Find all entire functions with $int_Bbb C |f(z)|, dz= 1$

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I know that an entire function with bounded $L^1$ norm is identically $0$, but I do not know how to attack this problem. Does this contradict the fact I stated about entire functions with bounded $L^1$ norm?




complex-analysis




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  • $dz$ should probably be $dA(z),$ where $A$ is area measure. What do you mean by $L^1$-norm?
    – zhw.
    Jul 31 at 22:31










  • $int int_Bbb C |f(x+iy)| dxdy$ is the $L^1$ norm
    –  apple
    Aug 1 at 1:30














up vote
1
down vote

favorite












I know that an entire function with bounded $L^1$ norm is identically $0$, but I do not know how to attack this problem. Does this contradict the fact I stated about entire functions with bounded $L^1$ norm?




complex-analysis




share|cite|improve this question





















  • $dz$ should probably be $dA(z),$ where $A$ is area measure. What do you mean by $L^1$-norm?
    – zhw.
    Jul 31 at 22:31










  • $int int_Bbb C |f(x+iy)| dxdy$ is the $L^1$ norm
    –  apple
    Aug 1 at 1:30












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I know that an entire function with bounded $L^1$ norm is identically $0$, but I do not know how to attack this problem. Does this contradict the fact I stated about entire functions with bounded $L^1$ norm?




complex-analysis




share|cite|improve this question













I know that an entire function with bounded $L^1$ norm is identically $0$, but I do not know how to attack this problem. Does this contradict the fact I stated about entire functions with bounded $L^1$ norm?





complex-analysis





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 31 at 18:32









Clayton

17.7k22781




17.7k22781









asked Jul 31 at 18:26









apple

133




133











  • $dz$ should probably be $dA(z),$ where $A$ is area measure. What do you mean by $L^1$-norm?
    – zhw.
    Jul 31 at 22:31










  • $int int_Bbb C |f(x+iy)| dxdy$ is the $L^1$ norm
    –  apple
    Aug 1 at 1:30
















  • $dz$ should probably be $dA(z),$ where $A$ is area measure. What do you mean by $L^1$-norm?
    – zhw.
    Jul 31 at 22:31










  • $int int_Bbb C |f(x+iy)| dxdy$ is the $L^1$ norm
    –  apple
    Aug 1 at 1:30















$dz$ should probably be $dA(z),$ where $A$ is area measure. What do you mean by $L^1$-norm?
– zhw.
Jul 31 at 22:31




$dz$ should probably be $dA(z),$ where $A$ is area measure. What do you mean by $L^1$-norm?
– zhw.
Jul 31 at 22:31












$int int_Bbb C |f(x+iy)| dxdy$ is the $L^1$ norm
–  apple
Aug 1 at 1:30




$int int_Bbb C |f(x+iy)| dxdy$ is the $L^1$ norm
–  apple
Aug 1 at 1:30










1 Answer
1






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up vote
2
down vote













There are no such functions.



For each $w in mathbb C$, we have
$$
|f(w)| le int_Bbb C |f(z)|, dz= 1
$$



Therefore, $f$ is bounded. Since $f$ is entire, $f$ is constant, by Liouville's theorem.



But then we cannot have $int_Bbb C |f(z)|, dz= 1$.






share|cite|improve this answer





















  • Why is $|f(w)| le int_Bbb C |f(z)|, dz$ valid?
    – md2perpe
    Jul 31 at 19:08










  • @md2perpe, because $displaystyle |f(w)| = lim_rto 0 int_D(w,r) |f(z)|, dz$.
    – lhf
    Aug 2 at 0:58











  • there is hint for the problem: to estimate the Taylor coefficients, but i still have no clear idea
    –  apple
    Aug 3 at 2:22










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













There are no such functions.



For each $w in mathbb C$, we have
$$
|f(w)| le int_Bbb C |f(z)|, dz= 1
$$



Therefore, $f$ is bounded. Since $f$ is entire, $f$ is constant, by Liouville's theorem.



But then we cannot have $int_Bbb C |f(z)|, dz= 1$.






share|cite|improve this answer





















  • Why is $|f(w)| le int_Bbb C |f(z)|, dz$ valid?
    – md2perpe
    Jul 31 at 19:08










  • @md2perpe, because $displaystyle |f(w)| = lim_rto 0 int_D(w,r) |f(z)|, dz$.
    – lhf
    Aug 2 at 0:58











  • there is hint for the problem: to estimate the Taylor coefficients, but i still have no clear idea
    –  apple
    Aug 3 at 2:22














up vote
2
down vote













There are no such functions.



For each $w in mathbb C$, we have
$$
|f(w)| le int_Bbb C |f(z)|, dz= 1
$$



Therefore, $f$ is bounded. Since $f$ is entire, $f$ is constant, by Liouville's theorem.



But then we cannot have $int_Bbb C |f(z)|, dz= 1$.






share|cite|improve this answer





















  • Why is $|f(w)| le int_Bbb C |f(z)|, dz$ valid?
    – md2perpe
    Jul 31 at 19:08










  • @md2perpe, because $displaystyle |f(w)| = lim_rto 0 int_D(w,r) |f(z)|, dz$.
    – lhf
    Aug 2 at 0:58











  • there is hint for the problem: to estimate the Taylor coefficients, but i still have no clear idea
    –  apple
    Aug 3 at 2:22












up vote
2
down vote










up vote
2
down vote









There are no such functions.



For each $w in mathbb C$, we have
$$
|f(w)| le int_Bbb C |f(z)|, dz= 1
$$



Therefore, $f$ is bounded. Since $f$ is entire, $f$ is constant, by Liouville's theorem.



But then we cannot have $int_Bbb C |f(z)|, dz= 1$.






share|cite|improve this answer













There are no such functions.



For each $w in mathbb C$, we have
$$
|f(w)| le int_Bbb C |f(z)|, dz= 1
$$



Therefore, $f$ is bounded. Since $f$ is entire, $f$ is constant, by Liouville's theorem.



But then we cannot have $int_Bbb C |f(z)|, dz= 1$.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 31 at 18:51









lhf

155k9160364




155k9160364











  • Why is $|f(w)| le int_Bbb C |f(z)|, dz$ valid?
    – md2perpe
    Jul 31 at 19:08










  • @md2perpe, because $displaystyle |f(w)| = lim_rto 0 int_D(w,r) |f(z)|, dz$.
    – lhf
    Aug 2 at 0:58











  • there is hint for the problem: to estimate the Taylor coefficients, but i still have no clear idea
    –  apple
    Aug 3 at 2:22
















  • Why is $|f(w)| le int_Bbb C |f(z)|, dz$ valid?
    – md2perpe
    Jul 31 at 19:08










  • @md2perpe, because $displaystyle |f(w)| = lim_rto 0 int_D(w,r) |f(z)|, dz$.
    – lhf
    Aug 2 at 0:58











  • there is hint for the problem: to estimate the Taylor coefficients, but i still have no clear idea
    –  apple
    Aug 3 at 2:22















Why is $|f(w)| le int_Bbb C |f(z)|, dz$ valid?
– md2perpe
Jul 31 at 19:08




Why is $|f(w)| le int_Bbb C |f(z)|, dz$ valid?
– md2perpe
Jul 31 at 19:08












@md2perpe, because $displaystyle |f(w)| = lim_rto 0 int_D(w,r) |f(z)|, dz$.
– lhf
Aug 2 at 0:58





@md2perpe, because $displaystyle |f(w)| = lim_rto 0 int_D(w,r) |f(z)|, dz$.
– lhf
Aug 2 at 0:58













there is hint for the problem: to estimate the Taylor coefficients, but i still have no clear idea
–  apple
Aug 3 at 2:22




there is hint for the problem: to estimate the Taylor coefficients, but i still have no clear idea
–  apple
Aug 3 at 2:22












 

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