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$|f(z)|leq e^z$ implies that a Holomorphic function is identically 0 on upper half plane
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This is a question from an exam in an undergraduate complex analysis course.
We denote $G=zin mathbbC$ - the upper half plane. Given a holomorphic function $fin Hol(G)$ such that $|f(z)|leq e^z$ for all $zin G$, prove that $f=0$ identically on G.
The question provides a hint, which is to take a large positive $R>0$, and a sufficiently large natural number $N$, and define the function:
$$g(z)=prod_k=0^N-1f(e^2pi ik/N*z+iR)$$
on the circle $D_R=$.
I tried to define a sequence of such functions $g_N(z)$ and managed to show that they converge uniformly to 0 when $N$ approaches infinity through:
$$0leq |g_N(z)|=prod_k=0^N-1|f(e^2pi ik/N*z+iR)|leq e^-N/rightarrow0$$However I got stuck afterwards, and couldn't find a way to apply the identitiy theorem as I wanted.
complex-analysis holomorphic-functions
asked Jul 28 at 17:51
Dean Gurvitz
287
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up vote
2
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This is a question from an exam in an undergraduate complex analysis course.
We denote $G=zin mathbbC$ - the upper half plane. Given a holomorphic function $fin Hol(G)$ such that $|f(z)|leq e^z$ for all $zin G$, prove that $f=0$ identically on G.
The question provides a hint, which is to take a large positive $R>0$, and a sufficiently large natural number $N$, and define the function:
$$g(z)=prod_k=0^N-1f(e^2pi ik/N*z+iR)$$
on the circle $D_R=$.
I tried to define a sequence of such functions $g_N(z)$ and managed to show that they converge uniformly to 0 when $N$ approaches infinity through:
$$0leq |g_N(z)|=prod_k=0^N-1|f(e^2pi ik/N*z+iR)|leq e^-N/rightarrow0$$However I got stuck afterwards, and couldn't find a way to apply the identitiy theorem as I wanted.
complex-analysis holomorphic-functions
asked Jul 28 at 17:51
Dean Gurvitz
287
I don't know an answer, I guess one should figure out where it is important that $R$ and $N$ are "sufficiently big"...how did you get that that product is $le e^-N/$?
– Calvin Khor
Jul 28 at 18:54
My guess at how a solution might go: for large enough $R,N$, $g_N,R(z)$ satisfies some kind of bound like that, which is an analytic function on $|z|<R$ that vanishes to infinite order at 0. Hence $g=0$, which means at least one of the terms forming the product is 0. Letting $N$ run through bigger and bigger primes, we see that the zero set of $f$ is not isolated, so $f$ is identically 0.
– Calvin Khor
Jul 28 at 19:00
I think $leq e^-N/$ which I wrote might actually be wrong. Your intuition seems correct but for a finite $N$ we don't necessarily vanish, and at the limit $g$ is no longer a finite product, so I don't know if a single term needs to be $0$ on its own.
– Dean Gurvitz
Jul 28 at 19:19
If a bound like that is correct, then you /do/ vanish at $z=0$, not talking about the limit $N→ ∞$. Note $e^ ≥ frac1z$ so $e^z = o(|z|^K)$ as $z→ 0$ for every $ K$
– Calvin Khor
Jul 28 at 19:35
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This is a question from an exam in an undergraduate complex analysis course.
We denote $G=zin mathbbC$ - the upper half plane. Given a holomorphic function $fin Hol(G)$ such that $|f(z)|leq e^z$ for all $zin G$, prove that $f=0$ identically on G.
The question provides a hint, which is to take a large positive $R>0$, and a sufficiently large natural number $N$, and define the function:
$$g(z)=prod_k=0^N-1f(e^2pi ik/N*z+iR)$$
on the circle $D_R=$.
I tried to define a sequence of such functions $g_N(z)$ and managed to show that they converge uniformly to 0 when $N$ approaches infinity through:
$$0leq |g_N(z)|=prod_k=0^N-1|f(e^2pi ik/N*z+iR)|leq e^-N/rightarrow0$$However I got stuck afterwards, and couldn't find a way to apply the identitiy theorem as I wanted.
complex-analysis holomorphic-functions
asked Jul 28 at 17:51
Dean Gurvitz
287
This is a question from an exam in an undergraduate complex analysis course.
We denote $G=zin mathbbC$ - the upper half plane. Given a holomorphic function $fin Hol(G)$ such that $|f(z)|leq e^z$ for all $zin G$, prove that $f=0$ identically on G.
The question provides a hint, which is to take a large positive $R>0$, and a sufficiently large natural number $N$, and define the function:
$$g(z)=prod_k=0^N-1f(e^2pi ik/N*z+iR)$$
on the circle $D_R=$.
I tried to define a sequence of such functions $g_N(z)$ and managed to show that they converge uniformly to 0 when $N$ approaches infinity through:
$$0leq |g_N(z)|=prod_k=0^N-1|f(e^2pi ik/N*z+iR)|leq e^-N/rightarrow0$$However I got stuck afterwards, and couldn't find a way to apply the identitiy theorem as I wanted.
complex-analysis holomorphic-functions
asked Jul 28 at 17:51
Dean Gurvitz
287
edited Jul 28 at 18:08
asked Jul 28 at 17:51
Dean Gurvitz
287
asked Jul 28 at 17:51
Dean Gurvitz
287
asked Jul 28 at 17:51
Dean Gurvitz
287
287
I don't know an answer, I guess one should figure out where it is important that $R$ and $N$ are "sufficiently big"...how did you get that that product is $le e^-N/$?
– Calvin Khor
Jul 28 at 18:54
My guess at how a solution might go: for large enough $R,N$, $g_N,R(z)$ satisfies some kind of bound like that, which is an analytic function on $|z|<R$ that vanishes to infinite order at 0. Hence $g=0$, which means at least one of the terms forming the product is 0. Letting $N$ run through bigger and bigger primes, we see that the zero set of $f$ is not isolated, so $f$ is identically 0.
– Calvin Khor
Jul 28 at 19:00
I think $leq e^-N/$ which I wrote might actually be wrong. Your intuition seems correct but for a finite $N$ we don't necessarily vanish, and at the limit $g$ is no longer a finite product, so I don't know if a single term needs to be $0$ on its own.
– Dean Gurvitz
Jul 28 at 19:19
If a bound like that is correct, then you /do/ vanish at $z=0$, not talking about the limit $N→ ∞$. Note $e^ ≥ frac1z$ so $e^z = o(|z|^K)$ as $z→ 0$ for every $ K$
– Calvin Khor
Jul 28 at 19:35
add a comment |Â
I don't know an answer, I guess one should figure out where it is important that $R$ and $N$ are "sufficiently big"...how did you get that that product is $le e^-N/$?
– Calvin Khor
Jul 28 at 18:54
My guess at how a solution might go: for large enough $R,N$, $g_N,R(z)$ satisfies some kind of bound like that, which is an analytic function on $|z|<R$ that vanishes to infinite order at 0. Hence $g=0$, which means at least one of the terms forming the product is 0. Letting $N$ run through bigger and bigger primes, we see that the zero set of $f$ is not isolated, so $f$ is identically 0.
– Calvin Khor
Jul 28 at 19:00
I think $leq e^-N/$ which I wrote might actually be wrong. Your intuition seems correct but for a finite $N$ we don't necessarily vanish, and at the limit $g$ is no longer a finite product, so I don't know if a single term needs to be $0$ on its own.
– Dean Gurvitz
Jul 28 at 19:19
If a bound like that is correct, then you /do/ vanish at $z=0$, not talking about the limit $N→ ∞$. Note $e^ ≥ frac1z$ so $e^z = o(|z|^K)$ as $z→ 0$ for every $ K$
– Calvin Khor
Jul 28 at 19:35
I don't know an answer, I guess one should figure out where it is important that $R$ and $N$ are "sufficiently big"...how did you get that that product is $le e^-N/$?
– Calvin Khor
Jul 28 at 18:54
I don't know an answer, I guess one should figure out where it is important that $R$ and $N$ are "sufficiently big"...how did you get that that product is $le e^-N/$?
– Calvin Khor
Jul 28 at 18:54
My guess at how a solution might go: for large enough $R,N$, $g_N,R(z)$ satisfies some kind of bound like that, which is an analytic function on $|z|<R$ that vanishes to infinite order at 0. Hence $g=0$, which means at least one of the terms forming the product is 0. Letting $N$ run through bigger and bigger primes, we see that the zero set of $f$ is not isolated, so $f$ is identically 0.
– Calvin Khor
Jul 28 at 19:00
My guess at how a solution might go: for large enough $R,N$, $g_N,R(z)$ satisfies some kind of bound like that, which is an analytic function on $|z|<R$ that vanishes to infinite order at 0. Hence $g=0$, which means at least one of the terms forming the product is 0. Letting $N$ run through bigger and bigger primes, we see that the zero set of $f$ is not isolated, so $f$ is identically 0.
– Calvin Khor
Jul 28 at 19:00
I think $leq e^-N/$ which I wrote might actually be wrong. Your intuition seems correct but for a finite $N$ we don't necessarily vanish, and at the limit $g$ is no longer a finite product, so I don't know if a single term needs to be $0$ on its own.
– Dean Gurvitz
Jul 28 at 19:19
I think $leq e^-N/$ which I wrote might actually be wrong. Your intuition seems correct but for a finite $N$ we don't necessarily vanish, and at the limit $g$ is no longer a finite product, so I don't know if a single term needs to be $0$ on its own.
– Dean Gurvitz
Jul 28 at 19:19
If a bound like that is correct, then you /do/ vanish at $z=0$, not talking about the limit $N→ ∞$. Note $e^ ≥ frac1z$ so $e^z = o(|z|^K)$ as $z→ 0$ for every $ K$
– Calvin Khor
Jul 28 at 19:35
If a bound like that is correct, then you /do/ vanish at $z=0$, not talking about the limit $N→ ∞$. Note $e^ ≥ frac1z$ so $e^z = o(|z|^K)$ as $z→ 0$ for every $ K$
– Calvin Khor
Jul 28 at 19:35
add a comment |Â
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I don't know an answer, I guess one should figure out where it is important that $R$ and $N$ are "sufficiently big"...how did you get that that product is $le e^-N/$?
– Calvin Khor
Jul 28 at 18:54
My guess at how a solution might go: for large enough $R,N$, $g_N,R(z)$ satisfies some kind of bound like that, which is an analytic function on $|z|<R$ that vanishes to infinite order at 0. Hence $g=0$, which means at least one of the terms forming the product is 0. Letting $N$ run through bigger and bigger primes, we see that the zero set of $f$ is not isolated, so $f$ is identically 0.
– Calvin Khor
Jul 28 at 19:00
I think $leq e^-N/$ which I wrote might actually be wrong. Your intuition seems correct but for a finite $N$ we don't necessarily vanish, and at the limit $g$ is no longer a finite product, so I don't know if a single term needs to be $0$ on its own.
– Dean Gurvitz
Jul 28 at 19:19
If a bound like that is correct, then you /do/ vanish at $z=0$, not talking about the limit $N→ ∞$. Note $e^ ≥ frac1z$ so $e^z = o(|z|^K)$ as $z→ 0$ for every $ K$
– Calvin Khor
Jul 28 at 19:35