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Let $\mathbf{D}$ be the open unit disc in $\mathbf{C}$ and let $f,g:\mathbf{D}\to \mathbf{C}$ be holomorphic functions such that the real valued function $\vert f\vert^2+\vert g\vert^2$ is bounded from above by some real number $c$ (everywhere on $\mathbf{D}$).

Question. Can we bound $$\left\vert \frac{df}{dz} \right\vert^2(0)+ \left\vert \frac{dg}{dz}\right\vert^2(0)$$ from above in terms of $c$?

Answer. Yes! See answer below.

Question. Let $x\in \mathbf{D}$. Can we bound $$\left\vert \frac{df}{dz} \right\vert^2(x)+ \left\vert \frac{dg}{dz}\right\vert^2(x)$$ from above in terms of $c$ and the norm of $x$?

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1 Answer 1

Differentiating the Cauchy's integral formula is enough: $$ |f'(0)| =\left| \frac{1}{2\pi i} \int_{|z|=r}\frac{f(z)}{z^2}\, dz\right|\le \frac{\sqrt c}{r^2}. $$ Here the radius $r\in(0,1)$ is arbitrary, so $|f'(0)|\le \sqrt c\;$ and $|f'(0)|^2+|g'(0)|^2\le 2c\ $.

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Great! Can we also replace the number $0$ by another point in $\mathbf{D}$ and use the Cauchy integral formula? I edited my question. –  Rayleigh Oct 8 '11 at 22:10
    
Wait a minute...Isn't the integral of $1/z^2$ over that contour zero? –  Rayleigh Oct 8 '11 at 22:34
    
Ow I understand what you're doing now. You're bounding your integral by just taking the supremum of $\vert f(z)/z^2\vert $ multiplied by the length of the contour. Great! –  Rayleigh Oct 8 '11 at 22:37
    
I think I can answer my own (other) question too now. The bound for $\vert f^\prime(a)\vert$ will be $\sqrt{c}/(r-\vert a \vert)^2$ following the same line of thought as above. –  Rayleigh Oct 8 '11 at 22:52
    
@Rayleigh yes, even $\sqrt{c}/(1-\vert a \vert)^2$. –  Andrew Oct 9 '11 at 6:33

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