Results & Lemmas (2)
Each result is auto-extracted from the paper via OCR - the math is shown verbatim and may carry occasional transcription artifacts. Click to expand the full statement.
THEOREM 1.
THEOREM 1. We have max A(r, 1/Ff) = 2*r2 (r2 + 2) for 0 < r ^ 1. For each r, 0 < r ^ 1, the maximum is attained only by Kg 's. PROOF: Given…
THEOREM 1. We have max A(r, 1/Ff) = 2*r2 (r2 + 2) for 0 < r ^ 1. For each r, 0 < r ^ 1, the maximum is attained only by Kg 's. PROOF: Given / E S, we can apply the area theorem [3, p.29] to n=l to obtain (2) n = l Since 1/-F/G0 = 1 ~ atz + z2 6»*"+1' z e D> n=l it follows from (2), together with |a2| ^ 2 , that
THEOREM 2
THEOREM 2. For 0 < r < 1 we Aave (5) 27rr(r4 + 4r2 + l)1/2(l - r2)"2 ^ L(r,K) < supL(r,f). /6S PROOF: This is a consequence of the…
THEOREM 2 . For 0 < r < 1 we Aave (5) 27rr(r4 + 4r2 + l)1/2(l - r2)"2 ^ L(r,K) < supL(r,f). /6S PROOF: This is a consequence of the expression of A(r, K) in (4), without appeal- ing to the expression of L[r,K) in terms of elliptic integrals (see [2]). We only apply to K the isoperimetric inequality: A(r,/) < 7r{i(r,/)/(27r)}2 for / G 5, which says that, of all rectifiable Jordan curves with the given perimeter L(r, / ) , (0 < r < 1), the circle has interior of maximum area. D Since inf (r4 + 4r2
Function classes studied:
Related Papers