April 13, 2011

Harmonic Analysis Lecture Notes 18

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 5:53 pm

continuation of the Rearrangement Discussions

Theorem Fix {1 \leq p, q < \infty}. If {f \in L^q} and {|\nabla f| \in L^p}, then {\nabla |f| \in L^p}. In fact, {\big| \nabla |f| \big| \leq |\nabla f|} a.e. Similarly, if {f,g \in L^p} with {\nabla f, \nabla g \in L^p}, then

\displaystyle  \begin{array}{rcl}  	|\nabla \max \{f,g\}| &\leq& \max \{ |\nabla f|, |\nabla g| \} \\ 	|\nabla \min \{f,g\}| &\leq& \max \{ |\nabla f|, |\nabla g| \} \\ 	\end{array}

Proof: It is easy once we figure out what {\nabla |f| \in L^p} means: There is an {\vec F : {\mathbb R}^d \rightarrow {\mathbb C}^d} with {|F| \in L^p} so that

\displaystyle  	\int \nabla \phi |f| \; dx = - \int \phi \vec F \; dx

for all {\phi \in C_c^\infty}.

Restart the proof: We wish to show that there is some {\vec F : {\mathbb R}^d \rightarrow {\mathbb C}^d} with {|F| \in L^p} so that

\displaystyle  	\int \nabla \phi |f| \; dx = - \int \phi \vec F \; dx

for all {\phi \in C_c^\infty} with {|\vec F| \leq |\nabla f|}. In fact, we will see that

\displaystyle  	\vec F = \left\{\begin{matrix} 	\tfrac{u \nabla u + v \nabla v}{\sqrt{u^2 + v^2}} & \hbox{ if } u^2 + v^2 \neq 0 \\ \\ 	0 & \hbox{ if } u^2 + v^2 = 0 	\end{matrix}\right.

does the job; here {f = u + iv}.

Choose {u_n, v_n \in C^\infty} converging to {u} and {v} in {L^q} a.e., and so that {\nabla u_n, \nabla v_n} converges to {\nabla u} and {\nabla v} a.e., and in {L^p} (e.g., mollify {u,v}: convolve {u} and {v} with a smooth bump {n^d \phi(nx)} with {\int \phi = 1}). We also smooth out the absolute-value function: {(u+iv) \mapsto \sqrt{\epsilon^2 + u^2 + v^2}}. As {\sqrt{\epsilon^2 + u_n^2 + v_n^2}} is smooth,

\displaystyle  	\int -\nabla \phi \sqrt{\epsilon^2 + u_n^2 + v_n^2} \; dx = \int \phi \frac{u_n \cdot \nabla u_n + v_n \cdot \nabla v_n}{\sqrt{\epsilon^2 + u_n^2 + v_n^2}}

That {(u+iv) \mapsto \sqrt{u^2 + v^2 +\epsilon^2}} is (uniformly) Lipschitz, so {\sqrt{\epsilon^2 + u_n^2 + v_n^2} \rightarrow \sqrt{\epsilon^2 + u^2 + v^2}} in {L^p}. Furthermore, noting {\nabla u_n = \nabla u + (\nabla u_n - \nabla u)}, where we treat the first term by the dominated convergence theorem with {|\phi| \; |\nabla u|} and treat the second term by {L^p} convergence and {L^{p'}} convergence of {\phi \frac{u_n}{\sqrt{\epsilon^2 + u_n^2 + v_n^2}}} by dominated convergence theorem with {|\phi}. Sending letting {n} tend to infinity on both sides of the above equality, we get

\displaystyle  	\int - \nabla \phi \sqrt{\epsilon^2 + u^2 + v^2} \; dx = \int \phi \frac{u \nabla u + v \nabla v}{\sqrt{\epsilon^2 + u^2 + v^2}}.

Sending {\epsilon} to {0} on the left side by DCT with {|\nabla \phi| \sqrt{1 + u^2 + v^2}} and sending {\epsilon} to {0} on the right side by DCT with {2|\phi| |\nabla f|}, we get

\displaystyle  	\int -\nabla \phi \sqrt{u^2 + v^2} = \int \phi \vec F.



\displaystyle  	\int |\nabla f^*|^2 \; dx \leq \int |\nabla f|^2 \; dx


  1. This is also true when we replace the exponent {2} above by some {p} such that {1 \leq p \leq \infty}, but the proof is more complicated.
  2. Equality can occur when {f \neq f^*(x- x_0)}, but only in special circumstances, exemplified perhaps by, say if we think of two bump functions on top of each other in the style of a submarine (viewed from the side).

Proof: Using Plancherel’s theorem

\displaystyle  \begin{array}{rcl}  	\text{RHS} &=& \displaystyle \int |\xi|^2 \; |\hat f (\xi)|^2 \; d\xi \\ \\ 		&=& \displaystyle \lim_{t \downarrow 0} \int \frac{1 - e^{-t|\xi|^2}}{t} |\hat f(\xi)|^2 \; d\xi \\ \\ 		&=& \displaystyle \lim_{t \downarrow 0} \frac{1}{t} \left[\int |f|^2 \; dx - \langle f, e^{t \nabla} f \rangle \right] \\ \\ 		&=& \displaystyle \lim_{t \downarrow 0} \frac{1}{t} \left[\int |f|^2 \; dx - \iint \overline{f(x)} \frac{e^{-|x-y|^2/(4t)}}{(4\pi t)^{d/2}} f(y) \; dy \; dx \right] \\ \\ 		&\geq& \displaystyle \lim_{t \downarrow 0} \frac{1}{t} \left[\underbrace{\int |f^*|^2 \; dx}_{\text{equimeasurability}} - \underbrace{\iint \overline{f^*(x)} \frac{e^{-|x-y|^2/(4t)}}{(4\pi t)^{d/2}} f^*(y) \; dy \; dx}_{\text{Riesz rearrangement ineq.}} \right], 	\end{array}

which is the left hand side of what we wanted.

Observed that this proof relied on {f \in L^2}. To see that this is enough, we treat non-negative {f} via

\displaystyle  	f_\epsilon = 	\left\{\begin{array}{ll} 	0 &\qquad f \leq \epsilon \\ 	f - \epsilon &\qquad \epsilon < f < \epsilon + \tfrac{1}{\epsilon} \\ 	\tfrac{1}{\epsilon} &\qquad f \geq \epsilon +\tfrac{1}{\epsilon} 	\end{array}\right.

which is defined via the minimum and maximum operations, noting {\max\{a,0\} = \tfrac{1}{2}(|a| + a)}, i.e., {f_\epsilon = \min\{ \tfrac{1}{\epsilon} , \max \{ f- \epsilon, 0\} \}}. This allows us to see that {|\nabla f_\epsilon | \rightarrow |\nabla f|} monotonically, using the previous theorem and similarly {|\nabla f_\epsilon^*| \rightarrow |\nabla f^*|} (note {(f^*)_\epsilon = (f_\epsilon)^*}). Sending {\epsilon \downarrow 0}, we get the result for non-negative for {f} that are not necessarily in {L^2}. For general {f} (possibly taking negative values), keep in mind that {\big|\nabla |f| \big| \leq |\nabla f|}. \Box

Theorem (Diamagnetic inequality) Let {A : {\mathbb R}^d \rightarrow {\mathbb R}^d} then,

\displaystyle  	\big| \nabla |f| \big| \leq |(\nabla + iA)f|, \hbox{ a.e.}

Proof: From above, we get {\partial_j |f| = \text{Re} \frac{\bar f \partial_j f}{|f|} = \text{Re} \frac{\bar f (\partial_j + i A_j) f}{|f|} \leq |(\partial_j + i A_j)f|}. Summing them and taking the square-root, we obtain the result. \Box

The energy of a quantum mechanical particle in an electromagnetic field is given by

\displaystyle  	E(\psi) = \int |(\nabla + i\vec{A})\psi|^2 + V |\psi|^2 \; dx

where {\psi} is the wave function, {\vec{A}}, {V} are the electromagnetic potentials:

\displaystyle  	\vec E = \nabla V, \hbox{\hskip 18pt} \vec B = \nabla \times \vec A

Note that {\vec E} does not determine {V} uniquely, but only up to an additive constant, but that is alright. More dramatically, {\nabla \times A_1 = \nabla \times A_2} just says {A_1 - A_2 = \nabla \phi} for general {\phi}. Change {A_1 \rightarrow A_2}, we replace {\psi} by {e^{i\phi} \psi},

\displaystyle  \begin{array}{rcl}  	(\nabla + iA_2) e^{i\phi} \psi &=& e^{i\phi} \nabla \psi + i \nabla \phi e^{i\phi} \psi + iA_2 e^{i\phi} \psi \\ 		&=& e^{i\phi} (\nabla + iA_1) \psi 	\end{array}

Equivalently, the operator {\psi \mapsto e^{i\phi} \psi} is a unitary mapping that conjugates {-(\nabla + i \vec A_2)^2 + V} to {-(\nabla + i \vec A_1)^2 + V}. By the diamagnetic inequality

\displaystyle  \begin{array}{rcl}  	\displaystyle \inf_{\scriptsize \begin{matrix} \psi \in \mathcal{S} \\ \|\psi\|_{L^2} = 1 \end{matrix}} E_{A,V}(\psi) &\geq& \displaystyle \inf_{\scriptsize \begin{matrix} \psi \in \mathcal{S} \\ \|\psi\|_{L^2} = 1 \end{matrix}} \int \big| \nabla |\psi| \big|^2 + V|\psi|^2 \\ \\ 	&=& \displaystyle \inf_{\scriptsize \begin{matrix} \psi \in \mathcal{S} \\ \|\psi\|_{L^2} = 1 \end{matrix}} \int |\nabla \psi|^2 + V|\psi|^2 \\ \\ 	&=& \displaystyle \inf_{\scriptsize \begin{matrix} \psi \in \mathcal{S} \\ \|\psi\|_{L^2} = 1 \end{matrix}} E_{O,V}(\psi) 	\end{array}

This shows that the lowest energy configuration has less (or equal) energy when the magnetic field is turned off. Naïvely, this says that it takes energy to push matter into a magnetic field. This property of a material is called diamagnetism (cf paramagnetism).

As an aside, {\psi : {\mathbb R}^3 \rightarrow {\mathbb C}}, actually not {{\mathbb C}}, but a one-dimensional vector space over {{\mathbb C}}, with an inner product. Note that this reduces to {{\mathbb C}} once we choose who {1} represents (a unit vector = orthonormal basis). Any pair of “frames” differs by a uni-modular complex number {e^{i\phi}}. Different “gauges” are just different choices of coordinates for such functions.

Now let us consider the idea that our notion of basis of our vector space depends upon our history. A “connection” is a means to perform “parallel transport”. The simplest version is: a function is “constant” if {\nabla \psi = 0}. More generally, we say that {f} is “unchanging” if {(\nabla + iA)\psi = 0}. When {\psi} traverses a circular path {\gamma :[0,1] \rightarrow {\mathbb C}} with {\gamma(0) = \gamma(1)}, we have {\psi \rightarrow e^{-i \int_0^1 A(\gamma(t)) \cdot \gamma'(t) \; dt} \psi}. So, no monodromy for all paths is equivalent to {\int_0^1 A(\gamma(t)) \cdot \gamma'(t) \; dt = 0} for all paths, which in turn is equivalent to {\nabla \times A = 0} in a simply connected domain, i.e., no magnetic field. Indeed, the total phase change

\displaystyle  	\oint_\gamma A \cdot d\vec r = \int_{\text{disk bounded by }\gamma} \nabla \times A \cdot d\vec S = \text{integral of ``curvature'' over this ``disk''}

cf Foccault pendulum experiment.


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