Multifractals

Dimension of the measure

The dimension of the measure mu is the smallest dimension of among all sets A with mu(A) > 0.

We shall show that the value of f(alpha) satisfying f(alpha) = alpha is the dimension of the measure mu. This will take several steps.

(1) The diagonal line intersects the graph of f(alpha) exactly at one point, and that point corresponds to q = 1. The proof involves showing tau(1) = 0.

Then for q = 1 from alpha = - dtau/dq we have

f(alpha) = alpha = (p1ln(p1) + ... + pNln(pN))/(p1ln(r1) + ... + pNln(rN))

We show that this expression is the smallest dimension of any set A with mu(A) > 0.

First, note the region S(i1...ik) with address i1...ik has measure

mu(S(i1...ik)) = p1n1(k) ... pNnN(k)

where n1(k) is the number of 1s in i1...ik, n2(k) is the number of 2s in i1...ik, and so on.

For almost all sequences i1i2..., nj(k)/k -> pj as k -> infinity. Consequently,

(1/k)ln(mu(S(i1...ik))/diam(S(i1...ik))s) = (1/k)ln((p1n1(k) ... pNnN(k))/(ri1...rik)s) = (n1(k)/k)ln(p1) + ... + (nN(k)/k)ln(pN) + (s/k)ln(1/(ri1...rik)) <= (n1(k)/k)ln(p1) + ... + (nN(k)/k)ln(pN) + (s/k)ln(1/(min{ri})k) = (n1(k)/k)ln(p1) + ... + (nN(k)/k)ln(pN) + s*ln(1/(min{ri})) -> p1ln(p1) + ... + pNln(pN) + s*ln(1/(min{ri}))

as k -> infinity. Consequently, if s < -(p1ln(p1) + ... + pNln(pN))/ln(1/min{ri}) then

limk -> infinity(1/k)ln(mu(S(i1...ik))/diam(S(i1...ik))s) < 0 so limk -> infinityln(mu(S(i1...ik))/diam(S(i1...ik))s) = -infinity and limk -> infinitymu(S(i1...ik))/diam(S(i1...ik))s = 0

From this we see Hs(A) >= mu(A)/c for all c > 0.

Because mu(A) > 0, we see Hs(A) = infinity and so d(A) >= s.

Then for all sets A with mu(A) > 0,

d(A) >= -(p1ln(p1) + ... + pNln(pN))/ln(1/min{ri}) >= (p1ln(p1) + ... + pNln(pN))/(p1ln(r1) + ... + pNln(rN))

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