Let $f$ be analytic on $D=\{z \in \mathbb{C}: |z| < 1 \}$ with $\lim_{|z| \to 1^{-}} f(z) =0.$ Prove $f=0$. I could not solve it . Any help will be appreciated. Thank you for your help.
Let $f$ be analytic on $D=\{z \in \mathbb{C}: |z| < 1 \}$ with $\lim_{|z| \to 1^{-}} f(z)=0.$ Prove $f=0$.
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complex-analysis
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5Use the maximum modulus principle. – 2017-01-05
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0On the boundary it takes maximum or minimum value – 2017-01-05
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0Use the open mapping theorem (my complex theorem of the month). – 2017-01-05
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0@ajotatxe it is the limit of $|z|$, not $z$. The terminology is standard and its meaning is clear. – 2017-01-05
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0it is limit of $|z|$ – 2017-01-05
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0From the limit can we say $0$ is maximum and minimum value of the function? – 2017-01-05
1 Answers
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Let $\hat{f}$ be the extension of $f$ to the closed unit disc $\bar{D} = \{z\;\;|\;\;|z|\leq 1\}$ obtained by setting $\hat{f}(z) = 0$ on the boundary of $D$. We see that $\hat{f}$ is continuous since $\lim_{|z|\to 1^-} f(z) = 0$. Moreover, $\bar{D}$ is compact, and therefore the image $\hat{f}(\bar{D})$ is compact as well (continuous functions map compact sets to compact sets), which in turn implies that $\hat{f}(\bar{D})$ is a closed and bounded subset of $\mathbb{C}$. This means that there is some $z_0$ in $\bar{D}$ such that $|\hat{f}(z_0)| \geq |\hat{f}(z)|$ for all $z\in \bar{D}$.
Now we have exactly two possibilities: Either $z_0\in D$ or $|z_0| = 1$.
If $|z_0| = 1$, then the maximum value of $|\hat{f}|$ is $0$, which can only happen if $\hat{f} \equiv 0$, which in turn would imply that $f\equiv 0$.
If $z_0\in D$, then there is a neighborhood $U$ of $z_0$ that is contained in $D$ since $D$ is open. We also know that $\hat{f} \equiv f$ on $D$. Using what we previously showed, we then see that $|f(z_0)| \geq |f(z)|$ for all $z\in U$. Since $f$ is analytic on $D$, the Maximum Modulus Principle implies that $f$ must be constant on $D$. Finally, since $\lim_{|z|\to 1^-}|f(z)| = 0$, we see that $f$ must be identically $0$ on $D$. QED.