A NOTE ON THE PROOF BY ADAMS AND CLARKE

best constants were also given in Ref.4 For Î½ = 1 and p = 2, Adams and Clarke 1 gave an elegant proof of this inequality using elementary calculus.

1

A NOTE ON THE PROOF BY ADAMS AND CLARKE OF GROSS’S LOGARITHMIC SOBOLEV INEQUALITY

¨ ¨ F. GUNG OR

1

and J. GUNSON

2

1

Department of Mathematics, Istanbul Technical University, 80626, Istanbul, Turkey 2

School of Mathematics and Statistics, University of Birmingham, Birmingham, B15 2TT, England 1

ABSTRACT For all continuously differentiable functions, we prove the inequality Z

−1

p

R

where kp =

Z

0 p

|u | dx ≥ 1 [1 p

R

|u|p ln |u| dx − kukpp ln kukp + kp kukpp ,

p>1

1

+ ln( p2 (p − 1) p Γ( p−1 ))] is the best possible. The method of p

proof follows Adams and Clarke. Key Words : logarithmic Sobolev inequality, best constant, minimising function

1. INTRODUCTION The logarithmic Sobolev inequality of Gross 3 is, in its Lebesgue measure form, the p = 2 case of the general inequality Z

p

−1

p

Z

ν

|∇u| d x ≥

|u|p ln |u| dν x − kukpν,p ln kukν,p + kν,p kukpν,p ,

(1)

where k.kν,p denotes the norm on the Banach space Lp (Rν , dν x), defined by kukpν,p =

Z

|u|p dν x.

Starting from the standard Sobolev inequalities, a version of the inequality (1) was derived by one of the authors in an earlier paper.4 Expressions for the 1

1991 Mathematics Subject Classification. Primary 26D10.

¨ ¨ and J.GUNSON F.GUNG OR

2

best constants were also given in Ref.4 For ν = 1 and p = 2, Adams and Clarke 1 gave an elegant proof of this inequality using elementary calculus. By converting to the Gaussian form and using induction on ν, the proof for general ν then follows. In this paper we prove (1) for ν = 1 and general p > 1, by adapting the method of Adams and Clarke. Unfortunately, the induction method fails to generalise when p 6= 2. A direct, but much more complicated, proof for the case of general ν and p will be given in a subsequent paper.

2. THE MAIN RESULTS Setting k.k1,p = k.kp and k1,p = kp , our main result is : Theorem 1. Let u : R → C be a continuously differentiable function. Then, for p > 1 Z −1

p

R

Z

0 p

|u | dx ≥

R

|u|p ln |u| dx − kukpp ln kukp + kp kukpp ,

(2)

1

where kp = p1 [1 + ln( p2 (p − 1) p Γ( p−1 ))]. p Proof of Theorem 1. In view of the pointwise inequality |(|u|)0 | ≤ |u0 |, which holds almost everywhere on R, we may assume that u is real and non-negative. If v is the symmetric decreasing rearrangement of u, then the rearrangement inequality Z 0

2

|v (x)| dx ≤

Z

2

|u0 (x)| dx

shows that we may also assume that u is even and decreasing in |x|. Following the ideas of Adams and Clarke, we first calculate inf{J[u]}, where Z∞

J[u] = 0

under the constraints

1 p F (u, u0 )dx and F (u, u0 ) = |u0 | − up ln u p

(3)

R∞ p u dx = r and u(0) = s, for any r, s > 0. A family 0

of extremals of the functional in (3) is readily obtained as the two parameter family of functions

1 u(x) = λ exp{− |x + µ|q }, q

(4)

A NOTE ON THE PROOF BY ADAMS AND CLARKE ...

3

where p−1 + q −1 = 1. Using the constraints to determine λ and µ, we get ∞

Z 1 q p s = λ exp{− |µ| } and r = λ exp{−(p − 1)|t + µ|q }dt. q

(5)

0

The function h(x) defined for x ∈ R by Z∞

q

exp {−(p − 1)|t + x|q }dt

h(x) = exp{(p − 1)|x| }

(6)

0

is strictly decreasing on R and satisfies h(µ) = r/sp . Hence we may write µ = h−1 (r/sp ) and ln λ = ln s + |µ|q /q. Moreover, by differentiating h(h−1 (x)) = x to give h0 (h−1 (x))(h−1 )0 (x) = 1 and using h0 (x) = ²(x)p|x|1/(p−1) h(x) − 1, we get −1

(h−1 )0 (x) = {p|h−1 (x)|1/(p−1) x − 1} .

(7)

Evaluating the functional J in (3) for an extremal (4) gives the expression Z∞ p

−1

λ µp

q

p

−1

exp {−(p − 1)|µ| } + λ (p

exp{−(p − 1)|t + µ|q }dt,

− ln λ) 0

which, when regarded as a function of r and s, can be written V (s, r) = p−1 {r(1 − p ln s − (p/q)|g|q ) + gsp },

(8)

where g(s, r) = µ = h−1 (r/sp ). The first partial derivatives of V can be calculated using (6) to give Vs (s, r) = g(s, r)sp−1

1 and Vr (s, r) = − ln s − |g(s, r)|q . q

(9)

Hence we have 

Z d  1 V u(x), u(t)p dt = Vs u0 − Vr up = gup−1 u0 + up ln u + up |g|q . dx q x

However, using the standard inequality

2

(10)

|x|p + px + p − 1 ≥ 0 , we get

1 1 u0 1 gup−1 u0 + up |g|q = (p − 1 + p|g|−q/p ²(g) )up |g|q ≥ − |u0 |p , q p u p

¨ ¨ and J.GUNSON F.GUNG OR

4 and so

Z∞

d  1 V u(x), (u(t))p dt ≥ up ln u − |u0 |p = −F (u, u0 ). dx p x

(11)

Integrating the inequality (11) over (0, ∞), we get 

Z∞

Z∞

F (u(x), u0 (x))dx ≥ V u(0), 0

Z∞

up dx − lim inf V u(x), x→∞

(u(t))p dt . (12)

x

0

In the following lemma it is shown that the limit on the right hand side of (12) is zero. Hence the inequality (12) shows that the extremals (4) are indeed minimising functions for the constrained minimisation problem (3). Moreover we get the best constant k(p) by minimising the right hand side of (12) with respect to u(0). For s > 0, the minimum of V (s, r) occurs when Vs = 0 or equivalently h−1 (r/sp ) = 0. But h−1 is strictly decreasing and so there is a unique minimum at the value of s given by r/sp = h(0). Thus we have V (s, r) ≥ V ((r/h(0))1/p , r) = (r/p)(1 − ln(r/h(0))) for s > 0,

(13)

where h(0) can be evaluated from (6) to be 1 1 p−1 (p − 1) p Γ( ). p p

From (12) and (13) we get Z∞ ( 0

) ∞ Z∞  1 0p 1Z p  p |u | − u ln u dx ≥ u dx − ln up dx + (1 + ln h(0)) . p p 0

(14)

0

Comparing with (2), we get kp = (1 + ln (2h(0)))/p. In particular, k2 = √ (1 + ln( π))/2. Lemma. If u(x) > 0,

R∞ p u dx < ∞ 0

and

R∞

|u0 |p dx < ∞, then

0

Z∞

lim inf V u(x), x→∞ 0

(u(t))p dt = 0.

(15)

A NOTE ON THE PROOF BY ADAMS AND CLARKE ...

5

Proof of Lemma. From the estimate h(x) < x−1/(p−1) on 0 < x < 2

∞, we get h−1 (x) < x1−p for x > 0, whence g(s, r)sp < r1−p sp . Similarly, if M=

R∞

−∞

exp(−(p − 1)tq )dt, then we have the estimate h(x) ≤ M exp((p − 1)xq )

for x ≤ 0 and consequently h−1 (x) ≤ −(ln(x/M )1/(p−1) )1/q for x ≥ M. Hence (p − 1)|g(s, r)|q ≥ ln r − p ln s − ln M for r ≥ M sp , which gives r(1 − p ln s − p|g|q /q) ≤ max{r(1 − ln r + ln M ), M sp (1 − p ln s)}. Using the estimate for gsp , we get 2

V (s, r) ≤ r1−p sp + p−1 max{r(1 − ln r + ln M ), M sp (1 − p ln s)}

(16)

for all r, s > 0. For u satisfying the hypotheses of the lemma, the expressions s = u(x), r =

R∞ p R∞ u dx and η = |u0 |p dx all tend to zero as x tends to infinity.

x

x

From H¨older’s inequality, the estimate  p2

s

Z∞

≤ p

∞  ∞ p−1 Z Z p |u0 |up−1 dx ≤ pp  |u0 | dx  up dx = pp ηrp−1

x

p

x

x

together with (16) then shows that (15) holds.

REFERENCES 1. R. Adams and F.H. Clarke, Amer. J. Math. , 101, 1265-1269, (1979). 2. E.F. Beckenbach and R.E. Bellmann, Inequalities, Springer-Verlag, N.Y. (1961). 3. L. Gross, Amer. J. Math., 97 , 1061-1083, (1976). 4. J. Gunson, Proceedings of the Inequalities Conference, Birmingham, (1987).