# THREE VARIANTS OF THE HARDY INEQUALITY 1. The ... THREE VARIANTS OF THE HARDY INEQUALITY 1. The ...

THREE VARIANTS OF THE HARDY INEQUALITY

RICARDO MENARES

Abstract. We explain a trick used to prove the classical Hardy inequality and we use it to

establish a weighted Poincaré inequality

Consider the following

1. The inequalities

Inequality 1.1. Let T = {f ∈ C ∞( [0, 1] ) such that f(1) = 0}. For every δ > 0 there exists

C δ > 0 such that

(1)

for all f ∈ T .

∫ 1

0

|f(x)| 2

x 1−δ dx ≤ C δ

∫ 1

0

x|f ′ (x)| 2 dx,

An elementary proof goes as follows: integrating by parts we have

Re

∫ 1

0

x δ f(x)f ′ (x)dx = − δ 2

∫ 1

0

|f(x)| 2

x 1−δ dx.

Writing x δ f(x)f ′ (x) = x δ−1/2 f(x)x 1/2 f ′ (x) and using the Cauchy-Schwarz inequality we get

δ 1

|f(x)| 2 ( ∫ 1

2 0 x 1−δ dx ≤ 0

( ∫ 1

) 1/2 ( ∫

x 2δ−1 |f(x)| 2 1

dx

(2)

0

0

If f is not the zero function, this implies (1) with C δ = (2/δ) 2 .

0

) 1/2

x|f ′ (x)| 2 dx

|f(x)| 2 ) 1/2 ( ∫ 1

1/2.

x 1−δ dx x|f ′ (x)| dx) 2

This trick can be ”sloganized” as: write the ”weighted” L 2 norm of f as an inner product

and then exploit the non homogeneity of inequality (2).

The trick is already found in the proof of the classical Hardy inequality ([HLP88], theorem

327). It can be adapted to the complex setting to show the following ”weighted” Poincaré

inequality:

Inequality 1.2. Let D := {|z| < 1} ⊂ C and let S = {f ∈ C ∞ (D) such that f = 0 on ∂D}.

For every δ > 0 there exists C δ > 0 such that

|f(z)| 2

(3) i

D |z| 2−δ dz ∧ d¯z ≤ C δi

for all f ∈ S

D

∣ ∂f

∣ 2 dz ∧ d¯z,

∂z

We remark that this inequality is a special case of the Caffarelli-Kohn-Nirenberg inequality,

which is also stronger ([CKN84], take n = p = r = 2, a = 1, γ = −1 + (δ/2), α = δ/2 in their

notation and use that |z| δ ≤ 1 when z ∈ D).

An application of this weighted Poincaré inequality is the following: let n be a positive integer

and let ϕ : D → D be the map ϕ(z) = z n . We consider the Sobolev space L 2 1 (D) 0 = {f ∈ L 2 (D)

1

such that ∂ ∂z f ∈ L2 (D) and f = 0 on ∂D}. Then inequality 1.2 is exactly what is needed to show

that ϕ ∗ : C0 ∞(D) → C∞ 0 (D) is a continuous linear map (with respect to the inherited Sobolev

topologies) and hence extends to a continuous functional ϕ ∗ : L 2 1 (D) 0 → L 2 1 (D) 0.

More generally, let ϕ : X → Y be a non constant holomorphic map between compact Riemann

surfaces. Then to prove that ϕ ∗ : L 2 1 (Y ) → L2 1 (X) is a continuous linear map, a difficulty

arises at the ramification points of ϕ. The weighted Poincaré inequality allows one to sort out

this problem (cf. [Men09]).

We can elaborate on the trick to get an inequality without assuming vanishing on the border:

Inequality 1.3. For all δ > 0 and for all f ∈ C ∞ (D), we have

( ∫

|f| 2 )

1/2 2

( ∫ 2π ) 1/2+

i

|z| 2−δ dz∧d¯z ≤ |f| 2 2

(( ∫

dθ i ∣ ∂f

∣ 2 ) 1/2+ ( ∫

dz∧d¯z i

δ

δ ∂z

D

0

2. Proofs of the weighted Poincaré inequalities

Lema 2.1. For all f ∈ S, we have

∣ ∂f

∣ 2 dz ∧ d¯z =

∂z

D

D

D

∣ ∂f

∣ 2 dz ∧ d¯z.

∂¯z

D

∣ ∂f

∣ 2 ) 1/2 )

dz∧d¯z .

∂¯z

Proof : the identity d(f ∧ ∂f) = ∂f ∧ ∂f + f∂∂f and the vanishing condition on the boundary

show that

(4)

Then

D

D

∣ ∂f

∣ 2 dz ∧ d¯z =

∂z

∣ ∂f

∣ 2 dz ∧ d¯z =

∂¯z

D

D

∂f ∧ ∂f =

∂f ∧ ∂f = − f∂∂f.

D

D

∂ ¯f ∧ ∂ ¯f = −

D

¯f∂∂f,

where the last equality is an application of (4) to ¯f. Finally, the identity d(f∂f + ¯f∂f) =

f∂∂f − ¯f∂∂f and the vanishing on the boundary allow us to conclude.

Proof of inequality 1.2: we first note that the integral on the left hand side is convergent.

Indeed,

(5)

|f(z)| 2

∫ 2π ∫ 1

i

D |z| 2−δ dz ∧ d¯z = 2 0 0

Let D ε = {ε < |z| < 1}. We have

|f| 2

r 1−δ drdθ ≤ 4π δ sup

D

|f| 2 .

2

( |z|

δ )

δ d ¯z |f|2 d¯z = |f|2

|z| 2−δ dz ∧ d¯z + 2 |z| δ ∂|f| 2

δ ¯z ∂z dz ∧ d¯z on D ε.

We have ∫ |z| δ

∂D ε ¯z |f|2 d¯z = −iε δ ∫ 2π

0

|f| 2 dθ → 0 as ε → 0. Then

But

|f| 2

i

D |z| 2−δ dz ∧ d¯z = −2 δ

D

|z| δ

¯z

∂|f| 2

dz ∧ d¯z.

∂z

(6)

D

|z| δ ∂|f| 2 ∣ ∣ ∣∣

¯z ∂z dz ∧ d¯z ≤ i |z|

∫D

δ−1 ∣∣ ∂|f| 2

∣dz ∧ d¯z

∂z

∫ ∣

≤ i |z| δ−1 ∣∣f ∂f

∣dz ∧ d¯z + i

∂z

D

2

D

|z| δ−1 ∣∣f ∂f

∣dz ∧ d¯z,

∂¯z

where we have used ∂ ∂z |f|2 = ¯f ∂ ∂z f + f ∂ ∂¯z

f. We bound the first integral using the Cauchy-

Schwarz inequality:

i

D

|z| δ−1 ∣∣f ∂f

∣dz ∧ d¯z

∂z

( ∫

i

( ∫

i

D

D

|f| 2 ) 1/2 ( ∫

|z| 2−2δ dz ∧ d¯z i ∣ ∂f

∣ 2 ) 1/2

dz ∧ d¯z

D ∂z

|f| 2 ) 1/2 ( ∫

|z| 2−δ dz ∧ d¯z i ∣ ∂f

∣ 2 1/2.

dz ∧ d¯z)

∂z

Because of lemma 2.1, the same bound applies to the second integral in (6). Putting everything

together, we get

( ∫

i

D

|f| 2

i

D |z| 2−δ dz ∧ d¯z ≤ 4 δ

This implies (3) with C δ = (4/δ) 2 .

|f| 2 ) 1/2 ( ∫

|z| 2−δ dz ∧ d¯z i

D

D

∣ ∂f

∣ 2 1/2.

dz ∧ d¯z)

∂z

Proof of inequality 1.3: starting, as before, from (5) and keeping track of the integral on the

border, we get

i

D

|f| 2

|z| 2−δ dz ∧ d¯z ≤ 2 δ

+ 2 δ

∫ 2π

0

(

i

D

|f| 2 dθ

|f| 2 ) 1/2 (( ∫

|z| 2−δ dz ∧ d¯z i

D

∣ ∂f

∣ 2 ) 1/2 ( ∫

dz ∧ d¯z + i

∂z

D

∣ ∂f

∣ 2 ) 1/2 )

dz ∧ d¯z .

∂¯z

We remark that if a, b, c, d are positive real numbers, then a ≤ b + √ a( √ c + √ d) implies

√ a ≤

b +

√ c +

d. Applying this to the above inequality gives the result.

References

[CKN84] L. Caffarelli, R. Kohn, and L. Nirenberg. First order interpolation inequalities with

weights. Compositio Math., 53(3):259–275, 1984. 1

[HLP88] G. H. Hardy, J. E. Littlewood, and G. Pólya. Inequalities. Cambridge Mathematical

Library. Cambridge University Press, Cambridge, 1988. Reprint of the 1952 edition.

1

[Men09] Ricardo Menares. Correspondences in Arakelov geometry. Applications to Hecke operators

on modular curves. arXiv:0911.0546v1, 2009. Submitted. 1

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