$f$ is analytic, prove that for all $z$ with $|z|=1$, $sum_{n=0}^infty|a_nz^n|leq 2max{|f(z)|:|z|=2}$ ...

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$f$ is analytic, prove that for all $z$ with $|z|=1$, $sum_{n=0}^infty|a_nz^n|leq 2max{|f(z)|:|z|=2}$



The Next CEO of Stack OverflowLet $f($z$)= sum _{n=0}^{infty } a_nz^n$ be the power series expansion of f about $0$. Prove that $|a_n| le (n+1)(1+1/n)^{n} < e(n+1)$Prove that $sum_{n=0}^{infty}e^{intheta}$ is boundedLet $f : mathbb{D} → mathbb{D},$ $f(z) = sum_{n = 0}^{infty} a_nz^n$ be a bounded analytic functionIf $f(z):=sum_{n=0}^infty a_nz^{-n}$ is compact convergent, then $f$ is holomorphicProve that if $f$ is entire and $|f(z)| leq |z|^2+12$ then $f$ is a polynomial of deg $leq 2$$sum_{k=1}^infty f(z^{n_k})$ is analyticHow do i prove that $Res(f(z)e^frac{1}{z};0)=sum_{n=0}^infty frac{a_n}{(n+1)!}$ with $f(z)=sum_{n=0}^infty a_nz^n$Prove $|f(z)| leq e^{|z|}$ for a complex, analytic $f$ with $|f^{(k)}(0)| leq 1$ for any $k$.Prove that there exists $hin V$, such that $|f'(frac{1}{2})|leq|h'(frac{1}{2})|$ for all $fin V$.Show that $ |a_n| leq 2(1-alpha)$ for every $n$












1












$begingroup$


Suppose that $f(z)=sum_{n=0}^infty a_nz^n$ for all $zinmathbb{C}$. Prove that for all $z$ with $|z|=1$, $sum_{n=0}^infty|a_nz^n|leq 2 max{|f(z)|:|z|=2}$.



My try: I have no idea about how to prove it. I guess it can proved by Cauchy estimates or use the relation
$$
f(z)=sum_{n=0}^inftyfrac{f^n(0)}{n!}z^n.
$$






complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you sure you transcribed the homework correctly?
    $endgroup$
    – copper.hat
    Mar 17 at 17:58










  • $begingroup$
    @copper.hat I am sure the HW is written like this. I totally have no idea how to bound the $sum_{n=0}^infty|a_nz^n|$.
    $endgroup$
    – whereamI
    Mar 17 at 18:00










  • $begingroup$
    Try Cauchy's integral formula and note that if $|z| le 1$ and $|w| = 2$ then $|z-w| ge 1$.
    $endgroup$
    – copper.hat
    Mar 17 at 18:18












  • $begingroup$
    Sorry, my comment was misleading and would only show that $|sum_n a_n|$ is bounded by the quantity in the question.
    $endgroup$
    – copper.hat
    Mar 17 at 18:48


















1












$begingroup$


Suppose that $f(z)=sum_{n=0}^infty a_nz^n$ for all $zinmathbb{C}$. Prove that for all $z$ with $|z|=1$, $sum_{n=0}^infty|a_nz^n|leq 2 max{|f(z)|:|z|=2}$.



My try: I have no idea about how to prove it. I guess it can proved by Cauchy estimates or use the relation
$$
f(z)=sum_{n=0}^inftyfrac{f^n(0)}{n!}z^n.
$$






complex-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you sure you transcribed the homework correctly?
    $endgroup$
    – copper.hat
    Mar 17 at 17:58










  • $begingroup$
    @copper.hat I am sure the HW is written like this. I totally have no idea how to bound the $sum_{n=0}^infty|a_nz^n|$.
    $endgroup$
    – whereamI
    Mar 17 at 18:00










  • $begingroup$
    Try Cauchy's integral formula and note that if $|z| le 1$ and $|w| = 2$ then $|z-w| ge 1$.
    $endgroup$
    – copper.hat
    Mar 17 at 18:18












  • $begingroup$
    Sorry, my comment was misleading and would only show that $|sum_n a_n|$ is bounded by the quantity in the question.
    $endgroup$
    – copper.hat
    Mar 17 at 18:48
















1












1








1





$begingroup$


Suppose that $f(z)=sum_{n=0}^infty a_nz^n$ for all $zinmathbb{C}$. Prove that for all $z$ with $|z|=1$, $sum_{n=0}^infty|a_nz^n|leq 2 max{|f(z)|:|z|=2}$.



My try: I have no idea about how to prove it. I guess it can proved by Cauchy estimates or use the relation
$$
f(z)=sum_{n=0}^inftyfrac{f^n(0)}{n!}z^n.
$$






complex-analysis







share|cite|improve this question











$endgroup$




Suppose that $f(z)=sum_{n=0}^infty a_nz^n$ for all $zinmathbb{C}$. Prove that for all $z$ with $|z|=1$, $sum_{n=0}^infty|a_nz^n|leq 2 max{|f(z)|:|z|=2}$.



My try: I have no idea about how to prove it. I guess it can proved by Cauchy estimates or use the relation
$$
f(z)=sum_{n=0}^inftyfrac{f^n(0)}{n!}z^n.
$$






complex-analysis




complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 17 at 17:55









Bernard

124k741118




124k741118










asked Mar 17 at 17:50









whereamIwhereamI

361115




361115












  • $begingroup$
    Are you sure you transcribed the homework correctly?
    $endgroup$
    – copper.hat
    Mar 17 at 17:58










  • $begingroup$
    @copper.hat I am sure the HW is written like this. I totally have no idea how to bound the $sum_{n=0}^infty|a_nz^n|$.
    $endgroup$
    – whereamI
    Mar 17 at 18:00










  • $begingroup$
    Try Cauchy's integral formula and note that if $|z| le 1$ and $|w| = 2$ then $|z-w| ge 1$.
    $endgroup$
    – copper.hat
    Mar 17 at 18:18












  • $begingroup$
    Sorry, my comment was misleading and would only show that $|sum_n a_n|$ is bounded by the quantity in the question.
    $endgroup$
    – copper.hat
    Mar 17 at 18:48




















  • $begingroup$
    Are you sure you transcribed the homework correctly?
    $endgroup$
    – copper.hat
    Mar 17 at 17:58










  • $begingroup$
    @copper.hat I am sure the HW is written like this. I totally have no idea how to bound the $sum_{n=0}^infty|a_nz^n|$.
    $endgroup$
    – whereamI
    Mar 17 at 18:00










  • $begingroup$
    Try Cauchy's integral formula and note that if $|z| le 1$ and $|w| = 2$ then $|z-w| ge 1$.
    $endgroup$
    – copper.hat
    Mar 17 at 18:18












  • $begingroup$
    Sorry, my comment was misleading and would only show that $|sum_n a_n|$ is bounded by the quantity in the question.
    $endgroup$
    – copper.hat
    Mar 17 at 18:48


















$begingroup$
Are you sure you transcribed the homework correctly?
$endgroup$
– copper.hat
Mar 17 at 17:58




$begingroup$
Are you sure you transcribed the homework correctly?
$endgroup$
– copper.hat
Mar 17 at 17:58












$begingroup$
@copper.hat I am sure the HW is written like this. I totally have no idea how to bound the $sum_{n=0}^infty|a_nz^n|$.
$endgroup$
– whereamI
Mar 17 at 18:00




$begingroup$
@copper.hat I am sure the HW is written like this. I totally have no idea how to bound the $sum_{n=0}^infty|a_nz^n|$.
$endgroup$
– whereamI
Mar 17 at 18:00












$begingroup$
Try Cauchy's integral formula and note that if $|z| le 1$ and $|w| = 2$ then $|z-w| ge 1$.
$endgroup$
– copper.hat
Mar 17 at 18:18






$begingroup$
Try Cauchy's integral formula and note that if $|z| le 1$ and $|w| = 2$ then $|z-w| ge 1$.
$endgroup$
– copper.hat
Mar 17 at 18:18














$begingroup$
Sorry, my comment was misleading and would only show that $|sum_n a_n|$ is bounded by the quantity in the question.
$endgroup$
– copper.hat
Mar 17 at 18:48






$begingroup$
Sorry, my comment was misleading and would only show that $|sum_n a_n|$ is bounded by the quantity in the question.
$endgroup$
– copper.hat
Mar 17 at 18:48












1 Answer
1






active

oldest

votes


















4












$begingroup$

You were on the right track with a Cauchy estimate approach. For $n=0,1,dots,$



$$a_n = frac{f^{(n)}(0)}{n!} = frac{1}{2pi i}int_{|z|=2}frac{f(z)}{z^{n+1}},dz.$$



Let $M= max_{|z|=2}|f(z)|.$ From the above, $|a_n| le dfrac{M}{2^n}.$ Thus



$$sum_{n=0}^{infty}|a_n| le Mcdot sum_{n=0}^{infty}2^{-n} = 2M.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – whereamI
    Mar 17 at 18:46










  • $begingroup$
    You're welcome. Notice that $f$ entire is not needed; we only need $f$ analytic on $D(0,2+epsilon).$
    $endgroup$
    – zhw.
    Mar 17 at 18:50












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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

You were on the right track with a Cauchy estimate approach. For $n=0,1,dots,$



$$a_n = frac{f^{(n)}(0)}{n!} = frac{1}{2pi i}int_{|z|=2}frac{f(z)}{z^{n+1}},dz.$$



Let $M= max_{|z|=2}|f(z)|.$ From the above, $|a_n| le dfrac{M}{2^n}.$ Thus



$$sum_{n=0}^{infty}|a_n| le Mcdot sum_{n=0}^{infty}2^{-n} = 2M.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – whereamI
    Mar 17 at 18:46










  • $begingroup$
    You're welcome. Notice that $f$ entire is not needed; we only need $f$ analytic on $D(0,2+epsilon).$
    $endgroup$
    – zhw.
    Mar 17 at 18:50
















4












$begingroup$

You were on the right track with a Cauchy estimate approach. For $n=0,1,dots,$



$$a_n = frac{f^{(n)}(0)}{n!} = frac{1}{2pi i}int_{|z|=2}frac{f(z)}{z^{n+1}},dz.$$



Let $M= max_{|z|=2}|f(z)|.$ From the above, $|a_n| le dfrac{M}{2^n}.$ Thus



$$sum_{n=0}^{infty}|a_n| le Mcdot sum_{n=0}^{infty}2^{-n} = 2M.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – whereamI
    Mar 17 at 18:46










  • $begingroup$
    You're welcome. Notice that $f$ entire is not needed; we only need $f$ analytic on $D(0,2+epsilon).$
    $endgroup$
    – zhw.
    Mar 17 at 18:50














4












4








4





$begingroup$

You were on the right track with a Cauchy estimate approach. For $n=0,1,dots,$



$$a_n = frac{f^{(n)}(0)}{n!} = frac{1}{2pi i}int_{|z|=2}frac{f(z)}{z^{n+1}},dz.$$



Let $M= max_{|z|=2}|f(z)|.$ From the above, $|a_n| le dfrac{M}{2^n}.$ Thus



$$sum_{n=0}^{infty}|a_n| le Mcdot sum_{n=0}^{infty}2^{-n} = 2M.$$






share|cite|improve this answer









$endgroup$



You were on the right track with a Cauchy estimate approach. For $n=0,1,dots,$



$$a_n = frac{f^{(n)}(0)}{n!} = frac{1}{2pi i}int_{|z|=2}frac{f(z)}{z^{n+1}},dz.$$



Let $M= max_{|z|=2}|f(z)|.$ From the above, $|a_n| le dfrac{M}{2^n}.$ Thus



$$sum_{n=0}^{infty}|a_n| le Mcdot sum_{n=0}^{infty}2^{-n} = 2M.$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 17 at 18:39









zhw.zhw.

74.8k43275




74.8k43275












  • $begingroup$
    Thank you very much!
    $endgroup$
    – whereamI
    Mar 17 at 18:46










  • $begingroup$
    You're welcome. Notice that $f$ entire is not needed; we only need $f$ analytic on $D(0,2+epsilon).$
    $endgroup$
    – zhw.
    Mar 17 at 18:50


















  • $begingroup$
    Thank you very much!
    $endgroup$
    – whereamI
    Mar 17 at 18:46










  • $begingroup$
    You're welcome. Notice that $f$ entire is not needed; we only need $f$ analytic on $D(0,2+epsilon).$
    $endgroup$
    – zhw.
    Mar 17 at 18:50
















$begingroup$
Thank you very much!
$endgroup$
– whereamI
Mar 17 at 18:46




$begingroup$
Thank you very much!
$endgroup$
– whereamI
Mar 17 at 18:46












$begingroup$
You're welcome. Notice that $f$ entire is not needed; we only need $f$ analytic on $D(0,2+epsilon).$
$endgroup$
– zhw.
Mar 17 at 18:50




$begingroup$
You're welcome. Notice that $f$ entire is not needed; we only need $f$ analytic on $D(0,2+epsilon).$
$endgroup$
– zhw.
Mar 17 at 18:50


















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