# Welcome.

## 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

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

$\Box$

Theorem

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

Remark

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.