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$ |f(z)-p_n(z)|le (n+2)|z|^n+1 $ with $ deg(p_n(z))le n $ over the unit disk

0

2

$begingroup$

Let $ f(z) $ be a holomorphic function over the unit disk $ Delta:=<1 $. Suppose that $ |f(z)|le 1 $ for $ zinDelta $. Show that for any positive integer $ n $, there is a polynomial $ p_n(z) $ of degree at most $ n $ such that $$ |f(z)-p_n(z)|le (n+2)|z|^n+1 $$
for any $ zin Delta $.

---

My attempt:

I notice that if we take $ p_n(z)=sum_k=0^nfracf^(k)(0)k!z^k $, then
beginalign
left|f(z)-sum_k=0^nfracf^(k)(0)k!z^kright|&=left| sum_k=n+1^inftyfracf^(k)(0)k!z^k right|\
&lesum_k=n+1^infty |z|^k\
&lefraczz
endalign
If $ |z|lefracn+1n+2 $, we have proved $ |f(z)-p_n(z)|le (n+2)|z|^n+1 $, but what about $ fracn+1n+2<|z|<1 $ ?

complex-analysis

share|cite|improve this question

edited Apr 8 at 16:01

user549397

asked Apr 1 at 14:07

user549397user549397

1,7111618

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get rid of $f^k(0)/k!$ in the first inequality (from first to second displayed line)?
  $endgroup$
  – uniquesolution
  Apr 1 at 14:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @uniquesolution By Cauchy inequalities.
  $endgroup$
  – user549397
  Apr 1 at 14:52
  

  
  
  
  

add a comment | 

0

2

$begingroup$

Let $ f(z) $ be a holomorphic function over the unit disk $ Delta:=<1 $. Suppose that $ |f(z)|le 1 $ for $ zinDelta $. Show that for any positive integer $ n $, there is a polynomial $ p_n(z) $ of degree at most $ n $ such that $$ |f(z)-p_n(z)|le (n+2)|z|^n+1 $$
for any $ zin Delta $.

---

My attempt:

I notice that if we take $ p_n(z)=sum_k=0^nfracf^(k)(0)k!z^k $, then
beginalign
left|f(z)-sum_k=0^nfracf^(k)(0)k!z^kright|&=left| sum_k=n+1^inftyfracf^(k)(0)k!z^k right|\
&lesum_k=n+1^infty |z|^k\
&lefraczz
endalign
If $ |z|lefracn+1n+2 $, we have proved $ |f(z)-p_n(z)|le (n+2)|z|^n+1 $, but what about $ fracn+1n+2<|z|<1 $ ?

complex-analysis

share|cite|improve this question

edited Apr 8 at 16:01

user549397

asked Apr 1 at 14:07

user549397user549397

1,7111618

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get rid of $f^k(0)/k!$ in the first inequality (from first to second displayed line)?
  $endgroup$
  – uniquesolution
  Apr 1 at 14:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @uniquesolution By Cauchy inequalities.
  $endgroup$
  – user549397
  Apr 1 at 14:52
  

  
  
  
  

add a comment | 

$begingroup$

Let $ f(z) $ be a holomorphic function over the unit disk $ Delta:=<1 $. Suppose that $ |f(z)|le 1 $ for $ zinDelta $. Show that for any positive integer $ n $, there is a polynomial $ p_n(z) $ of degree at most $ n $ such that $$ |f(z)-p_n(z)|le (n+2)|z|^n+1 $$
for any $ zin Delta $.

---

My attempt:

I notice that if we take $ p_n(z)=sum_k=0^nfracf^(k)(0)k!z^k $, then
beginalign
left|f(z)-sum_k=0^nfracf^(k)(0)k!z^kright|&=left| sum_k=n+1^inftyfracf^(k)(0)k!z^k right|\
&lesum_k=n+1^infty |z|^k\
&lefraczz
endalign
If $ |z|lefracn+1n+2 $, we have proved $ |f(z)-p_n(z)|le (n+2)|z|^n+1 $, but what about $ fracn+1n+2<|z|<1 $ ?

complex-analysis

share|cite|improve this question

edited Apr 8 at 16:01

user549397

asked Apr 1 at 14:07

user549397user549397

1,7111618

$endgroup$

share|cite|improve this question

edited Apr 8 at 16:01

user549397

asked Apr 1 at 14:07

user549397user549397

1,7111618

share|cite|improve this question

edited Apr 8 at 16:01

user549397

asked Apr 1 at 14:07

user549397user549397

1,7111618

asked Apr 1 at 14:07

user549397user549397

1,7111618

asked Apr 1 at 14:07

user549397user549397

1,7111618

1 Answer
1

active

oldest

votes

1

$begingroup$

The way to do it is like this: let $f(z)=suma_kz^k, p_n(z)=sum_0le k le na_kz^k$; note that $$frac12piint_0^2pif(re^itheta)dtheta=sum^2r^2k, 0<r<1, $$ so $|f(z)| le 1$ implies trivially $ |a_k|^2r^2kle 1, 0<r<1 $, so taking $r to 1$ we get $|a_k| le 1$ for all $ k $.

Let $$g(z)=fracf(z)-p_n(z)z^n+1,$$ $g$ analytic in the unit disc; fix $0<r<1$, hence by maximum modulus for $|z|le r,$ beginalign |g(z)| &le sup_fracr^n+1 \
&le sup_fracr^n+1\
& le fraca_kr^n+1 \
&le fracn+2r^n+1.endalign Letting $r to 1$ we get $|g(z)| le n+2$ in the whole (open) unit disc and we are done!

share|cite|improve this answer

edited Apr 9 at 1:42

user549397

1,7111618

answered Apr 8 at 22:28

ConradConrad

1,52745

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice! Thank you!
  $endgroup$
  – user549397
  Apr 9 at 1:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome
  $endgroup$
  – Conrad
  Apr 9 at 1:50
  

  
  
  
  

add a comment | 

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1

$begingroup$

The way to do it is like this: let $f(z)=suma_kz^k, p_n(z)=sum_0le k le na_kz^k$; note that $$frac12piint_0^2pif(re^itheta)dtheta=sum^2r^2k, 0<r<1, $$ so $|f(z)| le 1$ implies trivially $ |a_k|^2r^2kle 1, 0<r<1 $, so taking $r to 1$ we get $|a_k| le 1$ for all $ k $.

Let $$g(z)=fracf(z)-p_n(z)z^n+1,$$ $g$ analytic in the unit disc; fix $0<r<1$, hence by maximum modulus for $|z|le r,$ beginalign |g(z)| &le sup_fracr^n+1 \
&le sup_fracr^n+1\
& le fraca_kr^n+1 \
&le fracn+2r^n+1.endalign Letting $r to 1$ we get $|g(z)| le n+2$ in the whole (open) unit disc and we are done!

share|cite|improve this answer

edited Apr 9 at 1:42

user549397

1,7111618

answered Apr 8 at 22:28

ConradConrad

1,52745

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice! Thank you!
  $endgroup$
  – user549397
  Apr 9 at 1:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome
  $endgroup$
  – Conrad
  Apr 9 at 1:50
  

  
  
  
  

add a comment | 

1

$begingroup$

The way to do it is like this: let $f(z)=suma_kz^k, p_n(z)=sum_0le k le na_kz^k$; note that $$frac12piint_0^2pif(re^itheta)dtheta=sum^2r^2k, 0<r<1, $$ so $|f(z)| le 1$ implies trivially $ |a_k|^2r^2kle 1, 0<r<1 $, so taking $r to 1$ we get $|a_k| le 1$ for all $ k $.

Let $$g(z)=fracf(z)-p_n(z)z^n+1,$$ $g$ analytic in the unit disc; fix $0<r<1$, hence by maximum modulus for $|z|le r,$ beginalign |g(z)| &le sup_fracr^n+1 \
&le sup_fracr^n+1\
& le fraca_kr^n+1 \
&le fracn+2r^n+1.endalign Letting $r to 1$ we get $|g(z)| le n+2$ in the whole (open) unit disc and we are done!

share|cite|improve this answer

edited Apr 9 at 1:42

user549397

1,7111618

answered Apr 8 at 22:28

ConradConrad

1,52745

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice! Thank you!
  $endgroup$
  – user549397
  Apr 9 at 1:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are welcome
  $endgroup$
  – Conrad
  Apr 9 at 1:50
  

  
  
  
  

add a comment | 

$begingroup$

The way to do it is like this: let $f(z)=suma_kz^k, p_n(z)=sum_0le k le na_kz^k$; note that $$frac12piint_0^2pif(re^itheta)dtheta=sum^2r^2k, 0<r<1, $$ so $|f(z)| le 1$ implies trivially $ |a_k|^2r^2kle 1, 0<r<1 $, so taking $r to 1$ we get $|a_k| le 1$ for all $ k $.

Let $$g(z)=fracf(z)-p_n(z)z^n+1,$$ $g$ analytic in the unit disc; fix $0<r<1$, hence by maximum modulus for $|z|le r,$ beginalign |g(z)| &le sup_fracr^n+1 \
&le sup_fracr^n+1\
& le fraca_kr^n+1 \
&le fracn+2r^n+1.endalign Letting $r to 1$ we get $|g(z)| le n+2$ in the whole (open) unit disc and we are done!

share|cite|improve this answer

edited Apr 9 at 1:42

user549397

1,7111618

answered Apr 8 at 22:28

ConradConrad

1,52745

$endgroup$

share|cite|improve this answer

edited Apr 9 at 1:42

user549397

1,7111618

answered Apr 8 at 22:28

ConradConrad

1,52745

edited Apr 9 at 1:42

user549397

1,7111618

edited Apr 9 at 1:42

user549397

1,7111618

answered Apr 8 at 22:28

ConradConrad

1,52745

answered Apr 8 at 22:28

ConradConrad

1,52745