Weighted Moser-Trudinger inequalities

Posted on April 12, 2017 by Van Hoang Nguyen

Recently, M. Ishiwata, M. Nakamura and H. Wadade (see this link) proved the following weighted Moser–Trudinger inequalities of the scaling invariant form in whole space \mathbb R^N, N\geq 2

\displaystyle \int_{\mathbb R^N} \Phi_N(\alpha |u|^{\frac N{N-1}}) |x|^{-t} dx \leq C \left(\int_{\mathbb R^N} |u|^N |x|^{-s} dx\right)^{\frac{N(N-t)}{N-s}},     (1)

for any radial function u with \int_{\mathbb R^N} |\nabla u|^N dx \leq 1 and for any 0< \alpha < \alpha_{N,t}:= (N-t) \omega_{N-1}^{1/(N-1)} where \omega_{N-1} is the surface area of unit sphere in \mathbb R^N, -\infty < s \leq t < N, C is constant depending only on N,s,t and

\displaystyle \Phi_N(a) = e^a -\sum_{k=0}^{N-2} \frac{a^k}{k!},\qquad a \geq 0.

Note that (1) fails if \alpha \geq \alpha_{N,t}. M. Ishiwata, M. Nakamura and H. Wadade also proved the existence of extremal functions for (1) in the class of radial functions. A natural and interesting question arising is the validity of (1) for any function u with \int_{\mathbb R^N} |\nabla u|^N dx \leq 1 (i.e. we drop out the radiality assumption) and the existence of extremal functions for (1). The answer is positive if s =0 by the mean of rearrangement argument. In recent paper, M. Dong and G. Lu (see this link) solved  affirmatively this question in general. In their paper, they introduced a formula for changing functions which enables us to reduce the question to the case s =0. More precisely, if u is a function on \mathbb R^N with \int_{\mathbb R^N} |\nabla u|^N dx \leq 1, we define a new function v by

\displaystyle v(x) = \left(\frac{N-s}{N}\right)^{\frac{N-1}N} u(|x|^{\frac s{N-s}} x).   (*)

Denote by F the map F(x) = |x|^{\frac s{N-s}} x, by a simple calculations, we get its Jacobian matrix as

\displaystyle DF(x) = |x|^{\frac{s}{N-s}} \left(Id_{\mathbb R^N} + \frac s{N-s} \frac x{|x|} \otimes \frac x{|x|}\right),

where for a unit vector w in \mathbb R^N, w \otimes w denotes matrix of the orthogonal projection onto the line generated by w. Consequently, the Jacobian of F is

\displaystyle det(DF(x)) = \frac N{N-s} |x|^{\frac{Ns}{N-s}} = \frac N{N-s} |F(x)|^s.

Using the change of variables, we get

\displaystyle \int_{\mathbb R^N} |u|^N |x|^{-s} dx =\left(\frac N{N-s}\right)^N \int_{\mathbb R^N} |v|^N dx.   (2)

and

\displaystyle \int_{\mathbb R^N} \Phi_N(\alpha |u|^{\frac N{N-1}}) |x|^{-t} dx =\frac N{N-s} \int_{\mathbb R^N} \Phi_N\left(\alpha \frac{N}{N-s} |v|^{\frac N{N-1}}\right) |y|^{-\frac{N(t-s)}{N-s}} dy.   (3)

They also proved that

\displaystyle |\nabla v(x)|\leq \left(\frac N{N-s}\right)^{\frac1N} |x|^{\frac s{N-s}} |\nabla u(F(x))| = \left(det (DF(x))\right)^{\frac 1N} |\nabla u(F(x)|,     (4)

with equality if u is radial function. From (4) we get

\displaystyle \int |\nabla v|^N dx \leq \int |\nabla u|^N dx \leq 1.    (5)

Denote t^* =N(t-s)/(N-s) \in [0,N). Note that

\displaystyle \alpha_*: =\alpha \frac{N}{N-s} < \alpha_{N,t} \frac{N}{N-s} = \frac{N(N-t)}{N-s} \omega_{N-1}^{\frac 1{N-1}} = \alpha_{N,t^*}.   (6)

(2), (3), (5) and (6) enables us reducing the proof of (1) to the case s =0, t =t^* and \alpha =\alpha_* which is already known. Thus (1) holds in general. Let v be an extremal function of (1) in the case s =0, t= t^* \in [0,N) and \alpha =\alpha_*. The existence of v is well known. Moreover v is a radial function and \int |\nabla v|^N dx =1. Let u be defined from v by (*), we have equality in (4), hence \int |\nabla u|^N dx =1. Using (2) and (3), we see that u is an extremal functions of (1) for s and t in general, that is, the extremal function of (1) exists in general.

However, it seems that the inequality (4) does not holds in general. More precisely, in the case s < 0, (4) only holds for radial functions (in this case we have an equality) and if u is not radial function then a reverse inequality of (4) holds. Let us explain this fact. By a direct calculation, we have \displaystyle \nabla v(x) = \left(\frac{N-s}N\right)^{\frac{N-1}N} DF(x)^t(\nabla u(F(x))), hence

\displaystyle \nabla v(x) = \left(\frac{N-s}N\right)^{\frac{N-1}N} |x|^{\frac s{N-s}} \left(\nabla u(F(x)) + \frac{s}{N-s} \frac{x\cdot \nabla u(F(x))}{|x|} \frac{x}{|x|}\right)

and

\displaystyle |\nabla v(x)|^2 = \left(\frac{N-s}N\right)^{\frac{2(N-1)}N} |x|^{\frac {2s}{N-s}}\left[|\nabla u(F(x))|^2 + \left(\frac{s^2}{(N-s)^2} + \frac{2s}{N-s}\right) \frac{(x\cdot \nabla u(F(x)))^2}{|x|^2}\right].

Denote s(N) = \frac{s^2}{(N-s)^2} + \frac{2s}{N-s} < 0 since s < 0. Since \nabla u(F(x)) - \frac{x\cdot \nabla u(F(x))}{|x|} \frac x{|x|} is orthogonal to \frac x{|x|}, we then have

\displaystyle |\nabla v(x)|^2 = \left(\frac N{N-s}\right)^{\frac 2N} |x|^{\frac{2s}{N-s}} |\nabla u(F(x))|^2 - \left(\frac{N-s}N\right)^{\frac{2(N-1)}N} |x|^{\frac {2s}{N-s}} s(N) \left|\nabla u(F(x)) - \frac{x\cdot \nabla u(F(x))}{|x|} \frac x{|x|}\right|^2.

Since s(N) < 0, hence |\nabla v(x)| \geq (N/(N-s))^{2/N} |\nabla u(F(x))|, that is, a reverse inequality of (4) holds. Moreover, if u is not radial function, then there exists x such that

\displaystyle \left|\nabla u(F(x)) - \frac{x\cdot \nabla u(F(x))}{|x|} \frac x{|x|}\right| >0,

which implies |\nabla v(x)| > (N/(N-s))^{2/N} |\nabla u(F(x))|. Hence, a strictly reverse inequality of (4) holds if u is not radial.

 

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Hardy-Rellich inequality on functions which are orthogonal to radial functions.

Posted on January 7, 2017 by Van Hoang Nguyen

Hardy inequality asserts that

\displaystyle \frac{(n-2-2a)^2}4 \int_{\mathbb R^n} \frac{|f(x)|^2}{|x|^{2+2a}}dx \leq \int_{\mathbb R^n}\frac{ |\nabla f(x)|^2}{|x|^{2a}} dx

for any function f\in H^1(\mathbb R^n), n >2 + 2a. Rellich inequality asserts that

\displaystyle \frac{(n+2a)^2(n-2a-4)^2}{16} \int_{\mathbb R^n} \frac{|f(x)|^2}{|x|^{4+2a}} dx \leq \int_{\mathbb R^n}\frac{ |\Delta f(x)|^2}{|x|^{2a}} dx

for any function f\in H^2(\mathbb R^n), n > 4+2a. The constants (n-2-2a)^2/4 and (n+2a)^2(n-2a-4)^2/16 are sharp.

In this post, we will improve these inequalities when restricting to the functions f which is orthogonal to all radial function, i.e., the functions f such that

\displaystyle \int_{S^{n-1}} f(r\omega) d\omega = 0,\qquad\forall\, r >0.

A such function f can be expressed as

\displaystyle f(x) = \sum_{k=1}^\infty f_k(r) \phi_k(\omega),\quad x = r\omega,\, \omega \in S^{n-1},

where \phi_k is an eigenfunction of the Laplace-Beltrami operator on S^{n-1} with the eigenvalue c_k = k(n+k-2). For this expression, we get

\displaystyle -\Delta f(x) = \sum_{k=1}^\infty \left( -\Delta f_k + c_k \frac{f_k}{r^2}\right) \phi_k(\omega).

Theorem 1: If f is orthogonal to all radial functions, then

\displaystyle \frac{n^2}4 \int_{\mathbb R^n} \frac{|f(x)|^2}{|x|^2} dx \leq \int_{\mathbb R^n} |\nabla f(x)|^2 dx.

Proof: Writing f as above, we have

\displaystyle \int_{\mathbb R^n} \frac{|f(x)|^2}{|x|^{2}} dx = \sum_{k=1}^\infty \int_{\mathbb R^n} \frac{|f_k(|x|)|^2}{|x|^{2}} dx.

Using integration by parts, we get

\displaystyle \int_{\mathbb R^n} |\nabla f(x)|^2 dx = \int_{\mathbb R^n} -\Delta f(x) f(x) dx =\sum_{k=1}^\infty \int_{\mathbb R^n} \left( -\Delta f_k + c_k \frac{f_k}{r^2}\right) f_k dx.

By integration by parts and Hardy inequality, we have

\displaystyle \int_{\mathbb R^n} -\Delta f_k f_kdx = \int_{\mathbb R^n} |\nabla f_k|^2 dx \geq \frac{(n-2)^2}4 \int_{\mathbb R^n} \frac{|f_k|^2}{|x|^{2}}dx.

Since c_k \geq c_1 = n-1 for any k\geq 1, hence

\displaystyle \int_{\mathbb R^n} |\nabla f(x)|^2 dx \geq \sum_{k=1}^\infty \left(\frac{(n-2)^2}4 + n-1\right)\int_{\mathbb R^n} \frac{|f_k|^2}{|x|^2}dx = \frac{n^2}4 \int_{\mathbb R^n} \frac{|f(x)|^2}{|x|^2}dx,

as we desire.

Theorem 2: If f is orthogonal to all radial functions, then

\displaystyle \frac{(n^2 -4)^2}{16} \int_{\mathbb R^n} \frac{|f|^2}{|x|^4} dx \leq \int_{\mathbb R^n} |\Delta f|^2 dx.

Proof: Writing f as above, we have

\displaystyle \int_{\mathbb R^n} \frac{|f(x)|^2}{|x|^4} dx = \sum_{k=1}^\infty \int_{\mathbb R^n} \frac{|f_k|^2}{|x|^4} dx,

and

\displaystyle \int_{\mathbb R^n} |\Delta f|^2 dx = \sum_{k=1}^\infty \int_{\mathbb R^n} \left(-\Delta f_k + c_k \frac{f_k}{r^2} \right)^2 dx = \sum_{k=1}^\infty \int_{\mathbb R^n}\left( |\Delta f_k|^2-2c_k \frac{\Delta f_k f_k}{r^2} +c_k^2 \frac{f_k^2}{r^4}\right)  dx.

By Rellich inequality, we have

\displaystyle \int_{\mathbb R^n} |\Delta f_k|^2 dx \geq \frac{n^2(n-4)^2}{16} \int_{\mathbb R^n} \frac{|f_k|^2}{r^4} dx.

Integration by parts implies

\displaystyle -\int_{\mathbb R^n} \frac{\Delta f_k f_k}{r^2}dx = \int_{\mathbb R^n} \frac{|\nabla f_k|^2}{r^2}dx + (n-4) \int_{\mathbb R^n} \frac{f_k^2}{r^4} dx.

By Hardy inequality, we have

\displaystyle  \int_{\mathbb R^n} \frac{|\nabla f_k|^2}{r^2}dx \geq \frac{(n-4)^2}4 \int_{\mathbb R^n} \frac{f_k^2}{r^4} dx.

Summarizing, we have

\displaystyle \int_{\mathbb R^n} |\Delta f|^2 dx \geq \sum_{k=1}^\infty \left(\frac{n^2(n-4)^2}{16} + 2c_k \frac{n(n-4)}2 + c_k^2\right) \int_{\mathbb R^n} \frac{f_k^2}{r^4}dx= \sum_{k=1}^\infty \left(\frac{n(n-4)}4 +c_k\right)^2\int_{\mathbb R^n} \frac{f_k^2}{r^4}dx.

Since n(n-4)/4 +c_k \geq n(n-4)/4 + n-1 = (n^2 -4)/4, then

\displaystyle \int_{\mathbb R^n} |\Delta f|^2 dx \geq \frac{(n^2 -4)^2}{16} \sum_{k=1}^\infty \int_{\mathbb R^n} \frac{f_k^2}{r^4}dx = \frac{(n^2 -4)^2}{16} \int_{\mathbb R^n} \frac{f^2}{r^4}dx,

as our desire.

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Manifolds with nonnegative Ricci curvature and Sobolev inequalities

Posted on October 1, 2016 by Van Hoang Nguyen

The main object of this post is the following nice result due to Michel Ledoux:

Let (M,g) be a complete n-dimensional Riemannian manifold with nonnegative Ricci curvature. If the following Sobolev inequality

\displaystyle \left(\int_M |f|^p dv_g\right)^{\frac1p} \leq S(n,q) \left(\int_M |\nabla_g f|^q dv_g\right)^{\frac1q}

for some 1\leq q < n with p =nq/(n-q) and S(n,q) is the best constant in the Sobolev inequalities in Euclidean space (which was found independently by Talenti and Aubin). Then (M,g) is isometric to \mathbb R^n.

Note that S(n,1) = n^{-1} \omega_n^{-1/n} where \omega_n denotes the volume of the Euclidean unit ball in \mathbb R^n and for q\in (1,n)

\displaystyle S(n,q) = \frac1n \left(\frac{n(q-1)}{n-q}\right)^{\frac {q-1}q} \left(\frac{\Gamma(n+1)}{n \omega_n \Gamma(\frac nq) \Gamma(n-1 -\frac{n}q)}\right)^{\frac1n}.

The extremal functions for the Sobolev inequalities on \mathbb R^n are given by

\displaystyle u_\lambda(x) = (\lambda + |x|^{(q-1)/q})^{1 -n/q},\qquad \lambda >0.

In order to prove the result above, we need to use the Bishop’s comparison theorem for manifolds with nonnegative Ricci curvature: fix x_0\in M, denote B(x_0,r) the geodesic ball in M with center x_0 and radius r, denote V(r) the volume of B(x_0,r), then V(r) \leq \omega_n r^n = V_0(r) the volume of unit ball in \mathbb R^n. Moreover, if equality holds for any r >0 then (M,g) is isometric to \mathbb R^n.

Let us prove the result above. Fix a point x_0\in M, for any \theta >0, denote f(x) = \theta^{-1} d(x,x_0) with d denote the geodesic metric on M. Define the function

\displaystyle F(t) = \frac1{n-1} \int_M \frac1{(t + f^{q'})^{n-1}} dv_g,\quad t >0,

with q' =q/(q-1). Note that

\displaystyle \frac1{n-1} (t + f^{q'})^{-n+1} = \int_{f^{q'}}^\infty (t +s)^{-n} ds,

hence by Fubini theorem, we get

\displaystyle F(t) = q'\int_0^\infty V(\theta s) (t + s^{q'})^{-n} s^{q'-1} ds.

By the Bishop’s comparison theorem, F is well-defined on (0,\infty) and is differentiable, and

\displaystyle F'(t) = -\int_M (t+ f^{q'})^{-n} dv_g.

Since |\nabla_g d| =1, then applying the Sobolev inequalities to the function (t + f^{q'})^{1- n/q} implies that

\displaystyle \left(\int_M (t+ f^{q'})^{-n} dv_g\right)^{\frac1p} \leq S(n,q) \frac{n-q}{q-1} \left(\int_M \frac{f^{q'}}{(t+f^{q'})^n} dv_g\right)^{\frac 1q}.

Denote

\displaystyle A = \left(S(n,q) \frac{n-q}{q-1}\right)^{-q},

the previous inequality is equivalent to

\displaystyle A(-F'(t))^{q/p}  -t F'(t) \leq (n-1) F(t).

Considering the equation

\displaystyle A(-H'(t))^{q/p}  -t H'(t) = (n-1) H(t).

A solution H_0 of this equation is given by the optimal function for the Sobolev inequalities on \mathbb R^n, i.e,

\displaystyle H_0(t) = \frac1{n-1} \int_{\mathbb R^n} (t+ |x|^{q'})^{-n+1} dx = B t^{1- n/q},

with

\displaystyle B = H_0(1) = \frac q{n-q} \left(\frac{A (n-q)}{n(q-1)}\right)^{\frac p{p-q}}.

Our aim is to compare F and H_0. We have the following claim.

Claim: if F(t_0) < H_0(t_0) for some t_0 >0 then F(t) < H_0(t) for any 0< t < t_0.

If the claim does not hold, then there exists 0< t< t_0 such that F(t) = H_0(t). Define

\displaystyle t_1 = \sup\{t\in (0,t_0)\, :\, F(t) = H_0(t)\}.

By our contradiction assumption, then t_1 < t_0. Consequently, F(t) < H_0(t) on (t_1,t_0). Define the function \varphi_t(X) =A X^{q/p} + t X. This function is strictly increasing in X \geq 0, then there exists \varphi_t^{-1} (inversion function). Then

\displaystyle (F-H_0)'(t) = \varphi_t^{-1}((n-1)H_0(t)) - \varphi_t^{-1}((n-1)F(t)) \geq 0,\qquad\forall t\in (t_1,t_0).

Hence

\displaystyle 0 = (F-H_0)(t_1) \leq (F-H_0)(t_0) < 0,

which is impossible. The claim then is followed.

We next show that

\displaystyle \liminf_{t\to 0} \frac{F(t)}{H_0(t)} \geq \theta^n > 1.          (*)

Recall that

\displaystyle F(t) = q'\int_0^\infty V(\theta s) \frac{s^{q'-1}}{(t+ s^{q'})^n} ds = q' \theta^{(n-1)q'} \int_0^\infty V(s) \frac{s^{q'-1}}{(\theta^{q'}t+ s^{q'})^n} ds,

and

\displaystyle H_0(t) = q'\int_0^\infty V_0(s) \frac{s^{q'-1}}{(1+ s^{q'})^n} ds t^{1 -n/q}.

Since \lim_{s\to 0} \frac{V(s)} {V_0(s)} =1, then for any \epsilon >0, there exists \delta >0 such that V(s) \geq (1-\epsilon) V_0(s) for any 0< s < \delta. Thus, we have

\displaystyle \int_0^\infty V(s) \frac{s^{q'-1}}{(\theta^{q'}t+ s^{q'})^n} ds \geq (1-\epsilon) \int_0^\delta V_0(s) \frac{s^{q'-1}}{(\theta^{q'}t+ s^{q'})^n} ds.

Making change of variable, we obtain

\displaystyle \int_0^\infty V(s) \frac{s^{q'-1}}{(\theta^{q'}t+ s^{q'})^n} ds \geq (1-\epsilon) \theta^{q'(1-n/q)} t^{1-n/q} \int_0^{\delta/(\theta t^{1/q'})}V_0(s) \frac{s^{q'-1}}{(1+ s^{q'})^n} ds.

Thus, we have

\displaystyle \liminf_{t\to 0} \frac{F(t)}{H_0(t)} \geq (1-\epsilon) \theta^n.

Since \epsilon > 0 is arbitrary, then (*) holds. Using our claim, (*) and \theta >1 we can conclude that F(t) \geq H_0(t) for any t >0. Indeed, if F(t_0) < H_0(t_0) for some t_0 >0, then by our claim F(t) < H_0(t) for any t\in (0,t_0), hence

\displaystyle 1 < \theta^n \leq \liminf_{t\to 0} \frac{F(t)}{H_0(t) } \leq 1,

which is impossible.  From F(t) \geq H_0(t), we have

\displaystyle \int_{0}^{\infty} (V(\theta s) -V_0(s)) \frac{s^{q'-1}}{(t+ s^{q'})^n} ds \geq 0.

Letting \theta\to 1, we obtain

\displaystyle \int_{0}^{\infty} (V( s) -V_0(s)) \frac{s^{q'-1}}{(t+ s^{q'})^n} ds \geq 0.

This inequality and the Bishop’s comparison theorem implies that V(s) = V_0(s) for any s >0, thus (M,g) is isometric to \mathbb R^n. The proof is finished.

Note: The Bishop’s comparison theorem can be found in the book of Chavel “Riemannian Geometry: A modern introduction” which can be found in this . Recent years, the result of Ledoux have been generalized to the manifolds of nonnegative Ricci curvature which support a Sobolev-type inequality (such as, Gagliardo-Nirenberg inequality, Hardy-Sobolev inequality or logarithmic Soboblev inequality) with the sharps constant as in the case of Euclidean space. Very recently, Ledoux’s result has been proved for the Sobolev inequality of order 2 (for Laplace operator) by Kristaly et al with an extra assumption on the Laplace of the geodesic distance.

Question: Is there a similar statement for the sphere, that is, if (M,g) is a n-dimensional complete Riemannian manifolds with Ric \geq (n-1) g. Assume, in addition, that the Sobolev inequality

\displaystyle {\bf \left(\int_M |f|^{\frac{2n}{n-2}} dv_g\right)^{\frac{n-2}n} \leq \frac{4}{n(n-2)} \omega_n^{-\frac{2}n} \left(\int_M |\nabla_g f|^2 dv_g + \frac{n(n-2)}4 \int_M |f|^2 dv_g\right),}

holds on M. Then (M,g) is isometric to \mathbb S^n.

Posted in Analysis, Differential Geometry, Functional inequalities, Uncategorized | Tagged , , | 1 Comment

An exercise on the affine functions

Posted on September 29, 2016 by Van Hoang Nguyen

In this post, we solve the following exercise which gives a nice character of affine functions:

Let f be a C^1 smooth function on \mathbb R^n such that |\nabla f(x)|= 1 for any x\in \mathbb R^n. Then f is affine, i.e, there are a\in \mathbb R^n, \, |a| =1 and b\in \mathbb R such that f(x) =ax +b.

Here is my solution for it. Fix x\in \mathbb R^n, let us consider the equation

\displaystyle \gamma'(t) = \nabla f(\gamma(t)), \quad \gamma(0) =x.

Since f is C^1 and |\nabla f| =1 then \gamma is well-defined on \mathbb R. Define g(t) = f(\gamma(t)), we have

\displaystyle g'(t) = \nabla f(\gamma(t))  \gamma'(t) = |\nabla f(\gamma(t))|^2 =1.

Thus, g(t) = g(0) + t for any t, or equivalent

\displaystyle f(\gamma(t)) = f(x) + t,\qquad\forall \, t.

Since |\nabla f| =1 then |f(\gamma(t)) -f(x)| \leq |\gamma(t) -x|, hence

\displaystyle |\gamma(t) -x|\geq |t|,\quad \forall \, t\in \mathbb R.       (1)

In the other hand, for t\geq 0, by Cauchy-Schwartz inequality

\displaystyle |\gamma(t) -x| = |\int_0^t \gamma'(s) ds| \leq \int_0^t |\gamma'(s)|ds = t,       (2)

with equality holds iff \gamma'(s) is constant on [0,t] (The similar result also holds for t\leq 0). From (1) and (2) we have |\gamma(t) -x| = |t| and hence \gamma(t) = x + \omega t for some \omega\in \mathbb R^n, |\omega|=1. Since \omega =\gamma'(0) = \nabla f(\gamma(0)) = \nabla f(x), then

\displaystyle \gamma(t) = x + t \nabla f(x),\qquad t\in \mathbb R,

and

\displaystyle f(x + t \nabla f(x)) = f(x) + t.

Consequently, if f(y) = f(x) then

\displaystyle |x+ t\nabla f(x) -y| \geq |f(x+ t\nabla f(x)) - f(y)| = |t|, \qquad\forall t\in \mathbb R,

which is equivalent to

\displaystyle |x-y|^2 + t (x-y) \nabla f(x) \geq 0,\qquad\forall t\in \mathbb R

hence (y-x) \nabla f(x) =0. Thus, we have shown that

\displaystyle G:= \{y\, :\, f(y) = f(x)\} \subset \{y\, :\, (y-x)\nabla f(x) =0\}.    (*)

We next show that for any y\in \mathbb R^n, there exists an open subset U\subset \mathbb R^n containing y such that

\displaystyle U \cap\{z\,:\, f(z) = f(y)\} = U \cap\{z\, :\, (z-y)\nabla f(y) =0\}.   (**)

Without loss of generality, we can assume that y =0, and \nabla f(0) = (0,\ldots,0,1). Since \partial_n f(0) =1, then by implicit function theorem, there exist an open subset V\subset \mathbb R^{n-1} containing 0, an open subset W \subset \mathbb R containing 0 and a C^1 function g : V \to W such that

\displaystyle V\times W\cap \{z\,:\, f(z) = f(0)\} = \{(a, g(a))\, :\, a \in V\}.

Since f((a,g(a)) = f(0) then by (*) (for x =0) we have g(a) =0 for any a\in V. Denote U = V\times W, we have

\displaystyle U\cap \{z\,:\, f(z) =f(0) \}= \{(a,0)\,:\, a\in V\} = U \cap\{z\, :\, z_n =0\}.

For any y\in G, choosing U such that (**) holds. By (*) we have

\displaystyle U \cap \{z\, :\, (z-y)\nabla f(y) =0\} \subset U\cap \{z\, :\, (z-x) \nabla f(x) =0\}.

Since (y-x) \nabla f(x) =0, hence (z-y) \nabla f(x) =0 for any z \in U\cap\{z\,:\, (z-y) \nabla f(y) =0\}. This implies that \nabla f(x) =\pm \nabla f(y), hence

\displaystyle U \cap G = U\cap\{z\,:\, (z-x) \nabla f(x) =0\}.

Thus, G is an open subset of \{z\, :\, (z-x)\nabla f(x) =0\} with respect to the topology restricting from \mathbb R^n. It is obvious that G is a close subset, then G = \{z\, :\, (z-x)\nabla f(x) =0\} by the connectedness. The proof above also shows that \nabla f(y) = \pm \nabla f(x) for any y\in G. Since f is C^1 then \nabla f(y) =\nabla f(x) for any y\in G.

We finish our proof by showing that \nabla f is constant on \mathbb R^n. If not, there exist a,b \in \mathbb R^n such that \nabla f(a) \not= \nabla f(b). There are two cases:

Case 1: If \nabla f(a) and \nabla f(b) are linearly independent. Then we can find z in the subspace spanning by \nabla f(a) and \nabla f(b) such that

\displaystyle (z-a)\nabla f(a) = (z-b) \nabla f(b) =0.

Consequently, by (*) we get f(a) = f(z) = f(b) and hence \nabla f(a) = \nabla f(z) = \nabla f(b) which is impossible.

Case 2: If \nabla f(a) = -\nabla f(b), considering the function g(t) = \nabla f(a) \nabla f(a+ t(b-a)). This function is continuous on [0,1], g(0) =1, g(1) = -1, then there is t_0 \in (0,1) such that g(t_0) =0, hence \nabla f(a) and \nabla f(a + t_0(b-a)) are linearly independent. This is impossible by Case 1.

Thus, \nabla f is constant on \mathbb R^n, or f is affine.

 

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Volume inequality for even isotropic measures

Posted on September 5, 2016 by Van Hoang Nguyen

Let \mu be an even isotropic measure on the sphere S^{n-1}, where isotropic means that

\displaystyle \int_{S^{n-1}} v \otimes v d\mu(v) = Id_{\mathbb R^n},

here v\otimes v denotes the orthogonal projection to the line generated by v. Let f be an even, positive, continuous function in S^{n-1}. For 1\leq p\leq \infty, we denote Z_{p,f}(\mu) the convex body whose support function is given by

\displaystyle h_{Z_{p,f}(\mu)}(x) = \left( \int_{S^{n-1}} \frac{|x\cdot v|^p}{f(v)^p} d\mu(v)\right)^{\frac1p}.

Note that Z_{p,f}(\mu) is an origin-symmetric convex body.

In this post, we will prove the following inequality:

\displaystyle Vol(Z_{p,f}(\mu)^*) \leq 2^n \frac{\Gamma(1 + \frac1p)^n}{\Gamma(1 +\frac np)} \exp\Big(\int_{S^{n-1} }\ln f(v) d\mu(v)\Big), \quad\quad\quad\quad (1)

here Vol(K) denotes  the volume of a subset of \mathbb R^n, and K^* denote the polar convex body of the origin-symmetric convex body K. Moreover, there is equality in (1) iff there exists an orthogonal basis \{u_i\}_{i=1}^n such that 

\displaystyle \mu = \frac12 \sum_{i=1}^n \Big(\delta_{u_i} + \delta_{-u_i}\Big),

(here \delta_u denotes the Dirac measure concentrated at u), and f is constant on \{\pm u_i\}_{i=1}^n.

When f\equiv 1, the inequality (1) was proved by E. Lutwak, D. Yang and G. Zhang (see this paper) . When \mu is discrete measure, and f is constant on the support of \mu, the inequality (1) was proved by K. Ball (this paper) and the equality cases were characterized by F. Barthe (this paper).

Let us prove (1). Our proof is based on the mass transportation method. We also need the following inequality due to Ball and Barthe: If t: S^{n-1} \to (0,\infty) is continuous and \mu is an isotropic measure on S^{n-1}, then

\displaystyle det \Big(\int_{S^{n-1}} t(v) v\otimes v d\mu (v)\Big) \geq \exp\Big(\int_{S^{n-1}} \ln t(v) d\mu(v)\Big)\quad\quad\quad\quad (2)

with equality iff t(v_1) \cdots t(v_n) is constant for linearly independent v_1,\ldots, v_n in support of \mu (see this paper for a proof of this result).

Let \Phi :\mathbb R \to \mathbb R such that

\displaystyle \frac1{\Gamma(1 + \frac1p)} \int_{-\infty}^s e^{-|t|^p} dt = \frac2{\sqrt{\pi}} \int_{-\infty}^{\Phi(s)} e^{-t^2} dt.

Indeed, an expression of \Phi is as follows. Denote

\displaystyle \varphi_1(s) = \frac1{\Gamma(1 + \frac1p)} \int_{-\infty}^s e^{-|t|^p} dt,\quad \varphi_2(s) = \frac2{\sqrt{\pi}} \int_{-\infty}^{s} e^{-t^2} dt.

Then \varphi_1, \varphi_2 : \mathbb R \to (0,2) are strictly increasing, homeomorphism. Then \Phi = \varphi_2^{-1}\circ \varphi_1. Note that \Phi is strictly increasing, then \Phi'(s) > 0 for any s\in \mathbb R. Taking the differentiation, we get

\displaystyle |s|^p = -\ln \Phi'(s) - \ln c_{p,2} + |\Phi(s)|^2,\quad c_{p,2} = \frac{2\Gamma(1+ \frac1p)}{\sqrt{\pi}}.\quad\quad\quad\quad (3)

Define the map T:\mathbb R^n \to \mathbb R^n by

\displaystyle Tx = \int_{S^{n-1}} u \Phi(\frac{u\cdot x}{f(u)}) d\mu(u).

We have

\displaystyle dTx =  \int_{S^{n-1}} \frac{u\otimes u}{f(u)} \Phi'(\frac{u\cdot x}{f(u)}) d\mu(u).

From this we have

\displaystyle v \cdot dTx(v) = \int_{S^{n-1}} \frac{|u\cdot v|^2}{f(u)} \Phi'(\frac{u\cdot x}{f(u)}) d\mu(u).

Thus, dTx is a strict positive matrix, hence T is an injective. We have

\displaystyle \Gamma(1 + \frac np) Vol(Z_{p,f}(\mu)^*) = \int_{\mathbb R^n} \exp\Big(-h_{Z_{p,f}(\mu)}(x)^p\Big) dx = \int_{\mathbb R^n} \exp\Big(-\int_{S^{n-1}} \frac{|x\cdot v|^p}{f(v)^p} d\mu(v)\Big) dx.

This and (3) imply

\displaystyle \Gamma(1 + \frac np) Vol(Z_{p,f}(\mu)^*) = \int_{\mathbb R^n} \exp\Big(\int_{S^{n-1}}\Big(\ln c_{p,2} + \ln \Phi'(\frac{x\cdot u}{f(u)}) - \Phi(\frac{x\cdot u}{f(u)})^2\Big) d\mu(u)\Big) dx.\quad\quad\quad (4)

Using Ball-Barthe inequality (2), we obtain

\displaystyle det(dTx) \geq \int_{S^{n-1}} \ln \Big(\frac1{f(u)}\Phi'(\frac{x\cdot u}{f(u)}\Big) d\mu(u) =-\int_{S^{n-1}} \ln f(u) d\mu(u) + \int_{S^{n-1}} \ln \Phi'(\frac{x\cdot u}{f(u)}) d\mu(u),

or equivalently,

\displaystyle \int_{S^{n-1}} \ln \Phi'(\frac{x\cdot u}{f(u)}) d\mu(u) \leq det(dTx) + \int_{S^{n-1}} \ln f(u) d\mu(u). \quad\quad\quad (5)

Thank to Holder inequality, we have

\displaystyle |Tx|^2 = \int_{S^{n-1}} u\cdot Tx \Phi(\frac{x\cdot u}{f(u)}) d\mu(u) \leq \Big( \int_{S^{n-1}} (u\cdot Tx)^2 d\mu(u)\Big)^{\frac12} \Big( \int_{S^{n-1}}\Phi(\frac{x\cdot u}{f(u)})^2 d\mu(u)\Big)^{\frac12}.

Since \mu is isotropic, then

\displaystyle |x|^2 = \int_{S^{n-1}} |x\cdot u|^2 d\mu(u),\quad\quad\quad\quad (7)

for any x\in \mathbb R^n. Consequently, we have

\displaystyle |Tx|^2 \leq \int_{S^{n-1}}\Phi(\frac{x\cdot u}{f(u)})^2 d\mu(u). \quad\quad\quad\quad (6)

Denote c(f) = \exp\Big(\int_{S^{n-1}} \ln f(u) d\mu(u)\Big). Combining (4), (5) and (6) implies that

\displaystyle \Gamma(1 + \frac np) Vol(Z_{p,f}(\mu)^*) \leq c(f) c_{p,2}^n \int_{\mathbb R^n} e^{-|Tx|^2} det(dTx) dx = c(f) c_{p,2}^n \int_{T\mathbb R^n} e^{-|x|^2} dx.

Thus,

\displaystyle \Gamma(1 + \frac np) Vol(Z_{p,f}(\mu)^*) \leq c(f) c_{p,2}^n \int_{\mathbb R^n} \exp(-|x|^2) dx=  \pi^{\frac n2}c_{p,2}^n c(f).

This implies (1).

Suppose that there is equality in (1), thus we must have equality in Ball-Barthe inequality, hence for any x\in \mathbb R^n,

\displaystyle \frac1{f(u_1)} \Phi(\frac{x\cdot u_1}{f(u_1)})\cdots \frac1{f(u_n)} \Phi(\frac{x\cdot u_n}{f(u_n)})

are constant for any linearly independent u_1,\ldots,u_n in support of \mu. Since \mu is isotropic, then \mu is not concentrated in any great subsphere. Repeating the argument of Lutwak, Yang, Zhang (Lemma 4.1 and 4.3 in this paper), we obtain the condition of equality for (1).

The case p=\infty can be obtained by letting p\to\infty in (1). Another way to prove this inequality is using the mass transportation method as above. In this case, we replace the function e^{-|t|^p} /\Gamma(1 + \frac1p) by the function \chi_{[-1,1]}(t) and the function \Phi is defined in (-1,1).

Open problem: What happens if \mu is not even? In this case, we have an extra assumption that \mu is centered at origin, i.e, \int_{S^{n-1}} u d\mu(u) = 0. We denote Z_{p,f}^+(\mu) the convex body whose support function is given by

\displaystyle h_{Z_{p,f}^+(\mu)}(x) = \Big(2\int_{S^{n-1}} \frac{(x\cdot u)_+^p}{f(u)^p} d\mu(u)\Big)^{\frac1p},

where a_+ denotes the positive part of a. In this direction, we have the following conjecture. Let \nu denote the isotropic measure on S^{n-1} such that its support is the vertices of a regular simplex, then

\displaystyle Vol(Z_{p,f}^+(\mu)^*) \leq  Vol(Z_{p,f}^+(\nu)^*).

This conjecture was proved in the case p =\infty by E. Lutwak, D. Yang and G. Zhang (see this paper) for the function f\equiv 1 and by F. E. Schuster and M. Weberndorfer (see this paper) for any function f.

 

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Borell’s proof of Prékopa inequality

Posted on August 9, 2016 by Van Hoang Nguyen

Prékopa inequality says that if f,g,h are nonnegative, measurable functions on \mathbf{R}^n such that

\displaystyle \log h\left(\frac{x+y}2\right) \geq \frac12 \left(\log f(x) + \log g(y)\right)

holds for any x,y \in \mathbf{R}^n. Then

\displaystyle \log \left(\int h\right) \geq \frac12 \left(\log \left(\int f\right) + \log \left(\int g\right)\right).                            (1)

There are many proofs of (1). In this post, we explain a nice proof of (1) by Borell which is based on the heat flow (a generalization of this method can be found in this paper). Let f be a “good” nonnegative function on \mathbf R^n, we consider P_tf the solution of the heat equation

\displaystyle \partial_t P_tf = \Delta_x P_tf,\qquad P_0 f = f.

Note that P_tf has the nice representation as follows

\displaystyle P_tf(x) = (4\pi t)^{-n/2} \int f(y) e^{-|x-y|^2/4t} dy.

Then we have

\displaystyle \log P_t f(0) = -\frac n2 \log (4\pi t) + \log\left(\int f(y) e^{-|y|^2/4t} dy\right).               (2)

To prove (1), we will prove the following pointwise estimate

\displaystyle \log P_t h\left(\frac{x+y}2\right) \geq \frac 12\left(\log P_tf(x)+ \log P_t g(y)\right)              (3)

for any x,y \in \mathbf R^n. By taking x=y=0 in (3), using (2) and letting t\to\infty, we obtain (1).

Define the function C on (0,\infty)\times \mathbf R^{2n} by

\displaystyle C(t,x,y) = \log P_t h\left(\frac{x+y}2\right) - \frac 12\left(\log P_tf(x)+ \log P_t g(y)\right),

A simple computation shows that

\displaystyle \partial_t C(t,x,y) = \frac{\Delta P_th((x+y)/2)}{P_th((x+y)/2)} - \frac12\left(\frac{\Delta P_tf(x)}{P_tf(x)} + \frac{\Delta P_tg(y)}{P_tg(y)}\right).

We have

\displaystyle \partial_{x_i} C = \frac12 \frac{\partial_i P_th((x+y)/2)}{P_th((x+y)/2)} - \frac12 \frac{\partial_i P_t f(x)}{P_t f(x)},

\displaystyle \partial_{x_ix_j}C = \frac14\left(\frac{\partial_{ij}P_th(\frac{x+y}2)}{P_th(\frac{x+y}2)}-\frac{\partial_i P_th(\frac{x+y}2)\partial_j P_th(\frac{x+y}2)}{(P_th(\frac{x+y}2))^2}\right) -\frac12\left(\frac{\partial_{ij}P_t f(x)}{P_tf(x)} -\frac{\partial_i P_t f(x) \partial_j P_tf(x)}{(P_t f(x))^2}\right),

\displaystyle \partial_{y_i} C = \frac12 \frac{\partial_i P_th((x+y)/2)}{P_th((x+y)/2)} - \frac12 \frac{\partial_i P_t g(y)}{P_t g(y)},

\displaystyle \partial_{y_iy_j}C = \frac14\left(\frac{\partial_{ij}P_th(\frac{x+y}2)}{P_th(\frac{x+y}2)}-\frac{\partial_i P_th(\frac{x+y}2)\partial_j P_th(\frac{x+y}2)}{(P_th(\frac{x+y}2))^2}\right) -\frac12\left(\frac{\partial_{ij}P_t g(y)}{P_tg(y)} -\frac{\partial_i P_t g(y) \partial_j P_tg(y)}{(P_t g(y))^2}\right),

and

\displaystyle \partial_{x_i y_j} C = \frac14\left(\frac{\partial_{ij}P_th(\frac{x+y}2)}{P_th(\frac{x+y}2)}-\frac{\partial_i P_th(\frac{x+y}2)\partial_j P_th(\frac{x+y}2)}{(P_th(\frac{x+y}2))^2}\right).

Hence

\displaystyle \sum_{i=1}^{n} (\partial_{x_ix_i}C + 2\partial_{x_iy_i} C+ \partial_{y_iy_i}C) = \frac{\Delta P_th(\frac{x+y}2)}{P_th(\frac{x+y}2)} - \frac12\left(\frac{\Delta P_tf(x)}{P_tf(x)} + \frac{\Delta P_tg(y)}{P_tg(y)}\right)-\qquad\qquad\qquad\qquad\qquad\qquad\qquad -\frac{|\nabla P_th(\frac{x+y}2)|^2}{(P_t h(\frac{x+y}2))^2} + \frac12\left(\frac{|\nabla P_tf(x)|^2}{(P_tf(x)^2} + \frac{|\nabla P_tg(y)|^2}{(P_tg(y))^2}\right).

Denote the vector field b: \mathbf R^{2n}\to \mathbf R^{2n} by

\displaystyle b_i(x,y) = \frac{\partial_i P_th(\frac{x+y}2)}{P_th(\frac{x+y}2)} + \frac{\partial_i P_tf(x)}{P_tf(x)},\qquad b_{n+i}(x,y) =\frac{\partial_i P_th(\frac{x+y}2)}{P_th(\frac{x+y}2)} + \frac{\partial_i P_tg(y)}{P_tg(y)}.

Consequently, we have

\displaystyle \partial_t C = \mathcal{E} C + \langle b,\nabla C\rangle,

where \mathcal{E} is the degenerate elliptic operator,

\displaystyle \mathcal{E} = \sum_{i=1}^n (\partial_{x_ix_i}C + 2\partial_{x_iy_i} C+ \partial_{y_iy_i}C).

Since C(0,x,y) \geq 0, using the theory of parabolic-elliptic evolution equations, we then have C(t,x,y) \geq 0 which is exact (3).

Question: Borell-Brascamp-Lieb (BBL) inequality says that if f,g,h are nonnegative, measurable functions on \mathbf R^n such that

\displaystyle h^\alpha\left(\frac{x+y}2\right) \leq \frac12 \left( f(x)^\alpha + g(y)^\alpha\right),

with \alpha \in (-1/n,0). Then

\displaystyle \left(\int h\right)^{\gamma} \leq \frac12 \left( \left(\int f\right)^\gamma + \left(\int g\right)^\gamma\right),

where \gamma = \alpha/(1+ n\alpha).

Is there a proof of BBL by this way? Maybe, we need use the fast diffusion equation

\displaystyle \partial_t u = \Delta u^m,

for m\in (0,1).

 

 

 

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An improved Poincaré inequality for Gaussian measure.

Posted on July 11, 2015 by Van Hoang Nguyen

Let \gamma_n denote standard Gaussian measure on \mathbf{R}^n, i. e

\displaystyle d\gamma_n(x) = \frac{e^{-|x|^2/2}}{(2\pi)^{n/2}} \, dx.

The Poincaré inequality for \gamma_n states that for any Lipschitz function f in L^2(\gamma_n), it holds

\displaystyle {\rm Var}_{\gamma_n}(f) := \int f^2 d\gamma_n - \left(\int f d\gamma_n\right)^2 \leq \int |\nabla f|^2 d\gamma_n.

This inequality is sharp and equality holds iff f(x) = ax + b for some vector a\in \mathbf{R}^n and b \in \mathbf{R}.

An improvement of Poincaré inequality was recently proved by Goldstein, Nourdin and Peccati by using the spectral decomposition for the operator L = \Delta - x \nabla on L^2(\gamma_n) (see Section 6.2 in this link). This inequality is formulated as

\displaystyle {\rm Var}_{\gamma_n}(f) \leq \int |\nabla f|^2 d\gamma_n - \frac12 \int |\nabla f - \int \nabla f d\gamma_n|^2 d\gamma_n.           (1)

In this post, we give a new, short and simple proof of (1) based on the L^2-approach.  We first remark that we can assume \int f d\gamma_n =0. There exists u\in D(L) (domain of L) such that Lu =f. It is easy to check that

\displaystyle \int |Lu|^2 d\gamma_n = \int |\nabla u|^2 d\gamma_n + \int \|\nabla^2 u\|_{HS}^2 d\gamma_n,           (2)

where \nabla^2 u denotes Hessian of u. We have

\displaystyle {\rm Var}_{\gamma_n}(f) = 2\int f Lu d\gamma_n -\int (Lu)^2 d\gamma_n = -2\int \nabla u \nabla f d\gamma_n -\int (Lu)^2 d\gamma_n.     (3)

Thus, combining (2) and (3) yields

\displaystyle {\rm Var}_{\gamma_n}(f) = \int |\nabla f|^2 -\left(\int |\nabla u + \nabla f|^2 d\gamma_n + \int \|\nabla^2 u\|_{HS}^2 d\gamma_n\right).    (4)

Using Poincaré inequality, we get

\displaystyle \int \|\nabla^2 u\|_{HS}^2 d\gamma_n = \sum_{i=1}^n \int |\nabla \partial_i u|^2 d\gamma_n \geq \int |\nabla u - \int \nabla u d\gamma_n|^2 d\gamma_n.    (5)

Integration by parts implies that

\displaystyle \int \partial_i u d\gamma_n = \int u x_i d\gamma_n = -\int u L x_i d\gamma_n = -\int f x_i d\gamma_n = -\int \partial_i f d\gamma_n.   (6)

Combining (4), (5) and (6) and using inequality 2(|x|^2 + |y|^2) \geq |x+y|^2, we obtain (1).

Posted in Analysis, Functional inequalities, Probability | Tagged | 2 Comments

L_p affine isoperimetric inequalities

Posted on April 23, 2015 by Van Hoang Nguyen

In this post, we give an application of the shadow system (see here) to prove a L_p affine isoperimetric inequality due to Lutwak, Yang and Zhang (see this paper).  The proof we present here comes from the paper of Campi and Gronchi (see this paper).

Let K be a convex body in \mathbf{R}^n containing origin in its interior. Its L_p-centroid body is defined to be a convex body whose support function is given by

\displaystyle h(\Gamma_p K,x)^p = \frac1{c_{n,p} V(K)} \, \int_K | x\cdot y|^p dy,

where

\displaystyle c_{n,p} = \frac{\omega_{n+p}}{\omega_2\, \omega_n\, \omega_{p-1}},\quad \omega_q = \frac{\pi^{q/2}}{\Gamma(1 + k/2)},

is the normalized constant such that \Gamma_p B = B with B is the Euclidean unit ball.

The L_p Busemann-Petty centroid inequality states that 

\displaystyle V(\Gamma_p K) \geq V(K),

for any convex body K containing origin in its interior. Moreover, equality holds if and only if K is an origin symmetric ellipsoid.

In the following, we give a proof of this inequality by using the shadow system technique. We start with the following result

Theorem: Let (K_t)_t, t\in [0,1] be a parallel chord movement along the direction v, then (\Gamma_p K_t)_t is a shadow system in the same direction. 

Proof: Since \Gamma_p K_t is origin symmetric, we assume that

\displaystyle \Gamma_p K_t =\{x +y v\, :\, x\in \Gamma_p K_t\, | v^\bot,\, -g_t(-x) \leq y \leq g_t(x)\},

where g_t is a concave function on \Gamma_p K_t\, |v^\bot, and \Gamma_p K_t\, | v^\bot is the orthogonal projection of \Gamma_p K_t on v^\bot.

We first prove that \Gamma_p K_t \, | v^\bot is independent of t. Since (K_t)_t is a parallel chord movement along the direction v, then V(K_t) is independent of t, and there exists a speed constant \beta on v^\bot such that

\displaystyle K_t =\{z +t \beta(x) v\, :\, z\in K,\, x = z -(z\cdot v)v\},\quad t\in [0,1],

where K = K_0. Using the Fubini’s theorem, we obtain easily that if $u\in v^\bot$, then

\displaystyle h(\Gamma_p K_t\,|v^\bot,u) = h(\Gamma_p K_t,u) = h(\Gamma_p K,u) = h(\Gamma_p K\, | v^\bot, u),

where the first and the last equality come from the simple fact that h(L|v^\bot, u) = h(L,u) for any u \in v^\bot, and the second equality comes from Fubibi’s theorem. From these equalities, we obtain our assertion above.

Next we show that the function t\to g_t(x) is convex for every x\in \Gamma_p K_t| v^\bot. For a z\in \mathbf{R}^n, let us denote z| v^\bot for z -(z\cdot v) v.  Now using the Fubini’s theorem we have

\displaystyle h(\Gamma_p K_t,u)^p = \frac{1}{c_{n,p} V(K_0)} \int_K | z\cdot u + t\beta(z| v^\bot)|^p dz.     (1)

We have

\displaystyle g_t(x) =\max\{y\,:\, x +y v\in \Gamma_p K_t\} =\max\{y\,:\, (x+yv)\cdot u \leq h(\Gamma_p K_t,u),\, \forall u\},

hence

\displaystyle g_t(x) = \max\{y\, :\, y\, u\cdot v \leq h(\Gamma_p K_t,u) - x\cdot u,\, \forall\, u\}.

Since the support function is 1-homogeneous, and the vectors u such that u\cdot v \leq 0 give no bound for y, then we have

\displaystyle g_t(x) = \max\{y\,:\, y \leq h(\Gamma_p K_t, \omega + v)- x\cdot \omega,\, \omega\in v^\bot\},

or equivalent

\displaystyle g_t(x) =\inf\{h(\Gamma_p K_t, \omega +v) -x\cdot \omega,\, \omega\in v^\bot\}.     (2)

Using (1) and triangle inequality (or Minkowski inequality), we have for any \omega_1,\omega_2 \in v^\bot and a, t,s\in [0,1], denote b = at +(1-a)s,

\displaystyle h(\Gamma_p K_{b}, a\omega_1 +(1-a)\omega_2 +v) \leq a h(\Gamma_p K_t,\omega_1 +v)+ (1-a) h(\Gamma_p K_s,\omega_2 +v).          (3)

From (2), it is easy to show that

\displaystyle g_{b}(x) =\inf\{h(\Gamma_p K_{b}, a\omega_1 + (1-a)\omega_2 +v) -x\cdot(a\omega_1 +(1-a)\omega_2)\, :\, \omega_1, \omega_2 \in v^\bot\}.  (4)

From (2), (3) and (4), we obtain

\displaystyle g_{at + (1-a)s}(x) \leq a g_t(x) + (1-a) g_s(x).

We continue by showing that

\displaystyle -g_b(-x) \leq -a g_t(-x) + (1-a) g_s(x) \leq g_b(x),          (5)

with a,b,t,s\in [0,1] as above. In fact, we only have to prove the inequality on right hand side (indeed, changing x by -x we obtain the one on left hand side). As above, we have from (2) that

\displaystyle (1-a) g_s(x) = \inf\{h(\Gamma_p K_s, \omega_1 -a\omega_2 +(1-a)v) - x\cdot (\omega_1 -a \omega_2)\, \omega_1,\omega_2 \in v^\bot\}.    (6)

Using (1) and triangle inequality, we obtain

\displaystyle h(\Gamma_p K_s,\omega_1 -a\omega_2 +(1-a)v) \leq h(\Gamma_p K_b,\omega_1 +v) + a h(\Gamma_p K_t, \omega_2 +v),  (7)

From (2), (6) and (7) we obtain

\displaystyle (1-a) g_s(x) \leq g_{at +(1-a)s}(x) + a g_t(-x).

Hence (\Gamma_p K_t)_t is a shadow system (see the post). A result of Rogers and Shephard says that V(\Gamma_p K_t) is a convex function t\in [0,1].

To obtain the extremizers for the L_p-Busemann-Petty centroid inequality, we need the following result.

Theorem: V(\Gamma_p K_t) is strictly convex function unless \beta is a linear function on v^\bot.

Proof: Suppose that there exist t,s \in [0,1], a\in (0,1), t\not = s such that

\displaystyle V(\Gamma_p K_{at + (1-a)s}) = a V(\Gamma_p K_t) + (1-a) V(\Gamma_p K_s).     (*)

Since

\displaystyle V(\Gamma_p K_t) =\int_{v^\bot\cap \Gamma_p K} (g_t(x) + g_t(-x) dx = 2\int_{v^\bot \cap \Gamma_p K} g_t(x) dx.

From (*), the convexity of function t\to g_t(x) and the continuity of function g_t(\cdot), we must have

\displaystyle g_{at +(1-a) s}(x) = a g_t(x) + (1-a) g_s(x),\quad\forall\, x \in \Gamma_p K | v^\bot.    (**)

Fix a x in the interior of \Gamma_p K | v^\bot w.r.t topology on v^\bot, then there exist u_1,u_2 \in v^\bot such that

\displaystyle g_t(x) = h(\Gamma_pK_t,u_1 +v) -x\cdot u_1,\, g_s(x) = h(\Gamma_p K_s, u_2 +v).

Equality (**) for this x implies that

\displaystyle h(\Gamma_p K_{at +(1-a)s}, au_1 +(1-a)u_2 +v) = a h(\Gamma_p K_t, u_1 +v) - (1-a) h(\Gamma_p K_s, u_2 +v).

So we must have equality in the Minkowski inequality for integral, then

\displaystyle (u_1 +v) \cdot z +t\beta(z|v^\bot) = c (u_2 +v) \cdot z + cs\beta(z| v^\bot),

for every z \in K and for some constant c. Taking z =z' + \lambda v where z' in the interior of K| v^\bot, and \lambda is small. Differenting the equality in \lambda we obtain c =1. Since t\not =s then \beta is a linear function.

Come back to prove the L_p Busemann-Petty centroid inequality, suppose that in the direction v, the convex body K is given by

\displaystyle K =\{ x + y v\, :\, x \in K| v^\bot,\, f_K(x) \leq y \leq g_K(x)\},

for some suitable function f_K, g_K. Denote \beta(x) = -(f_K(x) + g_K(x)), and the family (K_t) is the parallel chord movement of K along the direction v with speed function \beta. Then K_{1/2}= S_vK is the Steiner symmetrization of K through the v^\bot and K_1 =K^v is the refection of K through v^\bot, then V(\Gamma_p K) =V(\Gamma_p K_1). From the convexity of V(\Gamma_p K_t) we have

\displaystyle V(\Gamma_p S_vK) \leq V(\Gamma_p K).       (***)

Choosing a sequence (u_i)_i \in S^{n-1} such that the sequence of convex bodies K_0 =K, K_i =S_{v_i} K_{i-1} converges (in the Hausdorff distance) to an origin-centered ball B with V(B) = V(K). Then \Gamma_p K_i converges to \Gamma_p B in the Hausdorff distance. From (***) we have

\displaystyle V(\Gamma_p K_i) \leq V(\Gamma_p K),

for any i. Letting i tend to infinity, we obtain the L_p Busemann-Petty centroid inequality.

Let us now assume that V(\Gamma_p K) = V(K), then for any direction v we have V(\Gamma_p S_v K) = V(\Gamma_p K). From the second theorem above, we must have \beta is a linear function, hence S_vK = K_{1/2} is a linear image of K.  If K is not an origin-symmetric ellipsoid, then a result of Petty says that there is a direction v such that S_v K is not a linear image of K. This is impossible, hence K must be an origin-symmetric ellipsoid.

 

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Đề thi Olympic sinh viên Toán 2015, bảng A

Posted on April 16, 2015 by Van Hoang Nguyen
de thi

Đề thi Olympic sinh viên 2015, bảng A

Câu A1: 1) Do a_{n+1} -a_n = -a_n^2/2 \leq 0 nên \{a_n\}_n là dãy giảm.

2) Do a_0 =1\in [0,2], và hàm f(x) = 2x -x^2 \geq 0 với x\in [0,2], do đó sử dụng qui nạp ta chứng minh được a_n \in [0,2] với mọi $n$. Từ đó suy ra dãy \{a_n\}_n hội tụ. Dễ dàng suy ra giới hạn bằng 0.

3) Nhận xét rẳng nếu dãy \{a_n\}_n hội tụ thì giới hạn phải bằng 0. Do dãy này giảm nên ta phải có a_n\geq 0 với mọi n. Do đó a_0 \geq 0. Điều kiện a_1 \geq 0 suy ra a_0 \in [0,2]. Do đó dãy hội tụ thì a_0\in [0,2]. Ta chứng minh đây cũng là điều kiện đủ. Thật vậy nếu a_0\in [0,2] sử dụng tính chất đơn điệu giảm của dãy và phương pháp qui nạp ta chứng minh được a_n \in [0,2] với mọi n. Từ đó suy ra dãy hộ tụ và giới hạn bằng 0.

Câu A2: Nếu \beta =0 thì f(x) = f(0) + x^2/\alpha^2.

Nếu \alpha =0 thì f(x) = f(0) -x^2/\beta^2.

Nếu \alpha \beta \not= 0, đặt r = \beta/\alpha. Ta có hai trường hợp sau:

TH1: Nếu |r| < 1, khi đó ta có

\displaystyle f(x) = f(rx) + \frac{x^2}{\alpha^2} = f(r^2x) + \frac{x^2}{\alpha^2} + \frac{r^2 x^2}{\alpha^2} = \cdots = f(r^n x) + \frac{x^2}{\alpha^2} \sum_{k=0}^n r^{2k},\quad \forall\, n.

Cho n\to \infty sử dụng tính liên tục tại 0 của hàm f suy ra

\displaystyle f(x) = f(0) + \frac{x^2}{\alpha^2} \sum_{k=0}^\infty r^{2k} = f(0) + \frac{x^2}{\alpha^2 -\beta^2}.

TH2: Nếu |r| > 1, đặt s = 1/r khi đó ta có (tương tự như trên bởi chuyển x^2 sang vế phải)

\displaystyle f(x) = f(sx) - \frac{x^2}{\beta^2} =\cdots = f(s^n x) - \frac{x^2}{\beta^2} \sum_{k=0}^n s^{2k},\quad\forall\, n.

Cho n\to \infty và sử dụng tính liên tục tại 0 của hàm f suy ra

\displaystyle f(x) = f(0) - \frac{x^2}{\beta^2} \sum_{k=0}^\infty s^{2x} = f(0) + \frac{x^2}{\alpha^2 -\beta^2}.

Vậy hàm f có dạng

\displaystyle f(x) = c + \frac{x^2}{\alpha^2 -\beta^2},

 với c là hằng số.

Nếu |\alpha| = |\beta| thì không tồn tại hàm số f. Thật vậy nếu \alpha = \beta thì kết luận là hiển nhiên, nếu \alpha = -\beta thì hàm f có tính chất này phải thỏa mãn

\displaystyle f(\alpha x) = f(-\alpha x) + x^2 = f(\alpha x) + 2 x^2,\quad \forall\, x \in \mathbf{R}.

Điều này là không thể.

Câu A3: Xét hàm g(x,y) = 6 y^2 f'(x) + 4 x f'(y) trên [0,1]^2. Ta có

\displaystyle \int_0^1\int_0^1 g(x,y) dxdy = 4 (f(1) -f(0)) = (f(1) - f(0)) \int_0^1\int_0^1 24 x y^2 dxdy.

Do hàm g liên tục trên [0,1]^2 nên theo định lý giá trị trung gian, có tồn tại x_1, x_2 \in (0,1)^2 sao cho

\displaystyle 6x_2^2 f'(x_1) + 4 x_1 f'(x_2) = 24 (f(1) -f(0))x_1 x_2^2.

Mặt khác tồn tại x_3 \in (0,1) sao cho f'(x_3) = f(1) -f(0) bởi định lý Lagrange. Từ đây suy ra điều phải chứng minh.

Câu A4: Đặt

\displaystyle F(x) = \int_0^x f(t)^2 dt.

Khi đó F là hàm tăng, khả vi và F'(x) = f(x)^2. Từ giả thiết suy ra

\displaystyle \lim_{x\to\infty} F'(x) F(x)^2 = a^2,

Do đó với mọi \epsilon \in (0,a^2), có tồn tại A = A(\epsilon) sao cho

\displaystyle a^2 -\epsilon < F'(x) F(x)^2 < a^2 +\epsilon,\quad\forall\, x\geq A.

Với mọi x > A, ta có

\displaystyle (a^2-\epsilon)(x-A) < \int_A^x F'(t) F(t)^2 dt < (a^2 + \epsilon)(x-A),

hay là

\displaystyle (a^2 -\epsilon)(1-\frac Ax) < \frac{F(x)^3-F(A)^3}{3x} < (a^2 +\epsilon) (1 -\frac Ax),\quad\forall\, x > A.

Do đó, từ bất đẳng thức vế trái suy ra

\displaystyle \liminf_{x\to\infty} \frac{F(x)^3}{3x} \geq a^2 -\epsilon,

và bất đẳng thức vế phải suy ra

\displaystyle \limsup_{x\to\infty} \frac{F(x)^3}{3x} \leq a^2 + \epsilon.

Do \epsilon \in (0,a^2) là túy ý nên

\displaystyle \liminf_{x\to\infty} \frac{F(x)^3}{3x} \geq a^2,\quad \limsup_{x\to\infty} \frac{F(x)^3}{3x} \leq a^2.

Từ đây suy ra

\displaystyle \lim_{x\to\infty} \frac{F(x)^3}{3x} = a^2.

Do đó

\displaystyle \lim_{x\to\infty} x^{\frac13} f(x) = \sqrt[3]{\frac a3}.

Câu A5: Đặt

\displaystyle S_n = \sum_{k =1}^n k (a_k -a_{k+1}).

Ta có \{S_n\}_n là dãy tăng. Ta cần chứng minh dãy này bị chặn.  Ta có

\displaystyle S_N = \sum_{k=1}^n k a_k - \sum_{k=2}^{n+1} (k-1) a_k = \sum_{k=1}^n a_k - n a_{n+1} \leq sum_{k=1}^\infty a_k.

Từ đây suy ra dãy \{S_n\}_n bị chặn, do đó chuỗi \sum_{n=1}^\infty n(a_n -a_{n-1}) hội tụ.

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An integral inequality from a competition for Vietnamese student in Maths

Posted on April 15, 2015 by Van Hoang Nguyen

Problem: Let f: [0,1]\to (-\infty, 1] be a continuous function such that \int_0^1 f(x) dx =0. Prove that

\displaystyle \int_0^1 f(x)^3 dx \leq \frac14.

Following is my proof whose method I learnt from the proof of Kullback-Pinsker-Csiszar inequality in the information theory (in the book “measure theory” of Bogachev). Put

\displaystyle A =\{x \, :\, 0\leq f(x)\leq 1\},

and denote a = |A| and b=\int_A f(x) dx. We have b\leq a. We have two cases:

Case 1: If a=1 then A^c must be empty set since it is an open subset of [0,1], then f \geq 0 on [0,1]. By our hypothesis \int_0^1 f(x) dx =0 then f(x) =0 for all x\in [0,1], hence

\displaystyle \int_0^1 f(x) dx = 0 \leq \frac14.

Case 2: $a < 1$, then we have

\displaystyle \int_A f(x)^3 dx \leq \int_A f(x) dx =b.

Using Jensen inequality, we have (note that |A^c| = 1-a > 0 and \int_{A^c} f(x) dx =-b)

\displaystyle \frac1{1-a} \int_{A^c} (-f(x))^3 dx \geq \left(\frac1{1-a} \int_{A^c} -f(x) dx\right)^3 = \frac{b^3}{(1-a)^3},

or equivalent

\displaystyle \int_{A^c} f(x)^3 dx \leq -(1-a)^{-2} b^3.

Therefore

\displaystyle \int_0^1 f(x)^3 dx \leq b - (1-a)^{-2} b^3\leq b -(1-b)^{-2} b^3,

here we use the fact b\leq a. It remains to prove that \displaystyle b -(1-b)^{-2}b^3 \leq \frac14. This inequality is equivalent to (3b-1)^2 \geq 0 which trivially holds. Our proof then is finished.

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