

Combining all of them, we eventually obtain

E[Φ_{2,(β+γ)/2}(|f|)] ≲ E[e + Φ_{2,β/2}(|f|)](1 + log^{γ}(e + E[Φ_{2,β/2}(|f|)]) + E[Φ_{2,β/2}(|∇f|)]. This is what we wanted.

By the above theorem, we can get ∥f∥_{Φ_{2,(β+γ)/2}} ≲ ∥f∥_{Φ_{2,β/2}} + ∥∇f∥_{Φ_{2,β/2}}. (3.18) The term ∥f∥_{Φ_{2,β/2}} in the right hand side is of no importance. To see this, we need the following proposition. Proposition 3.8. Take any p ≥ 1, β > 0, α ∈ ℝ. Then, for any ε > 0, there exists a constant K which depends on ε, p, β and α so that ∥f∥_{Φ_{p,α}} ≤ ε∥f∥_{Φ_{p,α+β}} + K_{ε}∥f∥_{1}. (3.19) Proof. Since

lim_{x→∞} \frac{Φ_{p,α+β}(εx)}{Φ_{p,α}(x)} = ∞,

there exists a constant C > 0 so that Φ_{p,α+β}(εx) ≥ Φ_{p,α}(x) for x ≥ C. When x ≤ C, we can take a constant K > 0 so that Φ_{p,α}(x) ≤ Kx.Therefore

E[Φ_{p,α}(|f|)] = E[Φ_{p,α}(|f|) ; |f| ≥ C] + E[Φ_{p,α}(|f|) ; |f| < C]
≤ E[Φ_{p,α+β}(ε|f|)] + E[K|f|].

For any λ > 0, we have

∥f∥_{Φ_{p,α}} ≤ \frac{1}{λ}\{E[Φ_{p,α}(λ|f|)] + 1\}

≤ \frac{1}{λ}\{E[Φ_{p,α+β}(ελ|f|)] + E[Kλ|f|] + 1\}

≤ \frac{ε}{λε}\{E[Φ_{p,α+β}(λε|f|)] + 1\} + KE[|f|].

Letting λ run over positive numbers and taking the infimum of the right hand side, we have

∥f∥_{Φ_{p,α}} ≤ ε∥f∥_{Φ_{p,α+β}} + K∥f∥_{1},

which is the desired result.

Taking into account the above result, we eventually obtain the following

Theorem 3.9. We assume p > 2, β ∈ ℝ or p = 2, β ≥ 0. Then we have

∥f∥_{Φ_{p,(β+γ)/2}} ≲ ∥∇f∥_{Φ_{p,β/2}} + ∥f∥_{1}. (3.20)

4. Beckner type inequality

Throughout this section, we again assume the following fractional logarithmic Sobolev inequality:

E [|f|^{2} log_{+}^{γ}(|f|^{2}/∥f∥_{2}^{2})] ≤ C_{1}∥∇f∥_{2}^{2} + C_{2}∥f∥_{2}^{2}. (4.1)

In Theorem 3.9, we take p so that pγ/2 = 1 and use Proposition 3.3 to obtain

E [|f|^{p} log_{+}(|f|^{p}/∥f∥_{p}^{p})] ≤ K_{1}∥∇f∥_{p}^{p} + K_{2}∥f∥_{p}^{p}.

We do not need to take positive part of log, so we formulate it in the following form:

E [|f|^{p} log(|f|^{p}/∥f∥_{p}^{p})] ≤ K_{1}∥∇f∥_{p}^{p} + K_{2}∥f∥_{p}^{p}. (4.2)

The left hand side is an entropy, i.e., E [|f|^{p} log(|f|^{p}/∥f∥_{p}^{p})] = Ent(|f|^{p}).

Let p be as above and take q ∈ [1, p). We are now interested in the following inequality:

\frac{E[|f|^{p}] − E[|f|^{q}]^{p/q}}{(p − q)^{δ}} ≤ L_{1}∥∇f∥_{p}^{p} + L_{2}∥f∥_{p}^{p}. (4.3)

The inequality of this type was discussed by Latała-Oleszkiewicz [5] in the case of p = 2. This inequality with p = 2 and δ = 1 holds for the Ornstein-Uhlenbeck process, which was first proved by Beckner [2]. So we call it the Beckner type inequality.

We first consider the case of δ = 1 for general p. It is just a slight modification of Latała-Oleszkiewicz’ argument.

Set α(t) = log ∥f∥_{1/t}. Clearly α(t) is convex. Take any p > 1 and fix it. β(t) = e^{pα(t)} = E[f^{1/t}]^{pt} is also convex. Therefore

\frac{β(t) − β(1/p)}{t − 1/p}

is non-decreasing on (1/p, 1] and

φ(q) = \frac{β(1/p) − β(1/q)}{1/q − 1/p}

is non-decreasing on [1, p). We now set

V_{p}(q) = \frac{β(1/p) − β(1/q)}{p − q} = \frac{E[|f|^{p}] − E[|f|^{q}]^{p/q}}{p − q}

and prove

(4.4)  lim_{q↑p} V_{p}(q) = \frac{1}{p} Ent(|f|^{p}),
(4.5)  V_{p}(q) ≤ Ent(|f|^{p}), q ∈ [1, p).

To show (4.4), by noting \frac{d}{dq}β(1/q)|_{q=p}, we get

\frac{d}{dq} E[|f|^{q}]^{p/q} = E[|f|^{q}]^{p/q} log E[|f|^{q}]^{p/q}(−\frac{p}{q^{2}} ) + \frac{p}{q}E[|f|^{q}]^{(p/q)−1}E[|f|^{p} log |f|]

and if, in particular, q = p, then

\frac{d}{dq} E[|f|^{q}]^{p/q}|_{q=p} = −\frac{1}{p}E[|f|^{p}] log E[|f|^{p}] + E[|f|^{p} log |f|]

= \frac{1}{p}E[|f|^{p} log |f|^{p}/∥f∥_{p}^{p}]

= \frac{1}{p} Ent(|f|^{p}),

which is (4.4).

Next let us see (4.5). Note

V_{p}(q) = \frac{1}{pq} \frac{β(1/p) − β(1/p)}{1/q − 1/p}
= \frac{1}{pq} φ(q) ≤ \frac{1}{p} lim_{q↑p} φ(q). (∵ φ is non-decreasing and q ≥ 1.)

Here

lim_{q↑p} φ(q) = lim_{q↑p} pqV_{p}(q) = p^{2} lim_{q↑p} V_{p}(q) = p Ent(|f|^{p}).

Combining these, we can easily show (4.5).

From these facts, we have the following Proposition 4.1. Take any p > 1 and fix it. If

Ent(|f|^{p}) ≤ K_{1}∥∇f∥_{p}^{p} + K_{2}∥f∥_{p}^{p}

holds, then it follows that

\frac{E[|f|^{p}] − E[|f|^{q}]^{p/q}}{p − q} ≤ K_{1}∥∇f∥_{p}^{p} + K_{2}∥f∥_{p}^{p}.

Conversely, if we assume

\frac{E[|f|^{p}] − E[|f|^{q}]^{p/q}}{p − q} ≤ L_{1}∥∇f∥_{p}^{p} + L_{2}∥f∥_{p}^{p},

then

Ent(|f|^{p}) ≤ pL_{1}∥∇f∥_{p}^{p} + pL_{2}∥f∥_{p}^{p}

follows.

Now we return to the Beckner type inequality. Theorem 4.2. We assume the fractional logarithmic Sobolev inequality (4.1). Then, putting p = 2/γ, there exist constants K_{1}, K_{2} so that for any q ∈ [1, p)

\frac{E[|f|^{p}] − E[|f|^{q}]^{p/q}}{p − q} ≤ K_{1}∥∇f∥_{p}^{p} + K_{2}∥f∥_{p}^{p}, q ∈ [1, p). (4.6)

Proof. From the assumption, the inequality (4.2) holds. Then Proposition 4.1 bears the result.

In the above theorem, we assumed p = \frac{2}{γ}. We now consider the case 2 ≤ p ≤ \frac{2}{γ}. The inequality (4.5) was crucial. Making use of (4.5), we will prove a little modified inequality to show (4.3). Proposition 4.3. For 1 ≤ q < p and 0 < δ < 1,

\frac{E[|f|^{p}] − E[|f|^{q}]^{p/q}}{(p − q)^{δ}} ≤ E [|f|^{p} log_{+}^{δ}(|f|^{p}/∥f∥_{p}^{p})] + \frac{(p − q)^{1−δ}}{e} ∥f∥_{p}^{p} (4.7)

Proof. From (4.5), it follows that E [|f|^{p}] ≤ E [|f|^{q}]^{p/q} + E [|f|^{p}(p − q) log(|f|^{p}/∥f∥_{p}^{p})] . (4.8) Define a set A by

A = {x; (p − q) log(|f|^{p}/∥f∥_{p}^{p}) ≤ 1}.

Then

|f|^{p}(p − q)^{δ} log_{+}^{δ}(|f|^{p}/∥f∥_{p}^{p}) ≥ |f|^{p}(p − q) log(|f|^{p}/∥f∥_{p}^{p}) on A

|f|^{p}(p − q)^{δ} log_{+}^{δ}(|f|^{p}/∥f∥_{p}^{p}) ≥ |f|^{p} on A^{c}.

Writing the indicator function of A by χ, we can see that

|fχ|^{p}(p − q)^{δ} log_{+}^{δ}(|fχ|^{p}/∥f∥_{p}^{p}) ≥ |fχ|^{p}(p − q) log(|fχ|^{p}/∥f∥_{p}^{p}),
|f(1 − χ)|^{p}(p − q)^{δ} log_{+}^{δ}(|f(1 − χ)|^{p}/∥f∥_{p}^{p}) ≥ |f(1 − χ)|^{p}.

Adding them and then integrating them, we get

E[|f|^{p}(p − q)^{δ} log_{+}^{δ}(|f|^{p}/∥f∥_{p}^{p})]
≥ E [|fχ|^{p}(p − q) log(|fχ|^{p}/∥f∥_{p}^{p})] + E [|f(1 − χ)|^{p}]
= E [|fχ|^{p}(p − q) log(|fχ|^{p}/∥fχ∥_{p}^{p})]
+ E [|fχ|^{p}(p − q) log(∥fχ∥^{p}/∥f∥_{p}^{p})] + E [|f(1 − χ)|^{p}]

≥ E [|fχ|^{p}] − E [|fχ|^{q}]^{p/q} + (p − q)∥fχ∥_{p}^{p} log(∥fχ∥^{p}/∥f∥_{p}^{p}) + E [|f(1 − χ)|^{p}]

= E [|f|^{p}] − E [|fχ|^{q}]^{p/q} + (p − q)∥fχ∥_{p}^{p} log(∥fχ∥^{p}/∥f∥_{p}^{p})
≥ E [|f|^{p}] − E [|f|^{q}]^{p/q} + (p − q)∥fχ∥_{p}^{p} log(∥fχ∥^{p}/∥f∥_{p}^{p}).

Thus we have E [|f|^{p}] − E [|f|^{q}]^{p/q} ≤ (p − q)^{δ}E [|f|^{p} log_{+}^{δ}(|f|^{p}/∥f∥_{p}^{p})] + (p − q)∥fχ∥_{p}^{p} log(∥f∥^{p}/∥fχ∥_{p}^{p}) and, by dividing the both hands by (p − q)^{δ},

\frac{E [|f|^{p}] − E [|f|^{q}]^{p/q}}{(p − q)^{δ}} ≤ E [|f|^{p} log_{+}^{δ}(|f|^{p}/∥f∥_{p}^{p})] + (p − q)^{1−δ}∥fχ∥_{p}^{p} log(∥f∥^{p}/∥fχ∥_{p}^{p}).

Now we assume ∥f∥_{p} = 1. Then ∥fχ∥_{p}^{p} ≤ 1 and hence

\frac{E [|f|^{p}] − E [|f|^{q}]^{p/q}}{(p − q)^{δ}} ≤ E [|f|^{p} log_{+}^{δ}(|f|^{p})] + (p − q)^{1−δ}∥fχ∥_{p}^{p} log(1/∥fχ∥_{p}^{p}).

Noting that the function −x log x (x > 0) takes its maximum \frac{1}{e} at x = \frac{1}{e}, we have

\frac{E [|f|^{p}] − E [|f|^{q}]^{p/q}}{(p − q)^{δ}} ≤ E [|f|^{p} log_{+}^{δ}(|f|^{p})] + \frac{(p − q)^{1−δ}}{e} .

Taking |f|/∥f∥_{p}, which satisfies ∥(|f|/∥f∥_{p})∥_{p} = 1, we obtain

\frac{E[|f|^{p}] − E[|f|^{q}]^{p/q}}{(p − q)^{δ}} ≤ E [|f|^{p} log_{+}^{δ}(|f|^{p}/∥f∥_{p}^{p})] + \frac{(p − q)^{1−δ}}{e} ∥f∥_{p}^{p}.

This is what we wanted.

Now we recall Theorem 3.9. Then, assuming the fractional logarithmic Sobolev inequality (4.1), we have, for p ≥ 2, E [|f|^{p} log_{+}^{pγ/2}(|f|^{p}/∥f∥_{p}^{p})] ≤ K_{1}∥∇f∥_{p}^{p} + K_{2}∥f∥_{p}^{p}. (4.9) Thus, from the above proposition, the following Beckner type inequality easily follows:

Theorem 4.4. Assume the fractional logarithmic Sobolev inequality (4.1). Then, for 2 ≤ p ≤ 2/γ, there exist constants L_{1} and L_{2} so that

\frac{E[|f|^{p}] − E[|f|^{q}]^{p/q}}{(p − q)^{pγ/2}} ≤ L_{1}∥∇f∥_{p}^{p} + L_{2}∥f∥_{p}^{p}, q ∈ [1, p). (4.10)

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