Sharp Inequalities and related problems85th Annual Meeting of the Indian Academy of Sciences
University of Hyderabad
K. Sandeep
Tata Institute of Fundamental Research
Center for Applicable Mathematics, Bangalore
November 8-10, 2019
K Sandeep 85th Annual Meeting
Inequalities : Discrete
2xy ≤ x2 + y2, for x , y ∈ R
K Sandeep 85th Annual Meeting
Inequalities : Discrete
2xy ≤ x2 + y2, for x , y ∈ R
xy ≤ xp
p+
yq
qwhere
1
p+
1
q= 1, x , y ∈ R
K Sandeep 85th Annual Meeting
Inequalities : Discrete
2xy ≤ x2 + y2, for x , y ∈ R
xy ≤ xp
p+
yq
qwhere
1
p+
1
q= 1, x , y ∈ R
For xi , yi ∈ R, i = 1, ..., n
n∑i=1
|xiyi | ≤ (n∑
i=1
|xi |p)1p (
n∑i=1
|yi |q)1q
K Sandeep 85th Annual Meeting
Poincare Inequality
Let −∞ < a < b < ∞ and u ∈ C 1([a, b]), u(a) = u(b) = 0
K Sandeep 85th Annual Meeting
Poincare Inequality
Let −∞ < a < b < ∞ and u ∈ C 1([a, b]), u(a) = u(b) = 0
|u(x)|2 = |∫ x
au′(t) dt|2 ≤ (b − a)
∫ b
a|u′|2dt
K Sandeep 85th Annual Meeting
Poincare Inequality
Let −∞ < a < b < ∞ and u ∈ C 1([a, b]), u(a) = u(b) = 0
|u(x)|2 = |∫ x
au′(t) dt|2 ≤ (b − a)
∫ b
a|u′|2dt
1
(b − a)2
∫ b
a|u|2dt ≤
∫ b
a|u′|2dt
K Sandeep 85th Annual Meeting
Poincare Inequality
There exists an optimal constant λ1 > 0 such that
λ1
∫ b
a|u|2dt ≤
∫ b
a|u′|2dt
holds for all functions u ∈ C 1([a, b]), u(a) = u(b) = 0.
K Sandeep 85th Annual Meeting
Poincare Inequality
Let Ω be a bounded domain in Rn, then there exists an oplimal
constant λ1(Ω) > 0 such that
λ1(Ω)
∫Ω|u(x)|2 dx ≤
∫Ω|∇u(x)|2 dx
holds for all u ∈ C 1c (Ω).
K Sandeep 85th Annual Meeting
Poincare Inequality
Let Ω be a bounded domain in Rn, then there exists an oplimal
constant λ1(Ω) > 0 such that
λ1(Ω)
∫Ω|u(x)|2 dx ≤
∫Ω|∇u(x)|2 dx
holds for all u ∈ C 1c (Ω).
λ1(Ω) is the first eigen value of −Δ
K Sandeep 85th Annual Meeting
Inequalities in Rn
Can we have this inequality in Rn ?:
C
∫Rn
|u(x)|2 dx ≤∫Rn
|∇u(x)|2 dx , u ∈ C 1c (R
n)
K Sandeep 85th Annual Meeting
Inequalities in Rn
Can we have this inequality in Rn ?:
C
∫Rn
|u(x)|2 dx ≤∫Rn
|∇u(x)|2 dx , u ∈ C 1c (R
n)
Answer : NO
K Sandeep 85th Annual Meeting
Inequalities in Rn
Can we have this inequality in Rn ?:
C
∫Rn
|u(x)|2 dx ≤∫Rn
|∇u(x)|2 dx , u ∈ C 1c (R
n)
Answer : NO
More generally can we have
C
[∫Rn
|u(x)|q dx
] pq
≤∫Rn
|∇u(x)|p dx , u ∈ C 1c (R
n)
K Sandeep 85th Annual Meeting
Inequalities in Rn
Suppose we have such an inequality for some p, q then
q =np
n − p:= p∗
K Sandeep 85th Annual Meeting
Sobolev Inequality
Sobolev Inequality : Let 1 ≤ p < n, there exists an optimal
constant Sp,n > 0 such that
Sp,n
[∫Rn
|u(x)|p∗ dx
] pp∗
≤∫Rn
|∇u(x)|p dx , u ∈ C 1c (R
n)
K Sandeep 85th Annual Meeting
Sobolev Inequality
Sobolev Inequality : Let 1 ≤ p < n, there exists an optimal
constant Sp,n > 0 such that
Sp,n
[∫Rn
|u(x)|p∗ dx
] pp∗
≤∫Rn
|∇u(x)|p dx , u ∈ C 1c (R
n)
Morrey’s Inequality : Let p > n, there exists an optimal constant
Sp,n > 0 such that
supx �=y
u(x)− u(y)
|x − y |1− np
≤ Sn,p
[∫Rn
|∇u(x)|p dx
] 1p
K Sandeep 85th Annual Meeting
Best Constants and Extremals
First consider the case p = 1. i.e we have the inequality
S1,n
[∫Rn
|u(x)| nn−1 dx
] n−1n
≤∫Rn
|∇u(x)| dx , u ∈ C 1c (R
n)
K Sandeep 85th Annual Meeting
Best Constants and Extremals
First consider the case p = 1. i.e we have the inequality
S1,n
[∫Rn
|u(x)| nn−1 dx
] n−1n
≤∫Rn
|∇u(x)| dx , u ∈ C 1c (R
n)
S1,n = n1−1n [ωn−1]
1n
where ωn−1 = Hn−1(Sn−1), the surface measure of the boundary
of unit ball in Rn.
K Sandeep 85th Annual Meeting
Best Constants and Extremals
First consider the case p = 1. i.e we have the inequality
S1,n
[∫Rn
|u(x)| nn−1 dx
] n−1n
≤∫Rn
|∇u(x)| dx , u ∈ C 1c (R
n)
S1,n = n1−1n [ωn−1]
1n
where ωn−1 = Hn−1(Sn−1), the surface measure of the boundary
of unit ball in Rn.
S1,n is not achieved
K Sandeep 85th Annual Meeting
Isoperimetric Inequality
Isoperimetric problem :Determine the shape of the closed plane
curve having a given length and enclosing the maximum area
K Sandeep 85th Annual Meeting
Isoperimetric Inequality
Let Ω be a bounded open set in Rn with smooth boundary then
Hn−1(∂Ω) ≥ n1−1n [ωn−1]
1n |Vol Ω| n
n−1
Iquality holds iff Ω is a ball
K Sandeep 85th Annual Meeting
Equivalence
Federer-Fleming : Sobolev inequality with p = 1 is equivalent to
the Isoperimetric Inequality
K Sandeep 85th Annual Meeting
Cartan-Hadamard Manifold
Conjecture : Let (M, g) be a Cartan-Hadamard manifold then the
isoperimetric inequality :
Hn−1(∂Ω) ≥ n1−1n [ωn−1]
1n |Vol Ω| n
n−1
holds in (M, g).
• n=2, Weil, 1926
• n=3, Kleiner, 1992
• n=4, Croke,1984
• n ≥ 5, Joel Spruck,et al...2019 (???)
K Sandeep 85th Annual Meeting
The case p = 2
S2,n
[∫Rn
|u(x)| 2nn−2 dx
] n−2n
≤∫Rn
|∇u(x)|2 dx , u ∈ C 1c (R
n)
K Sandeep 85th Annual Meeting
The case p = 2
S2,n
[∫Rn
|u(x)| 2nn−2 dx
] n−2n
≤∫Rn
|∇u(x)|2 dx , u ∈ C 1c (R
n)
Extremal functions exist
K Sandeep 85th Annual Meeting
The case p = 2
S2,n
[∫Rn
|u(x)| 2nn−2 dx
] n−2n
≤∫Rn
|∇u(x)|2 dx , u ∈ C 1c (R
n)
Extremal functions exist
If u is an Extremal then u solves
−Δu = un+2n−2 , u > 0 ,
∫Rn
|∇u|2 < ∞.
K Sandeep 85th Annual Meeting
Extremals
• Rotational Symmetry
K Sandeep 85th Annual Meeting
Extremals
• Rotational Symmetry
• Conformal invariance
Let K : Rn → Rn be a conformal map and u is a solution of the
PDE, the
u := J(K )n−22 u(K (x))
is also a solution
K Sandeep 85th Annual Meeting
Extremals
• All solutions are of the from
U(x) =
[√N(N − 2)ε
ε2 + |x − x0|2]N−2
2
for some ε > 0 and x0 ∈ RN
K Sandeep 85th Annual Meeting
Equivalent Geometric problem
Obata : Let g be a metric on the unit sphere Sn conformal to the
standard metric on Sn, then g is of constant scalar curvature 1 iff
g is of constant sectional curvature 1.
K Sandeep 85th Annual Meeting
Yamabe Problem
Let (M, g) be a compact Riemannian manifold of dimension N and
scalar curvature Kg , can we find a metric g conformal to g such
that g has constant scalar curvature.?
If g = u4
N−2 g , then this is equivalent to solving the PDE
−4(N − 1)
N − 2Δgu + Kgu = ku
N+2N−2
where k is a constant.
K Sandeep 85th Annual Meeting
Hardy-Sobolev -Maz’ya Inequality.
Let 2 ≤ k < N, RN = R
k × Rh. Denote a point x ∈ R
N by
x = (y , z) ∈ Rk × R
h, then for t ∈ [0, 2) there exists an optimal
constant S = St,N,k > 0 such that
S
⎛⎝∫RN
|u|2∗(t)|y |t dx
⎞⎠
22∗(t)
≤∫RN
|∇u|2dx − λ
∫RN
|u|2|y |2 dx
holds for all u ∈ D1,2(RN) where 2∗(t) = 2(N−t)N−2 and
0 ≤ λ ≤ (k−2)2
4 .
K Sandeep 85th Annual Meeting
Euler Lagrange Equation.
−Δu − λu
|y |2 =up(t)−1
|y |t , u > 0, u ∈ D1,2(RN)
where x = (y , z) ∈ Rk × R
h = RN
K Sandeep 85th Annual Meeting
Classification of Solution
Question : Uniqueness of Solution
Difficulty : Lack of rotational symmetry
K Sandeep 85th Annual Meeting
Classification of Solution.
• (G Mancini , Sandeep) The problem has Hyperbolic symmetry
and showed that the Solution space is of N − k + 1 dimensional
K Sandeep 85th Annual Meeting
Classification of solutions when t = 1.
Theorem
(Fabri, Mancini, S) Let u0 be the function given by
u0(x) = u0(y , z) = cN,k
((1 + |y |)2 + |z |2)−N−2
2
where cN,k = {(N − 2)(k − 1)}N−22 . Then u is a solution of
−Δu =u
NN−2
|y | in RN , u > 0, u ∈ D1,2(RN)
if and only if u(y , z) = λN−22 u0(λy , λz + z0) for some λ > 0 and
z0 ∈ RN−k
K Sandeep 85th Annual Meeting
Thank You
K Sandeep 85th Annual Meeting