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Prove that a square-integrable entire function is identically zero
Clash Royale CLAN TAG#URR8PPP
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Suppose $f$ is entire and
$$iint_mathbbC|f(z)|^2dxdy < infty$$Prove that $fequiv 0.$
So far I have:
Suppose $f$ is bounded. Then $f$ is constant by virtue of Liouville and so the conclusion is obvious. Thus, assume $f$ non-bounded. Then
$$Big|iint_mathbbCf^2(z)dABig| leq iint_mathbbC|f(z)|^2dxdy < infty$$
We can parameterize
beginalign* iint_mathbbCf^2(z)dA &= int_0^inftyint_0^2pif^2(re^itheta);dtheta;rdr \
&= int_0^inftyint_0^2pif^2(re^itheta);ire^itheta[-ifrac1re^-itheta]dtheta;rdr \
&= int_0^inftyoint_C_Rf^2(z);[-ifrac1z]dz;rdr \
&= int_0^infty2pi r drfrac12pi ioint_C_Rfracf^2(z)zdz \ \
&= 2pi f^2(0)int_0^infty rdr
endalign*
Whence f(0) = 0. And I have no idea where to go from there.
The "entire" bit seems to suggest Liouville but I have already dealt with the bounded case. Please advise ...
complex-analysis contour-integration complex-integration entire-functions
edited Jul 22 at 21:17
user357151
13.8k31140
asked Jun 12 '16 at 1:51
matty_k_walrus
1637
add a comment |Â
up vote
7
down vote
favorite
Suppose $f$ is entire and
$$iint_mathbbC|f(z)|^2dxdy < infty$$Prove that $fequiv 0.$
So far I have:
Suppose $f$ is bounded. Then $f$ is constant by virtue of Liouville and so the conclusion is obvious. Thus, assume $f$ non-bounded. Then
$$Big|iint_mathbbCf^2(z)dABig| leq iint_mathbbC|f(z)|^2dxdy < infty$$
We can parameterize
beginalign* iint_mathbbCf^2(z)dA &= int_0^inftyint_0^2pif^2(re^itheta);dtheta;rdr \
&= int_0^inftyint_0^2pif^2(re^itheta);ire^itheta[-ifrac1re^-itheta]dtheta;rdr \
&= int_0^inftyoint_C_Rf^2(z);[-ifrac1z]dz;rdr \
&= int_0^infty2pi r drfrac12pi ioint_C_Rfracf^2(z)zdz \ \
&= 2pi f^2(0)int_0^infty rdr
endalign*
Whence f(0) = 0. And I have no idea where to go from there.
The "entire" bit seems to suggest Liouville but I have already dealt with the bounded case. Please advise ...
complex-analysis contour-integration complex-integration entire-functions
edited Jul 22 at 21:17
user357151
13.8k31140
asked Jun 12 '16 at 1:51
matty_k_walrus
1637
3
Try to consider the translation $g(z)=f(z-z_0)$.
– Xiang Yu
Jun 12 '16 at 2:05
How do you get the expression in the second line following the words "we can parametrize"?
– DanielWainfleet
Jun 12 '16 at 6:33
@user254665 I made an error with parentheses, now corrected.
– matty_k_walrus
Jun 13 '16 at 16:37
@XiangYu But of course. Thank you!
– matty_k_walrus
Jun 13 '16 at 16:39
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Suppose $f$ is entire and
$$iint_mathbbC|f(z)|^2dxdy < infty$$Prove that $fequiv 0.$
So far I have:
Suppose $f$ is bounded. Then $f$ is constant by virtue of Liouville and so the conclusion is obvious. Thus, assume $f$ non-bounded. Then
$$Big|iint_mathbbCf^2(z)dABig| leq iint_mathbbC|f(z)|^2dxdy < infty$$
We can parameterize
beginalign* iint_mathbbCf^2(z)dA &= int_0^inftyint_0^2pif^2(re^itheta);dtheta;rdr \
&= int_0^inftyint_0^2pif^2(re^itheta);ire^itheta[-ifrac1re^-itheta]dtheta;rdr \
&= int_0^inftyoint_C_Rf^2(z);[-ifrac1z]dz;rdr \
&= int_0^infty2pi r drfrac12pi ioint_C_Rfracf^2(z)zdz \ \
&= 2pi f^2(0)int_0^infty rdr
endalign*
Whence f(0) = 0. And I have no idea where to go from there.
The "entire" bit seems to suggest Liouville but I have already dealt with the bounded case. Please advise ...
complex-analysis contour-integration complex-integration entire-functions
edited Jul 22 at 21:17
user357151
13.8k31140
asked Jun 12 '16 at 1:51
matty_k_walrus
1637
Suppose $f$ is entire and
$$iint_mathbbC|f(z)|^2dxdy < infty$$Prove that $fequiv 0.$
So far I have:
Suppose $f$ is bounded. Then $f$ is constant by virtue of Liouville and so the conclusion is obvious. Thus, assume $f$ non-bounded. Then
$$Big|iint_mathbbCf^2(z)dABig| leq iint_mathbbC|f(z)|^2dxdy < infty$$
We can parameterize
beginalign* iint_mathbbCf^2(z)dA &= int_0^inftyint_0^2pif^2(re^itheta);dtheta;rdr \
&= int_0^inftyint_0^2pif^2(re^itheta);ire^itheta[-ifrac1re^-itheta]dtheta;rdr \
&= int_0^inftyoint_C_Rf^2(z);[-ifrac1z]dz;rdr \
&= int_0^infty2pi r drfrac12pi ioint_C_Rfracf^2(z)zdz \ \
&= 2pi f^2(0)int_0^infty rdr
endalign*
Whence f(0) = 0. And I have no idea where to go from there.
The "entire" bit seems to suggest Liouville but I have already dealt with the bounded case. Please advise ...
complex-analysis contour-integration complex-integration entire-functions
edited Jul 22 at 21:17
user357151
13.8k31140
asked Jun 12 '16 at 1:51
matty_k_walrus
1637
edited Jul 22 at 21:17
user357151
13.8k31140
edited Jul 22 at 21:17
user357151
13.8k31140
edited Jul 22 at 21:17
user357151
13.8k31140
13.8k31140
asked Jun 12 '16 at 1:51
matty_k_walrus
1637
asked Jun 12 '16 at 1:51
matty_k_walrus
1637
asked Jun 12 '16 at 1:51
matty_k_walrus
1637
1637
3
Try to consider the translation $g(z)=f(z-z_0)$.
– Xiang Yu
Jun 12 '16 at 2:05
How do you get the expression in the second line following the words "we can parametrize"?
– DanielWainfleet
Jun 12 '16 at 6:33
@user254665 I made an error with parentheses, now corrected.
– matty_k_walrus
Jun 13 '16 at 16:37
@XiangYu But of course. Thank you!
– matty_k_walrus
Jun 13 '16 at 16:39
add a comment |Â
3
Try to consider the translation $g(z)=f(z-z_0)$.
– Xiang Yu
Jun 12 '16 at 2:05
How do you get the expression in the second line following the words "we can parametrize"?
– DanielWainfleet
Jun 12 '16 at 6:33
@user254665 I made an error with parentheses, now corrected.
– matty_k_walrus
Jun 13 '16 at 16:37
@XiangYu But of course. Thank you!
– matty_k_walrus
Jun 13 '16 at 16:39
3
3
Try to consider the translation $g(z)=f(z-z_0)$.
– Xiang Yu
Jun 12 '16 at 2:05
Try to consider the translation $g(z)=f(z-z_0)$.
– Xiang Yu
Jun 12 '16 at 2:05
How do you get the expression in the second line following the words "we can parametrize"?
– DanielWainfleet
Jun 12 '16 at 6:33
How do you get the expression in the second line following the words "we can parametrize"?
– DanielWainfleet
Jun 12 '16 at 6:33
@user254665 I made an error with parentheses, now corrected.
– matty_k_walrus
Jun 13 '16 at 16:37
@user254665 I made an error with parentheses, now corrected.
– matty_k_walrus
Jun 13 '16 at 16:37
@XiangYu But of course. Thank you!
– matty_k_walrus
Jun 13 '16 at 16:39
@XiangYu But of course. Thank you!
– matty_k_walrus
Jun 13 '16 at 16:39
add a comment |Â
1 Answer
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Fix $z_0in Bbb C$. For all $r > 0$, $$f(z_0) = frac12pi int_0^2pi f(z_0 + re^itheta), dtheta$$ By the Cauchy-Schwarz inequality, $$lvert f(z_0)rvert^2 le frac12piint_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta$$ for all $r > 0$. Hence, for every $R > 0$,
$$int_0^R lvert f(z_0)rvert^2 r, dr le frac12piint_0^R int_0^2pi lvert f(z_0 + re^itheta)rvert^2 r, dtheta, dr $$ or $$lvert f(z_0)rvert^2 fracR^22 le frac12piiint_D(z_0;R) lvert f(z)rvert^2, dx, dy$$ Hence, $$lvert f(z_0)rvert^2 le frac1pi R^2iint_D(z_0;R) lvert f(z)rvert^2, dx, dyle frac1pi R^2iint_Bbb C lvert f(z)rvert^2, dx, dy to 0quad textas quad Rto infty$$
So $f(z_0) = 0$. Since $z_0$ was arbitrary, $f equiv 0$.
answered Jun 12 '16 at 2:34
kobe
33.9k22146
Okay. How is Cauchy-Schwarz applicable? What is the inner product? Is it $$langle u, vrangle = int_0^2pi uv$$ Since the functions are complex-valued it is not necessarily true that $$langle v,urangle = overlinelangle u, vrangle$$. If $$langle u, vrangle = int_0^2pi |uv|$$ then we don't have linearity in the first argument.
– matty_k_walrus
Jun 13 '16 at 16:33
Think simpler than that. We have $4pi^2 lvert f(z_0)rvert^2 le left(int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dthetaright)^2$, and by the Cauchy-Schwarz inequality, $$left(int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dthetaright)^2 le int_0^2pi 1^2, dthetacdot int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta = 2pi int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta$$ Dividing through by $4pi^2$ yields the second line in the above answer.
– kobe
Jun 13 '16 at 17:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
Fix $z_0in Bbb C$. For all $r > 0$, $$f(z_0) = frac12pi int_0^2pi f(z_0 + re^itheta), dtheta$$ By the Cauchy-Schwarz inequality, $$lvert f(z_0)rvert^2 le frac12piint_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta$$ for all $r > 0$. Hence, for every $R > 0$,
$$int_0^R lvert f(z_0)rvert^2 r, dr le frac12piint_0^R int_0^2pi lvert f(z_0 + re^itheta)rvert^2 r, dtheta, dr $$ or $$lvert f(z_0)rvert^2 fracR^22 le frac12piiint_D(z_0;R) lvert f(z)rvert^2, dx, dy$$ Hence, $$lvert f(z_0)rvert^2 le frac1pi R^2iint_D(z_0;R) lvert f(z)rvert^2, dx, dyle frac1pi R^2iint_Bbb C lvert f(z)rvert^2, dx, dy to 0quad textas quad Rto infty$$
So $f(z_0) = 0$. Since $z_0$ was arbitrary, $f equiv 0$.
answered Jun 12 '16 at 2:34
kobe
33.9k22146
Okay. How is Cauchy-Schwarz applicable? What is the inner product? Is it $$langle u, vrangle = int_0^2pi uv$$ Since the functions are complex-valued it is not necessarily true that $$langle v,urangle = overlinelangle u, vrangle$$. If $$langle u, vrangle = int_0^2pi |uv|$$ then we don't have linearity in the first argument.
– matty_k_walrus
Jun 13 '16 at 16:33
Think simpler than that. We have $4pi^2 lvert f(z_0)rvert^2 le left(int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dthetaright)^2$, and by the Cauchy-Schwarz inequality, $$left(int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dthetaright)^2 le int_0^2pi 1^2, dthetacdot int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta = 2pi int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta$$ Dividing through by $4pi^2$ yields the second line in the above answer.
– kobe
Jun 13 '16 at 17:03
add a comment |Â
up vote
10
down vote
Fix $z_0in Bbb C$. For all $r > 0$, $$f(z_0) = frac12pi int_0^2pi f(z_0 + re^itheta), dtheta$$ By the Cauchy-Schwarz inequality, $$lvert f(z_0)rvert^2 le frac12piint_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta$$ for all $r > 0$. Hence, for every $R > 0$,
$$int_0^R lvert f(z_0)rvert^2 r, dr le frac12piint_0^R int_0^2pi lvert f(z_0 + re^itheta)rvert^2 r, dtheta, dr $$ or $$lvert f(z_0)rvert^2 fracR^22 le frac12piiint_D(z_0;R) lvert f(z)rvert^2, dx, dy$$ Hence, $$lvert f(z_0)rvert^2 le frac1pi R^2iint_D(z_0;R) lvert f(z)rvert^2, dx, dyle frac1pi R^2iint_Bbb C lvert f(z)rvert^2, dx, dy to 0quad textas quad Rto infty$$
So $f(z_0) = 0$. Since $z_0$ was arbitrary, $f equiv 0$.
answered Jun 12 '16 at 2:34
kobe
33.9k22146
Okay. How is Cauchy-Schwarz applicable? What is the inner product? Is it $$langle u, vrangle = int_0^2pi uv$$ Since the functions are complex-valued it is not necessarily true that $$langle v,urangle = overlinelangle u, vrangle$$. If $$langle u, vrangle = int_0^2pi |uv|$$ then we don't have linearity in the first argument.
– matty_k_walrus
Jun 13 '16 at 16:33
Think simpler than that. We have $4pi^2 lvert f(z_0)rvert^2 le left(int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dthetaright)^2$, and by the Cauchy-Schwarz inequality, $$left(int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dthetaright)^2 le int_0^2pi 1^2, dthetacdot int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta = 2pi int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta$$ Dividing through by $4pi^2$ yields the second line in the above answer.
– kobe
Jun 13 '16 at 17:03
add a comment |Â
up vote
10
down vote
up vote
10
down vote
Fix $z_0in Bbb C$. For all $r > 0$, $$f(z_0) = frac12pi int_0^2pi f(z_0 + re^itheta), dtheta$$ By the Cauchy-Schwarz inequality, $$lvert f(z_0)rvert^2 le frac12piint_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta$$ for all $r > 0$. Hence, for every $R > 0$,
$$int_0^R lvert f(z_0)rvert^2 r, dr le frac12piint_0^R int_0^2pi lvert f(z_0 + re^itheta)rvert^2 r, dtheta, dr $$ or $$lvert f(z_0)rvert^2 fracR^22 le frac12piiint_D(z_0;R) lvert f(z)rvert^2, dx, dy$$ Hence, $$lvert f(z_0)rvert^2 le frac1pi R^2iint_D(z_0;R) lvert f(z)rvert^2, dx, dyle frac1pi R^2iint_Bbb C lvert f(z)rvert^2, dx, dy to 0quad textas quad Rto infty$$
So $f(z_0) = 0$. Since $z_0$ was arbitrary, $f equiv 0$.
answered Jun 12 '16 at 2:34
kobe
33.9k22146
Fix $z_0in Bbb C$. For all $r > 0$, $$f(z_0) = frac12pi int_0^2pi f(z_0 + re^itheta), dtheta$$ By the Cauchy-Schwarz inequality, $$lvert f(z_0)rvert^2 le frac12piint_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta$$ for all $r > 0$. Hence, for every $R > 0$,
$$int_0^R lvert f(z_0)rvert^2 r, dr le frac12piint_0^R int_0^2pi lvert f(z_0 + re^itheta)rvert^2 r, dtheta, dr $$ or $$lvert f(z_0)rvert^2 fracR^22 le frac12piiint_D(z_0;R) lvert f(z)rvert^2, dx, dy$$ Hence, $$lvert f(z_0)rvert^2 le frac1pi R^2iint_D(z_0;R) lvert f(z)rvert^2, dx, dyle frac1pi R^2iint_Bbb C lvert f(z)rvert^2, dx, dy to 0quad textas quad Rto infty$$
So $f(z_0) = 0$. Since $z_0$ was arbitrary, $f equiv 0$.
answered Jun 12 '16 at 2:34
kobe
33.9k22146
answered Jun 12 '16 at 2:34
kobe
33.9k22146
answered Jun 12 '16 at 2:34
kobe
33.9k22146
answered Jun 12 '16 at 2:34
kobe
33.9k22146
33.9k22146
Okay. How is Cauchy-Schwarz applicable? What is the inner product? Is it $$langle u, vrangle = int_0^2pi uv$$ Since the functions are complex-valued it is not necessarily true that $$langle v,urangle = overlinelangle u, vrangle$$. If $$langle u, vrangle = int_0^2pi |uv|$$ then we don't have linearity in the first argument.
– matty_k_walrus
Jun 13 '16 at 16:33
Think simpler than that. We have $4pi^2 lvert f(z_0)rvert^2 le left(int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dthetaright)^2$, and by the Cauchy-Schwarz inequality, $$left(int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dthetaright)^2 le int_0^2pi 1^2, dthetacdot int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta = 2pi int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta$$ Dividing through by $4pi^2$ yields the second line in the above answer.
– kobe
Jun 13 '16 at 17:03
add a comment |Â
Okay. How is Cauchy-Schwarz applicable? What is the inner product? Is it $$langle u, vrangle = int_0^2pi uv$$ Since the functions are complex-valued it is not necessarily true that $$langle v,urangle = overlinelangle u, vrangle$$. If $$langle u, vrangle = int_0^2pi |uv|$$ then we don't have linearity in the first argument.
– matty_k_walrus
Jun 13 '16 at 16:33
Think simpler than that. We have $4pi^2 lvert f(z_0)rvert^2 le left(int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dthetaright)^2$, and by the Cauchy-Schwarz inequality, $$left(int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dthetaright)^2 le int_0^2pi 1^2, dthetacdot int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta = 2pi int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta$$ Dividing through by $4pi^2$ yields the second line in the above answer.
– kobe
Jun 13 '16 at 17:03
Okay. How is Cauchy-Schwarz applicable? What is the inner product? Is it $$langle u, vrangle = int_0^2pi uv$$ Since the functions are complex-valued it is not necessarily true that $$langle v,urangle = overlinelangle u, vrangle$$. If $$langle u, vrangle = int_0^2pi |uv|$$ then we don't have linearity in the first argument.
– matty_k_walrus
Jun 13 '16 at 16:33
Okay. How is Cauchy-Schwarz applicable? What is the inner product? Is it $$langle u, vrangle = int_0^2pi uv$$ Since the functions are complex-valued it is not necessarily true that $$langle v,urangle = overlinelangle u, vrangle$$. If $$langle u, vrangle = int_0^2pi |uv|$$ then we don't have linearity in the first argument.
– matty_k_walrus
Jun 13 '16 at 16:33
Think simpler than that. We have $4pi^2 lvert f(z_0)rvert^2 le left(int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dthetaright)^2$, and by the Cauchy-Schwarz inequality, $$left(int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dthetaright)^2 le int_0^2pi 1^2, dthetacdot int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta = 2pi int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta$$ Dividing through by $4pi^2$ yields the second line in the above answer.
– kobe
Jun 13 '16 at 17:03
Think simpler than that. We have $4pi^2 lvert f(z_0)rvert^2 le left(int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dthetaright)^2$, and by the Cauchy-Schwarz inequality, $$left(int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dthetaright)^2 le int_0^2pi 1^2, dthetacdot int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta = 2pi int_0^2pi lvert f(z_0 + re^itheta)rvert^2, dtheta$$ Dividing through by $4pi^2$ yields the second line in the above answer.
– kobe
Jun 13 '16 at 17:03
add a comment |Â
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3
Try to consider the translation $g(z)=f(z-z_0)$.
– Xiang Yu
Jun 12 '16 at 2:05
How do you get the expression in the second line following the words "we can parametrize"?
– DanielWainfleet
Jun 12 '16 at 6:33
@user254665 I made an error with parentheses, now corrected.
– matty_k_walrus
Jun 13 '16 at 16:37
@XiangYu But of course. Thank you!
– matty_k_walrus
Jun 13 '16 at 16:39