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 Incoming links: Integral mean bound for an odd square-root transform

For

f (z) = z + \sum_{n \geq 2} a_{n} z^{n}

(2 π)^{- 1} \int ∣ f (r e^{i θ}) ∣ d θ \leq r / (1 - r)

∣ a_{n} ∣ \leq r^{1 - n} / (1 - r)

. Choosing

r = (n - 1) / n

and applying the logarithm tangent bound gives the displayed bound for every

n \geq 2

Because

∣ h^{2} ∣ = ∣ h ∣^{2}

M (r) := \frac{1}{2 π} \int_{0}^{2 π} ∣ h^{2} (r e^{i θ}) ∣ d θ = n \geq 1 \sum ∣ c_{n} ∣^{2} r^{2 n} .

Differentiate this power series for

r < 1

and use iii:

M^{'} (r) = \frac{2}{r} \sum_{n \geq 1} n ∣ c_{n} ∣^{2} r^{2 n} = \frac{2 Area ( h ( r D ))}{π r} \leq \frac{2 r}{( 1 - r ^{2} ) ^{2}} .

Since

M (0) = 0

M (r) \leq \int_{0}^{r} \frac{2 s}{( 1 - s ^{2} ) ^{2}} d s = \frac{r ^{2}}{1 - r ^{2}} .