Mixing
Holomorphic but not a Dirichlet series
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
I want an example of holomorphic map $f : Omega to Bbb C$ where $Omega$ is the half-plane $mathrmRe(s) > 1$, such that there is no sequence $(a_n)$ of complex numbers with
$$f(s) = sum_n geq 1 a_n/n^s, quadforall s in Omega$$
I've read that $(s − 1)zeta(s)$ is such an example, but I'm not sure how to prove it.
riemann-zeta holomorphic-functions dirichlet-series
asked Jul 24 at 10:07
Alphonse
1,767622
add a comment |Â
up vote
1
down vote
favorite
I want an example of holomorphic map $f : Omega to Bbb C$ where $Omega$ is the half-plane $mathrmRe(s) > 1$, such that there is no sequence $(a_n)$ of complex numbers with
$$f(s) = sum_n geq 1 a_n/n^s, quadforall s in Omega$$
I've read that $(s − 1)zeta(s)$ is such an example, but I'm not sure how to prove it.
riemann-zeta holomorphic-functions dirichlet-series
asked Jul 24 at 10:07
Alphonse
1,767622
Maybe $f(s)=e^s$ would work?
– Alphonse
Jul 24 at 10:10
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want an example of holomorphic map $f : Omega to Bbb C$ where $Omega$ is the half-plane $mathrmRe(s) > 1$, such that there is no sequence $(a_n)$ of complex numbers with
$$f(s) = sum_n geq 1 a_n/n^s, quadforall s in Omega$$
I've read that $(s − 1)zeta(s)$ is such an example, but I'm not sure how to prove it.
riemann-zeta holomorphic-functions dirichlet-series
asked Jul 24 at 10:07
Alphonse
1,767622
I want an example of holomorphic map $f : Omega to Bbb C$ where $Omega$ is the half-plane $mathrmRe(s) > 1$, such that there is no sequence $(a_n)$ of complex numbers with
$$f(s) = sum_n geq 1 a_n/n^s, quadforall s in Omega$$
I've read that $(s − 1)zeta(s)$ is such an example, but I'm not sure how to prove it.
riemann-zeta holomorphic-functions dirichlet-series
asked Jul 24 at 10:07
Alphonse
1,767622
asked Jul 24 at 10:07
Alphonse
1,767622
asked Jul 24 at 10:07
Alphonse
1,767622
asked Jul 24 at 10:07
Alphonse
1,767622
1,767622
Maybe $f(s)=e^s$ would work?
– Alphonse
Jul 24 at 10:10
add a comment |Â
Maybe $f(s)=e^s$ would work?
– Alphonse
Jul 24 at 10:10
Maybe $f(s)=e^s$ would work?
– Alphonse
Jul 24 at 10:10
Maybe $f(s)=e^s$ would work?
– Alphonse
Jul 24 at 10:10
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
If a function can be represented by a Dirichlet series,
$$f(s) = sum_n = 1^infty fraca_nn^s$$
for $operatornameRe s > sigma_0$, then we have
$$lim_operatornameRe s to +infty f(s) = a_1 tag$ast$$$
uniformly in $operatornameIm s$. (And the convergence to $a_1$ is exponentially fast.) If the Dirichlet series converges absolutely for $operatornameRe s = sigma_1$, then for $sigma = operatornameRe s > sigma_1$ we can estimate
beginalign
lvert f(s) - a_1rvert
&= Biggllvert sum_n = 2^infty fraca_nn^sBiggrrvert \
&leqslant sum_n = 2^infty fraclvert a_nrvertn^sigma \
&= sum_n = 2^infty fraclvert a_nrvertn^sigma_1cdot frac1n^sigma - sigma_1 \
&leqslant frac12^sigma - sigma_1sum_n = 2^infty fraclvert a_nrvertn^sigma_1,.
endalign
So a function that does not satisfy $(ast)$, be it that there are paths (with real part tending to $+infty$) along which $f(s)$ oscillates, or that $lvert f(s)rvert$ is unbounded along such a path, cannot be represented by a Dirichlet series.
Since
begingather
lim_operatornameRe s to +infty lvert (s-1)zeta(s)rvert = +infty quad textand\
lim_operatornameRe s to +infty lvert e^srvert = +infty
endgather
we see that neither of those functions can be represented by a Dirichlet series.
The function defined by $f(s) = 1/s$ for $operatornameRe s > 1$ satisfies $(ast)$ - with $a_1 = 0$ - uniformly in $operatornameIm s$, but since the convergence to $0$ is too slow - it's not exponential - this function cannot be represented by a Dirichlet series either.
answered Jul 24 at 10:48
Daniel Fischer♦
171k16154274
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If a function can be represented by a Dirichlet series,
$$f(s) = sum_n = 1^infty fraca_nn^s$$
for $operatornameRe s > sigma_0$, then we have
$$lim_operatornameRe s to +infty f(s) = a_1 tag$ast$$$
uniformly in $operatornameIm s$. (And the convergence to $a_1$ is exponentially fast.) If the Dirichlet series converges absolutely for $operatornameRe s = sigma_1$, then for $sigma = operatornameRe s > sigma_1$ we can estimate
beginalign
lvert f(s) - a_1rvert
&= Biggllvert sum_n = 2^infty fraca_nn^sBiggrrvert \
&leqslant sum_n = 2^infty fraclvert a_nrvertn^sigma \
&= sum_n = 2^infty fraclvert a_nrvertn^sigma_1cdot frac1n^sigma - sigma_1 \
&leqslant frac12^sigma - sigma_1sum_n = 2^infty fraclvert a_nrvertn^sigma_1,.
endalign
So a function that does not satisfy $(ast)$, be it that there are paths (with real part tending to $+infty$) along which $f(s)$ oscillates, or that $lvert f(s)rvert$ is unbounded along such a path, cannot be represented by a Dirichlet series.
Since
begingather
lim_operatornameRe s to +infty lvert (s-1)zeta(s)rvert = +infty quad textand\
lim_operatornameRe s to +infty lvert e^srvert = +infty
endgather
we see that neither of those functions can be represented by a Dirichlet series.
The function defined by $f(s) = 1/s$ for $operatornameRe s > 1$ satisfies $(ast)$ - with $a_1 = 0$ - uniformly in $operatornameIm s$, but since the convergence to $0$ is too slow - it's not exponential - this function cannot be represented by a Dirichlet series either.
answered Jul 24 at 10:48
Daniel Fischer♦
171k16154274
add a comment |Â
up vote
2
down vote
accepted
If a function can be represented by a Dirichlet series,
$$f(s) = sum_n = 1^infty fraca_nn^s$$
for $operatornameRe s > sigma_0$, then we have
$$lim_operatornameRe s to +infty f(s) = a_1 tag$ast$$$
uniformly in $operatornameIm s$. (And the convergence to $a_1$ is exponentially fast.) If the Dirichlet series converges absolutely for $operatornameRe s = sigma_1$, then for $sigma = operatornameRe s > sigma_1$ we can estimate
beginalign
lvert f(s) - a_1rvert
&= Biggllvert sum_n = 2^infty fraca_nn^sBiggrrvert \
&leqslant sum_n = 2^infty fraclvert a_nrvertn^sigma \
&= sum_n = 2^infty fraclvert a_nrvertn^sigma_1cdot frac1n^sigma - sigma_1 \
&leqslant frac12^sigma - sigma_1sum_n = 2^infty fraclvert a_nrvertn^sigma_1,.
endalign
So a function that does not satisfy $(ast)$, be it that there are paths (with real part tending to $+infty$) along which $f(s)$ oscillates, or that $lvert f(s)rvert$ is unbounded along such a path, cannot be represented by a Dirichlet series.
Since
begingather
lim_operatornameRe s to +infty lvert (s-1)zeta(s)rvert = +infty quad textand\
lim_operatornameRe s to +infty lvert e^srvert = +infty
endgather
we see that neither of those functions can be represented by a Dirichlet series.
The function defined by $f(s) = 1/s$ for $operatornameRe s > 1$ satisfies $(ast)$ - with $a_1 = 0$ - uniformly in $operatornameIm s$, but since the convergence to $0$ is too slow - it's not exponential - this function cannot be represented by a Dirichlet series either.
answered Jul 24 at 10:48
Daniel Fischer♦
171k16154274
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If a function can be represented by a Dirichlet series,
$$f(s) = sum_n = 1^infty fraca_nn^s$$
for $operatornameRe s > sigma_0$, then we have
$$lim_operatornameRe s to +infty f(s) = a_1 tag$ast$$$
uniformly in $operatornameIm s$. (And the convergence to $a_1$ is exponentially fast.) If the Dirichlet series converges absolutely for $operatornameRe s = sigma_1$, then for $sigma = operatornameRe s > sigma_1$ we can estimate
beginalign
lvert f(s) - a_1rvert
&= Biggllvert sum_n = 2^infty fraca_nn^sBiggrrvert \
&leqslant sum_n = 2^infty fraclvert a_nrvertn^sigma \
&= sum_n = 2^infty fraclvert a_nrvertn^sigma_1cdot frac1n^sigma - sigma_1 \
&leqslant frac12^sigma - sigma_1sum_n = 2^infty fraclvert a_nrvertn^sigma_1,.
endalign
So a function that does not satisfy $(ast)$, be it that there are paths (with real part tending to $+infty$) along which $f(s)$ oscillates, or that $lvert f(s)rvert$ is unbounded along such a path, cannot be represented by a Dirichlet series.
Since
begingather
lim_operatornameRe s to +infty lvert (s-1)zeta(s)rvert = +infty quad textand\
lim_operatornameRe s to +infty lvert e^srvert = +infty
endgather
we see that neither of those functions can be represented by a Dirichlet series.
The function defined by $f(s) = 1/s$ for $operatornameRe s > 1$ satisfies $(ast)$ - with $a_1 = 0$ - uniformly in $operatornameIm s$, but since the convergence to $0$ is too slow - it's not exponential - this function cannot be represented by a Dirichlet series either.
answered Jul 24 at 10:48
Daniel Fischer♦
171k16154274
If a function can be represented by a Dirichlet series,
$$f(s) = sum_n = 1^infty fraca_nn^s$$
for $operatornameRe s > sigma_0$, then we have
$$lim_operatornameRe s to +infty f(s) = a_1 tag$ast$$$
uniformly in $operatornameIm s$. (And the convergence to $a_1$ is exponentially fast.) If the Dirichlet series converges absolutely for $operatornameRe s = sigma_1$, then for $sigma = operatornameRe s > sigma_1$ we can estimate
beginalign
lvert f(s) - a_1rvert
&= Biggllvert sum_n = 2^infty fraca_nn^sBiggrrvert \
&leqslant sum_n = 2^infty fraclvert a_nrvertn^sigma \
&= sum_n = 2^infty fraclvert a_nrvertn^sigma_1cdot frac1n^sigma - sigma_1 \
&leqslant frac12^sigma - sigma_1sum_n = 2^infty fraclvert a_nrvertn^sigma_1,.
endalign
So a function that does not satisfy $(ast)$, be it that there are paths (with real part tending to $+infty$) along which $f(s)$ oscillates, or that $lvert f(s)rvert$ is unbounded along such a path, cannot be represented by a Dirichlet series.
Since
begingather
lim_operatornameRe s to +infty lvert (s-1)zeta(s)rvert = +infty quad textand\
lim_operatornameRe s to +infty lvert e^srvert = +infty
endgather
we see that neither of those functions can be represented by a Dirichlet series.
The function defined by $f(s) = 1/s$ for $operatornameRe s > 1$ satisfies $(ast)$ - with $a_1 = 0$ - uniformly in $operatornameIm s$, but since the convergence to $0$ is too slow - it's not exponential - this function cannot be represented by a Dirichlet series either.
answered Jul 24 at 10:48
Daniel Fischer♦
171k16154274
edited Jul 24 at 11:18
answered Jul 24 at 10:48
Daniel Fischer♦
171k16154274
answered Jul 24 at 10:48
Daniel Fischer♦
171k16154274
answered Jul 24 at 10:48
Daniel Fischer♦
171k16154274
171k16154274
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861180%2fholomorphic-but-not-a-dirichlet-series%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Maybe $f(s)=e^s$ would work?
– Alphonse
Jul 24 at 10:10