Let $f$ be a continuous real-valued function in a circle $|z|\leq1$ and $|f|\leq1$. Prove that $| \oint_{|z|=1} f(z)dz |\leq4 $.

Any hint, please.


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  • 1
    $\begingroup$ "From top to integral"? What does that mean? $\endgroup$ – DonAntonio Feb 24 at 13:26
  • $\begingroup$ My bad. I am not native speaker $\endgroup$ – Shorty12319 Feb 24 at 13:43
  • $\begingroup$ You can’t do a contour integral on a ball! $\endgroup$ – Mindlack Feb 24 at 14:01
  • $\begingroup$ Corrected the condition $\endgroup$ – Shorty12319 Feb 24 at 14:05
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    $\begingroup$ Isn't $ \oint_{|z|=1} dz = 2 \pi$ ? $\endgroup$ – N74 Feb 24 at 15:28

Let real $x,y$ (here we use the real nature of $f$ and by a slight abuse of notation we denote $f(e^{i\theta}) = f(\theta)$ on the unit circle) be $\int_{0}^{2\pi} f(\theta)\cos(\theta)d\theta$ and $\int_{0}^{2\pi} f(\theta)\sin(\theta)d\theta$ respectively; by parametrization, the problem reduces to showing that |$x+iy$| $\leq 4$, or that $\sqrt{x^2+y^2} \leq 4$; we can assume $\sqrt{x^2+y^2}$ not zero as there is nothing to prove then and let $\alpha$ s.t. $\cos(\alpha) = \frac{x}{\sqrt{x^2+y^2}}$, $\sin(\alpha) = \frac{y}{\sqrt{x^2+y^2}}$.

Then $\sqrt{x^2+y^2}$ = $\int_{0}^{2\pi}{f(\theta)\cos(\theta)\cos(\alpha)d\theta + \int_{0}^{2\pi}f(\theta)\sin(\theta)\sin(\alpha)}d\theta$=$\int_{0}^{2\pi} f(\theta)\cos(\theta-\alpha)d\theta$ and by taking absolute values and using $|f(\theta)| \leq 1$, RHS is at most $\int_{0}^{2\pi} |\cos(\theta-\alpha)|d\theta$ = $4$, so we are done.

(by periodicity $\int_{0}^{2\pi} |\cos(\theta-\alpha)|d\theta$ = $\int_{0}^{2\pi} |\cos(\theta)|d\theta$ = $4\int_{0}^{\frac{\pi}{2}} \cos\theta d\theta = 4$)

  • $\begingroup$ Where are you using real-value of function? $\endgroup$ – Shorty12319 Feb 24 at 22:20
  • $\begingroup$ Sorry, and why the last integral is exactly equal to 4? $\endgroup$ – Shorty12319 Feb 24 at 22:36
  • $\begingroup$ And what does it mean RHS? $\endgroup$ – Shorty12319 Feb 24 at 22:39
  • $\begingroup$ In the fact that $x,y$ are real. RHS means right hand side $\endgroup$ – Conrad Feb 24 at 22:50
  • $\begingroup$ Note that it is enough for $f$ to be defined real, measurable and bounded on the unit circle only $\endgroup$ – Conrad Feb 24 at 23:14

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