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March 30, 2011

Harmonic Analysis Lecture Notes 17

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 5:55 pm

Rearrangement Inequalities (cf Lieb–Loss, Ch 3)

Definition Given a Borel set {A \subseteq {\mathbb R}^d} of finite measure, we define {A^* = B(0,r)}, an open ball in {{\mathbb R}^d}, with {r} chosen so that {|B(x,r)| = |A|}. Similarly, we denote the (radially) symmetric decreasing rearrangement of a Borel function {f : {\mathbb R}^d \rightarrow {\mathbb C}} as follows: Suppose {\{x : |f| > \lambda\}} has finite measure for every {\lambda > 0} and define

\displaystyle  	f^*(x) = \int_0^\infty \chi_{\{x' : |f(x')| > \lambda\}^*}(x) \; d\lambda

Remark

  1. {|f(x)| = \int_0^\infty \chi_{\{|f| > \lambda\}}(x) \; d\lambda}.
  2. As a positive linear combination of spherically symmetric functions that are decreasing in radius, so {f^*} has both of these properties.
  3. {|f|} and {f^*} are equi-measurable, that is,

    \displaystyle  		\big| \{|f| > t\} \big| = \big| \{ f^* > t\} \big|.

    To see this, note that

    \displaystyle  		\{ f^* > t \} = \{ |f| > t\}^*.

    So in particular

    \displaystyle  | \{ x : f^* > t \}| = | \{ x : |f| > t\}|

    which is to say {|f|} and {f^*} are equi-measurable.

  4. From the above,

    \displaystyle  \int \underbrace{\phi(|f|)}_{\phi(0) = 0} \; dx = \int_0^\infty | \{ |f| > \lambda \}| \underbrace{\phi'(\lambda)}_{\tiny \begin{matrix} \text{want bounded variation} \\ \text{on any interval} \end{matrix}} = \int \phi(f^*) \; dx

Proposition Fix {f,g} measurable, then

\displaystyle  (1) \hbox{\hskip 28pt} \Big| \int fg \; dx \Big| \leq \int |fg| \; dx \leq \int f^* g^* \; dx.

For any convex increasing function {\phi [0, \infty) \rightarrow [0,\infty)} with {\phi(0) = 0} (e.g., {\phi(\lambda) = \lambda^p})

\displaystyle  (2) \hbox{\hskip 28pt} \int \phi(|f-g|) \; dx \geq \int \phi( \Big| |f| - |g| \Big| ) \; dx \geq \int \phi(f^* - g^*) \; dx.

Moreover, if {g = g^*} is strictly radially decreasing (that is, for all {r} there is {\lambda} such that {\{ g > \lambda\} = B(0,r)} — note that {g} cannot be compactly supported, nor have plateaus) and {\phi} is strictly convex then equality occurs in either case if and only if {f = f^*}.

Example If {g = \chi_{(-1,1)}}, then both {f_1 = \chi_{(-4/5,-3/5)} + \chi_{(-1/5,1/5)} + \chi_{(3/5, 4/5)}} and {f_2 = \text{``Gaussian-like"}} give the same answer as {f_j^*} without {f_j = f_j^*}.

Proof: Suppose {F : [0,\infty) \times [0,\infty) \rightarrow {\mathbb R}} obey

\displaystyle  (*) \hbox{\hskip 28pt} F(\lambda, \mu) - F(\lambda, 0) - F(0,\mu) + F(0,0) \geq 0

and {F(0,0) =0}. This is implied by

\displaystyle  \frac{\partial^2 F}{\partial \lambda \partial \mu} \geq 0

and integrating over {[0,\lambda] \times [0,\mu]}. For example, consider {F(\lambda, \mu) = \lambda \mu} or {F(\lambda, \mu) = -\phi(|\lambda - \mu|)}. Note that in the second case we get

\displaystyle  \begin{array}{rcl}  \displaystyle \frac{\partial F}{\partial \lambda} &=& - \phi'(|\lambda - \mu|) \text{signum}( \lambda - \mu) = \phi'(|\mu - \lambda|) \text{signum}( \lambda - \mu) \\ \displaystyle \frac{\partial^2 F}{\partial \mu \partial \lambda} = \phi''(|\lambda - \mu|) + \underbrace{\phi'(|\lambda - \mu|)}_{\geq 0 \text{, }\phi\text{ inc}} \delta(\lambda - \mu) \end{array}

We will show that when {F(\lambda, \mu) = F(\lambda,0) + F(0,\mu) + \int_0^\lambda \int_0^\mu \underbrace{\frac{\partial^2 F}{\partial \lambda \partial \mu}}_{\text{positive measure}}},

\displaystyle  \int F(|f|, |g|) \; dx \leq \int F(f^*, g^*) \; dx.

Now

\displaystyle  \begin{array}{rcl}  \text{LHS} &=& \displaystyle \int F(|f|, 0) \; dx + \int F(0, |g|) \; dx+ \iiint \frac{\partial^2 F}{\partial \lambda \partial \mu}(\lambda, \mu) \chi_{\{|f| > \lambda\}} (x) \chi_{\{ |g| > \mu \}}(x) \; d\lambda \; d\mu \; dx \\ \\ &=& \displaystyle \int F(f^*, 0) + F(0, g^*) \; dx + \iint \underbrace{| \{|f| > \lambda\} \cap \{ |g| > \mu\} |}_{\scriptsize \begin{matrix} \leq \hbox{min of meas} \\ = |\{ |f| > \lambda\}^* \cap \{ |g| > \mu \}^*| \\ = |\{ f^* > \lambda \} \cap \{ g^* > \mu \}| \end{matrix}} \frac{\partial^2 F}{\partial \lambda \partial \mu} \; d\lambda \; d\mu \\ \\ &\leq& \displaystyle \int F(f^*, g^*) \; dx = \text{RHS} \end{array}

Note that for equality we must have { |\{ |f| > \lambda\}^* \cap \{ |g| > \mu \}^*| = |\{ f^* > \lambda \} \cap \{ g^* > \mu \}| } for {\frac{\partial^2 F}{\partial \lambda \partial \mu} \; d\lambda \; d\mu} almost everywhere and so in our example for a.e. {\lambda, \mu}. \Box

Theorem (Riesz Rearrangement Inequality) For measurable {f,g,h} non-negative

\displaystyle  	\iint_{{\mathbb R}^d \times {\mathbb R}^d} f(x) g(x-y) h(y) \; dy \; dx \leq \iint f^*(x) g^*(x-y) h^*(y) \; dx \; dy.

Moreover, if {g = g^*} is strictly radially decreasing and equality eoccurs, then {f(x) = f^*(x-x_0)} and {h(x) = h^*(x-x_0)} for some {x_0 \in {\mathbb R}^d}.

Remark By linear changes of variables, one can essentially permute {f,g,} and {h} modulo possible additional reflections.

Application (Non-rotating) stars are spherically symmetric. Minimize

\displaystyle  	F = \int \underbrace{\Phi}_{\tiny \begin{matrix} \text{complicated thermodynamic} \\ \text{energy contribution} \end{matrix}}\big(\underbrace{\rho}_{\text{density}}(x) \big) \; dx - \iint \frac{\rho(x) \rho(y)}{|x-y|} \; dx \; dy.

Passing from {\rho} to {\rho^*} preserves {\int \Phi\big(\rho(x)\big) \; dx} but will decrease {F} (because of the second term, cf theorem above) unless {\rho} is already spherically symmetric about some point {x_0 \in {\mathbb R}^3}.

Proof: (of the 1D case, without cases of equality) First write {f(x) = \int_0^\infty \chi_{\{ |f| > \lambda\}}(x) \; d\lambda} and similarly for {g,h} to see that it suffices to treat the case where {f,g,} and {h} are characteristic functions of measurable sets.

By standard approximation arguments, we can pretend that our our open sets are finite collections of open intervals, e.g., {f(x) = \sum_{j=1}^J \chi_{(-\frac{l_j}{2}, \frac{l_j}{2})} (x - a_j)} with mutually disjoint closures.

Now add a new `time’ parameter

\displaystyle  	f_t(x) = \sum_{j=1}^J \chi_{(-\frac{l_j}{2}, \frac{l_j}{2})} (x - ta_j)

and similarly for {g_t,h_t}. Note that when {t = 1}, then {f_t = f, g_t = g, h_t = h}. We claim that {\int f_t(x) g_t(x-y) h_t(y) \; dx \; dy} is a decreasing function of {t}, at least up until the first collision of a pair of intervals, for which it suffices to treat one summand from each of {f,g,} and {h}. In this way, we are claiming that

\displaystyle  \begin{array}{rcl}  	0 &\geq& \displaystyle \frac{d}{dt} \iint \chi_{(-\frac{l}{2}, \frac{l}{2})} (x - ta) \chi_{(-\frac{w}{2}, \frac{w}{2})} (x - y - tb) \chi_{(-\frac{l'}{2}, \frac{l'}{2})}(y - tc) \\ \\ 		&=& \frac{d}{dt} \int_{-\frac{l}{2}}^\frac{l}{2} \int_{-\frac{l'}{2}}^\frac{l'}{2} \chi_{(-\frac{w}{2}, \frac{w}{2})} \big(x' - y' + t(a-b-c) \big) \; dx' \; dy' \\ \\ 		&=& \frac{d}{dt} (pic) 	\end{array}

Combining this with the natural amalgamation on collision algorithm yields the claim.

Before going to the multi-dimensional case, we first concretize a notion. \Box

Definition (Steiner symmetrization) Given {f : {\mathbb R}^d \mapsto [0,\infty)} and a direction {e \in S^{d-1} \subseteq {\mathbb R}^d} we define, writing {x = x'' e + x^\perp} with {x^\perp \perp e},

\displaystyle  	f^{*e}(x) = [t \mapsto f(te + x^\perp)]^* (x'')

Note that this is actually measurable! Oops I missed a picture. By Fubini, the {(d-1)}-dimensional measure of {F \cap \{{\mathbb R} e + x^\perp\}} is a measurable function of {x^\perp}, and this measurable function completely describes the set {F^{*e}}.

Proof: (of the 2D case, as a model for general dimension, without cases of equality) Let {X} and {Y} denote the Steiner symmetrization in the {e_1} and {e_2} directions, respectively: {(Xf)(x) = f^{*e_1}(x)}.

Let {R_\alpha} denote rotation by some {\alpha \not \in {\mathbb Q} \pi}. As in the 1D cases, it suffices to treat the case where {f,g}, and {h} are characteristic functions of bounded measurable sets, say {F, G,H}. Now define {F_{k+1} = XYR_\alpha F_k} with {F_0 = F} and similarly for {G_k} and {H_k}. By the 1D Riesz inequality (and the fact that {R_\alpha} is linear and measure preserving)

\displaystyle  	\iint \chi_{F_{k+1}}(x) \chi_{G_{k+1}}(x-y) \chi_{H_{k+1}}(y) \; dx \; dy \geq \iint \chi_{F_k}(x) \chi_{G_k}(x-y) \chi_{H_k}(y) \; dx \; dy.

We just need to show that {F_k \stackrel{subseq}{\longrightarrow} F^*} say a.e. (and DCT applies because all of the {F_k}‘s live in the same big ball {B(0,R)} that contain {F}) or {L^2} or others, and same for {G_k, H_k}.

Oops there is another picture here. Now, letting {K} be some measurable subset of {{\mathbb R}^2}, we have {XY K = \{ |y| < w(|x|)\}} with {w} decreasing (lower semicontinuous).

We have sets {F_k = \{ |y| < w_k(|x|)\}} with {w_k} decreasing and uniformly bounded (by the radius {R} above) with support contained in a common, compact set (e.g., {[0,R]}). By the Cantor diagonal slash trick we can find a subsequence of our {w_k} that converge at each rational {x}. Because our functions are monotone, we obtain pointwise convergence everywhere except a countable set where the limit function has jumps (there are at most countably many of them). Thus, along this sequence, {\chi_{F_k}} converge almost everywhere (and so also in {L^2}) to {\chi_{\tilde F}} for some {\tilde F = \{ |y| < w(|x|)\}} with {w} decreasing. We need to show {\tilde F} is actually {F^*}; by the Dominated Convergence Theorem, we obtain {| \tilde F| = |F| = |F^*|}. So, it suffices to show {\tilde F} is actually a ball. Fix {\gamma(x) = e^{-|x|^2}}, which is strictly radially decreasing (in the technical sense that every {B(0,r)} appears as a super-level set). By the {L^2}-contracting property of rearrangements (which incidentally is equivalent to the enhancement-of-dot-products property, noting {\| f - g\|^2 = \|f\|^2 + \|g\|^2 - 2 \text{Re}\langle f,g \rangle}), therefore {\| \gamma - (XYR_\alpha)^k\chi_F\|_{L^2}} is decreasing in {k} and so converges (even without passing to our subsequence). In particular,

\displaystyle  	\|\gamma - (XY R_\alpha) \chi_{\tilde F}\|_{L^2} = \lim_{\text{subseq}} \| \gamma - (XYR_\alpha)^{k+1} \chi_F\| = \lim_{\text{subseq}} \|\gamma - (XYR_\alpha)^k F\|_{L^2} = \|\gamma - \chi_{\tilde F}\|_{L^2}.

From the case of equality for our first arrangement inequalities,

\displaystyle  	\big\| X[ YR_\alpha \chi_{\tilde F} - \gamma] \big\|_{L^2} = \|YR_\alpha \chi_{\tilde F} - \gamma \|_{L^2}

By our first rearrangement inequalities, we have

\displaystyle  	\| XYR_\alpha \chi_{\tilde F} - \gamma\|_{L^2} = \|XYR_\alpha \chi_{\tilde F} - \underbrace{X\gamma}_{= \gamma}\| \leq \|YR_\alpha \chi_{\tilde F} - \gamma\|_{L^2} = \|YR_{\alpha} \chi_{\tilde F} - \underbrace{Y\gamma}_{= \gamma}\|_{L^2} \leq \|R_\alpha \chi_{\tilde F} - \gamma \|_{L^2} = \|\chi_{\tilde F} - \gamma\|_{L^2}

By our previous computation, therefore equality holds.

So now let us recall when equality holds in the 1D case: as {\gamma} is already symmetrical, so must the other functions be symmetrical. Therefore, {|YR_\alpha \tilde F \triangle R_\alpha \tilde F| = 0}, and hence {R_\alpha \tilde F} is symmetrical across the {x}-axis. Undoing the rotation, we see that {\tilde F} has two a.e. reflection axes that are at an angle {\alpha \not \in \pi{\mathbb Q}}. Thus, the group generated by these two reflections, which is dense in {O(2)}, leaves {|R\tilde F \triangle \tilde F|= 0} for all {R} in this group. Note that {R \mapsto |R \tilde F \triangle \tilde F|} acts on {O(2)}, indeed {|R \tilde F \triangle \tilde F| = \|\chi_{\tilde F} \circ R^{_1}(\cdot) - \chi_{\tilde F}\|_{L^2}} and consequently {|R\tilde F \triangle \tilde F| = 0} for all {R \in O(2)} and so {\tilde F} is almost every a ball.

Recall that if {g = g^*} is strictly radially decreasing, then equality requires {f(x) = f^*(x - x_0)} and {h(x) = h^*(x-x_0)}. So, for the proof of cases of equality, we use induction on dimension. Moreover we just need to consider the case when {f = \chi_F} and {h = \chi_H} for {F} and {H} being sets of finite measure. Similarly, writing {g = g^*} as an integral over the characteristic functions of its super-level sets, our problem reduces to the following: if for a.e. {r>0},

\displaystyle  	(*)\hbox{\hskip 28pt}\iint \chi_F \chi_{B(0,r)}(x-y) \chi_H(y) \; dx \; dy = \iint \chi_{F^*}(x) \chi_{B(0,r)}(x-y) \chi_{H^*}(y) \; dx \; dy,

then {\int |\chi_F(x) - \chi_{F^*}(x-x_0)| + |\chi_H(x) - \chi_{H^*}(x-x_0)| \; dx = 0}.

In the one-dimensional case, if {r > \tfrac{1}{2}|F^*| + \tfrac{1}{2}|H^*|= \tfrac{1}{2}|F| + \tfrac{1}{2}|H|}, then {|x - y| < r} for almost {x \in F^*} and {y \in H^*} and so {\text{RHS}(*) = |F| \cdot |H|} and thus {\text{LHS}(*) = |F| \cdot |H|} for a.e. {r > \tfrac{1}{2}|F| + \tfrac{1}{2}|H|}. Sending {r \downarrow \tfrac{1}{2}|F| + \tfrac{1}{2}|H|} within this set, we see that every pair of Lebesgue points of {F} and {H} are separated by no more than {\tfrac{1}{2} |F| + \tfrac{1}{2}|H|}. From this and elementary geometry, we see that {F} and {H} are a.e. intervals with a common center.

For the higher dimensional case,

\displaystyle  	\text{LHS}(*) = \iint \left[\iint \chi_F(x_1, x') \chi_{B(0,\sqrt{r^2 - |x_1 - y_1|^2})}(x'-y') \chi_H(y_1,y') \; dx' \; dy' \right] \; dx_1 \; dy_1

and similarly for the right hand side. By Riesz, we have inequality of inner integrals for all {x_1, y_1}, so equality of {\text{LHS}(*) = \text{RHS}(*)} requires equality for a.e. {x_1} and {y_1}. Thus for a.e. {x_1} and {y_1},

\displaystyle  	F|_{x_1} := \{ x' : (x_1, x') \in F\} = \big[(F|_{x_1})^* \text{ translated by some } x_0'(x_1, y_1) \big]

and similarly for the slices of {H}, with the same translation parameter {x_0'}. Because {F|_{x_1}} does not depend on {y_1} nor {H|_{y_1}} on {x_1}, we see that {x_0'} is a.e. constant. Thus, there is an axis parallel to {e_1} and passing through {(0,x_0')} so that {F} and {H} both are {O(d-1)} symmetric about this axis. We can make the same argument for a second axis, say one with an angle {\alpha \not \in \pi {\mathbb Q}}, and so see that {F} and {H} are a.e. invariant for a dense set of rotations about the intersection of the axes (If the intersections do not meet, we could use successive rotations to march our set to infinite, contradicting the fact that {F} is invariant and of finite measure.) To finish, we use the fact that {g\mapsto |gF \triangle F|} is a continuous function on the group of Euclidean motions (affine group, or {O(d) \rtimes {\mathbb R}^d}). \Box

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