Remark 3 above gives a very simple proof that the exact expression found for the (necessarily single) eigenvalue of is in fact its norm .

]]>Let be positive and square integrable on .

By taking (for )

We get , and

Which (by the above C-S bounding argument) gives the simple bound

(1)

Remarks:

1. The bound (1) may be generalized (for not necessarily positive) by taking the supremum norm over the support of .

2. This upper bound should be good for sufficiently close to an eigenfunction of (e.g. using giving good lower bound for ).

3. If is an eigenfunction of with corresponding eigenvalue , the bound (1) shows that

– implying that has at most one(!) eigenvalue (which must be ).

4. It seems that the above upper bound derivation can be generalized for .

]]>Let us imagine we are entering a train station and instead of taking the train, we’re flying by a helicopter and viewing the railway from a bird’s eye. From the helicopter, we see the starting point of the railway. The end can not be seen. Our railway is endless. Sitting in the helicopter, we can change the color of the railroad stripes as we wish.

First, we shall paint with red color every second stripe.

Thereafter we shall paint with yellow color every third stripe that is not painted red.

Thereafter we shall paint with blue color every fifth stripe that is not painted red or yellow.

Thereafter we shall paint with green color every seventh stripe that is not painted red or yellow or blue.

And so on according to the following prime numbers.

The question about the existence of an infinite number of twin primes, can be illustrated as follows: a coloring stage (coloring stage n) leaves behind an infinite number of pairs of stripes unpainted, which are separated by one red stripe. Is it possible to determine that the next coloring step (coloring stage n +1) will fail to “eliminate” these pairs ?

If the number of twin primes is finite, then there is a coloring stage which “eliminates” the pairs of white stripes which are left by its predecessor (from a certain point on the number line). Thus, if it appears that each coloring stage – no matter how advanced it is – leaves behind an infinite number of pairs of white stripes, then we can assume the existence of an infinite number of pairs of twin primes.

Read more

https://drive.google.com/file/d/0B3YBCg8YHwOjcmthUTlVSTdvZ1k/edit?usp=sharing

]]>A simple lower bound on :

Suppose that for some .

Hence (by the crude bound above – with )

, i.e.

Examples:

:

:

:

:

:

:

:

:

:

Remark: these (somewhat crude) numerical bounds show that can be much larger than expected!

]]>It should be remarked that the last bound (with ) is already optimal! to see that, let (for fixed and denote for by the above upper bound for . Since

It follows (e.g. by taking derivatives under the integral sign) that is convex (and therefore has a minimum at if .

It remains only to find for which .

A simple calculation shows that for

Remark: this is quite large! ().

]]>Regarding the amount of precision of the two input matrices needed for a target precision for the max generalized eigenvalue, I guess a full answer would involve something like a ‘generalized eigenvalue problem condition number’ which I don’t know how to do. Googling shows that Maple and MATLAB have possibly related functions named EigenConditionNumbers and condeig.

On the other hand, a half-answer is easy because if low precision matrices give you a candidate rational vector whose associated ratio is > 4 then you are done.

For Maynard’s example the problem seems strangely stable. For example lowering the precision by changing N[M, 150] to N[M] gives a ratio that is equal to >12 places, even though the input matrices are badly conditioned with respect to matrix inversion.

ev = Eigenvectors[{N[M2], N[M1]}, 1][[1]];

rv = Rationalize[ev, 10^-40];

Ratio = k*(rv.M2.rv)/(rv.M1.rv);

N[Ratio, 20]

4.0020697619380349324

]]>