Prove that minimum of $ f $ on $ |z| = 1 $ is no greater than $ |a_0| + cdots + |a_m| $
$begingroup$
Let $ D $ be the open unit disk. Suppose $ f in H(D) cap C(bar{D}) $ and write $ f(z) = sum a_n z^n $ for $ |z| < 1 $. If $ f $ has exactly $ m $ zeros in $ D $, prove that
$$ min_{|z|=1} |f(z) | leq |a_0| + cdots + |a_m|. $$
My original idea was to prove by contradiction: suppose $$ |f(z)| > |a_0| + cdots + |a_m| geq |a_0 + a_1z + cdots + a_mz^m|. $$
Then if we can use Rouché's theorem, this implies that $ f $ has the same number of zeros as $ f -g $, which is $ m + 1 $, a contradiction. However I don't think I can apply Rouché's theorem since $ f $ is not known to be holomorphic on $ bar{D} $. How do I solve this problem?
complex-analysis
asked Jan 5 at 11:23
calmcalm
1387
$endgroup$
add a comment |
$begingroup$
Let $ D $ be the open unit disk. Suppose $ f in H(D) cap C(bar{D}) $ and write $ f(z) = sum a_n z^n $ for $ |z| < 1 $. If $ f $ has exactly $ m $ zeros in $ D $, prove that
$$ min_{|z|=1} |f(z) | leq |a_0| + cdots + |a_m|. $$
My original idea was to prove by contradiction: suppose $$ |f(z)| > |a_0| + cdots + |a_m| geq |a_0 + a_1z + cdots + a_mz^m|. $$
Then if we can use Rouché's theorem, this implies that $ f $ has the same number of zeros as $ f -g $, which is $ m + 1 $, a contradiction. However I don't think I can apply Rouché's theorem since $ f $ is not known to be holomorphic on $ bar{D} $. How do I solve this problem?
complex-analysis
asked Jan 5 at 11:23
calmcalm
1387
$endgroup$
add a comment |
$begingroup$
Let $ D $ be the open unit disk. Suppose $ f in H(D) cap C(bar{D}) $ and write $ f(z) = sum a_n z^n $ for $ |z| < 1 $. If $ f $ has exactly $ m $ zeros in $ D $, prove that
$$ min_{|z|=1} |f(z) | leq |a_0| + cdots + |a_m|. $$
My original idea was to prove by contradiction: suppose $$ |f(z)| > |a_0| + cdots + |a_m| geq |a_0 + a_1z + cdots + a_mz^m|. $$
Then if we can use Rouché's theorem, this implies that $ f $ has the same number of zeros as $ f -g $, which is $ m + 1 $, a contradiction. However I don't think I can apply Rouché's theorem since $ f $ is not known to be holomorphic on $ bar{D} $. How do I solve this problem?
complex-analysis
asked Jan 5 at 11:23
calmcalm
1387
$endgroup$
Let $ D $ be the open unit disk. Suppose $ f in H(D) cap C(bar{D}) $ and write $ f(z) = sum a_n z^n $ for $ |z| < 1 $. If $ f $ has exactly $ m $ zeros in $ D $, prove that
$$ min_{|z|=1} |f(z) | leq |a_0| + cdots + |a_m|. $$
My original idea was to prove by contradiction: suppose $$ |f(z)| > |a_0| + cdots + |a_m| geq |a_0 + a_1z + cdots + a_mz^m|. $$
Then if we can use Rouché's theorem, this implies that $ f $ has the same number of zeros as $ f -g $, which is $ m + 1 $, a contradiction. However I don't think I can apply Rouché's theorem since $ f $ is not known to be holomorphic on $ bar{D} $. How do I solve this problem?
complex-analysis
complex-analysis
asked Jan 5 at 11:23
calmcalm
1387
asked Jan 5 at 11:23
calmcalm
1387
asked Jan 5 at 11:23
calmcalm
1387
asked Jan 5 at 11:23
calmcalm
1387
asked Jan 5 at 11:23
calmcalm
1387
1387
add a comment |
add a comment |
1 Answer
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$begingroup$
Just apply the same argument to the $|z|=1-epsilon$. $max |f(z)| >|a_0|+|a_1|+...+|a_m|$ implies $|f(z)| >|a_0+a_1z+...+a_mz^{m}|$ for $|z|=1-epsilon$ of $epsilon $ is small enough. Als0, number of zeros of a polynomial on the open disk equals the number of zeros on $|z|<1-epsilon$ for $epsilon $ is small enough.
answered Jan 5 at 11:34
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
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$begingroup$
Just apply the same argument to the $|z|=1-epsilon$. $max |f(z)| >|a_0|+|a_1|+...+|a_m|$ implies $|f(z)| >|a_0+a_1z+...+a_mz^{m}|$ for $|z|=1-epsilon$ of $epsilon $ is small enough. Als0, number of zeros of a polynomial on the open disk equals the number of zeros on $|z|<1-epsilon$ for $epsilon $ is small enough.
answered Jan 5 at 11:34
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
$endgroup$
add a comment |
$begingroup$
Just apply the same argument to the $|z|=1-epsilon$. $max |f(z)| >|a_0|+|a_1|+...+|a_m|$ implies $|f(z)| >|a_0+a_1z+...+a_mz^{m}|$ for $|z|=1-epsilon$ of $epsilon $ is small enough. Als0, number of zeros of a polynomial on the open disk equals the number of zeros on $|z|<1-epsilon$ for $epsilon $ is small enough.
answered Jan 5 at 11:34
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
$endgroup$
add a comment |
$begingroup$
Just apply the same argument to the $|z|=1-epsilon$. $max |f(z)| >|a_0|+|a_1|+...+|a_m|$ implies $|f(z)| >|a_0+a_1z+...+a_mz^{m}|$ for $|z|=1-epsilon$ of $epsilon $ is small enough. Als0, number of zeros of a polynomial on the open disk equals the number of zeros on $|z|<1-epsilon$ for $epsilon $ is small enough.
answered Jan 5 at 11:34
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
$endgroup$
Just apply the same argument to the $|z|=1-epsilon$. $max |f(z)| >|a_0|+|a_1|+...+|a_m|$ implies $|f(z)| >|a_0+a_1z+...+a_mz^{m}|$ for $|z|=1-epsilon$ of $epsilon $ is small enough. Als0, number of zeros of a polynomial on the open disk equals the number of zeros on $|z|<1-epsilon$ for $epsilon $ is small enough.
answered Jan 5 at 11:34
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
edited Jan 5 at 11:39
answered Jan 5 at 11:34
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
answered Jan 5 at 11:34
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
answered Jan 5 at 11:34
Kavi Rama MurthyKavi Rama Murthy
72.9k53170
72.9k53170
add a comment |
add a comment |
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