Multifractals

The graph of tau(q) is concave up

Differentiating

p_i^qr_i^tau(q) = 1

twice with respect to q gives

p_i^qr_i^tau(q)((d²tau/dq²)ln(r_i) + (ln(p_i) + ln(r_i) dtau/dq)²) = 0 (*)

Solving for d²tau/dq² we find

d²tau/dq² = -( p_i^qr_i^tau(q)(ln(p_i) + (dtau/dq)ln(r_i))²)/( p_i^qr_i^tau(q)(ln(r_i)))

Because each p_i^qr_i^tau(q) > 0 and each ln(r_i) < 0, we see d²tau/dq² >= 0.

Finally, suppose d²tau/dq² = 0. Then equation (*) becomes

p_i^qr_i^tau(q)( ln(p_i) + ln(r_i) dtau/dq)² = 0

Consequently, for each i, dtau/dq = -ln(p_i)/ln(r_i). That is, all the ln(p_i)/ln(r_i) are equal.

That is, if not all the ln(p_i)/ln(r_i) are equal, then d²tau/dq² > 0.

Return to Multifractals from IFS.