** continuation of the Rearrangement Discussions **

TheoremFix . If and , then . In fact, a.e. Similarly, if with , then

*Proof:* It is easy once we figure out what means: There is an with so that

for all .

Restart the proof: We wish to show that there is some with so that

for all with . In fact, we will see that

does the job; here .

Choose converging to and in a.e., and so that converges to and a.e., and in (e.g., mollify : convolve and with a smooth bump with ). We also smooth out the absolute-value function: . As is smooth,

That is (uniformly) Lipschitz, so in . Furthermore, noting , where we treat the first term by the dominated convergence theorem with and treat the second term by convergence and convergence of by dominated convergence theorem with . Sending letting tend to infinity on both sides of the above equality, we get

Sending to on the left side by DCT with and sending to on the right side by DCT with , we get

Theorem

Remark

- This is also true when we replace the exponent above by some such that , but the proof is more complicated.
- Equality can occur when , 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

which is the left hand side of what we wanted.

Observed that this proof relied on . To see that this is enough, we treat non-negative via

which is defined via the minimum and maximum operations, noting , i.e., . This allows us to see that monotonically, using the previous theorem and similarly (note ). Sending , we get the result for non-negative for that are not necessarily in . For general (possibly taking negative values), keep in mind that .

Theorem (Diamagnetic inequality)Let then,

*Proof:* From above, we get . Summing them and taking the square-root, we obtain the result.

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

where is the wave function, , are the electromagnetic potentials:

Note that does not determine uniquely, but only up to an additive constant, but that is alright. More dramatically, just says for general . Change , we replace by ,

Equivalently, the operator is a unitary mapping that conjugates to . By the diamagnetic inequality

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, , actually not , but a one-dimensional vector space over , with an inner product. Note that this reduces to once we choose who represents (a unit vector = orthonormal basis). Any pair of “frames” differs by a uni-modular complex number . 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 . More generally, we say that is “unchanging” if . When traverses a circular path with , we have . So, no monodromy for all paths is equivalent to for all paths, which in turn is equivalent to in a simply connected domain, i.e., no magnetic field. Indeed, the total phase change

cf Foccault pendulum experiment.

## Leave a Reply