# continuous real-valued function [closed]

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

## closed as off-topic by GNUSupporter 8964民主女神 地下教會, Alexander Gruber♦Feb 25 at 7:13

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• "From top to integral"? What does that mean? – DonAntonio Feb 24 at 13:26
• My bad. I am not native speaker – Shorty12319 Feb 24 at 13:43
• You can’t do a contour integral on a ball! – Mindlack Feb 24 at 14:01
• Corrected the condition – Shorty12319 Feb 24 at 14:05
• Isn't $\oint_{|z|=1} dz = 2 \pi$ ? – 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 $$\int_{0}^{2\pi} |\cos(\theta-\alpha)|d\theta$$ = $$\int_{0}^{2\pi} |\cos(\theta)|d\theta$$ = $$4\int_{0}^{\frac{\pi}{2}} \cos\theta d\theta = 4$$)
• In the fact that $x,y$ are real. RHS means right hand side – Conrad Feb 24 at 22:50
• Note that it is enough for $f$ to be defined real, measurable and bounded on the unit circle only – Conrad Feb 24 at 23:14