Function holomorphic in the units disk with different bound
Suppose $f$ is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Moreover suppose that for $|z|=1$ we have:
$Re(z)leq0Rightarrow |f(z)|leq 1$
$Re(z)>0Rightarrow |f(z)|leq 2$
and I Need to prove $|f(0)|leq sqrt{2}$ I know from the maximum modulus principle we have that:
$$1leq max_{|z|=1}|f|=max_{bar{D}(0,1)}|f|leq 2$$
but I can't really see where the square root come from so I cannot go any further.
complex-analysis holomorphic-functions maximum-principle
asked Dec 10 at 9:54
Renato Faraone
2,32911627
add a comment |
Suppose $f$ is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Moreover suppose that for $|z|=1$ we have:
$Re(z)leq0Rightarrow |f(z)|leq 1$
$Re(z)>0Rightarrow |f(z)|leq 2$
and I Need to prove $|f(0)|leq sqrt{2}$ I know from the maximum modulus principle we have that:
$$1leq max_{|z|=1}|f|=max_{bar{D}(0,1)}|f|leq 2$$
but I can't really see where the square root come from so I cannot go any further.
complex-analysis holomorphic-functions maximum-principle
asked Dec 10 at 9:54
Renato Faraone
2,32911627
out of curiosity, is the bound tight?
– AccidentalFourierTransform
Dec 10 at 18:11
@AccidentalFourierTransform As far as the text of my exercises says no
– Renato Faraone
Dec 11 at 10:04
add a comment |
Suppose $f$ is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Moreover suppose that for $|z|=1$ we have:
$Re(z)leq0Rightarrow |f(z)|leq 1$
$Re(z)>0Rightarrow |f(z)|leq 2$
and I Need to prove $|f(0)|leq sqrt{2}$ I know from the maximum modulus principle we have that:
$$1leq max_{|z|=1}|f|=max_{bar{D}(0,1)}|f|leq 2$$
but I can't really see where the square root come from so I cannot go any further.
complex-analysis holomorphic-functions maximum-principle
asked Dec 10 at 9:54
Renato Faraone
2,32911627
Suppose $f$ is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Moreover suppose that for $|z|=1$ we have:
$Re(z)leq0Rightarrow |f(z)|leq 1$
$Re(z)>0Rightarrow |f(z)|leq 2$
and I Need to prove $|f(0)|leq sqrt{2}$ I know from the maximum modulus principle we have that:
$$1leq max_{|z|=1}|f|=max_{bar{D}(0,1)}|f|leq 2$$
but I can't really see where the square root come from so I cannot go any further.
complex-analysis holomorphic-functions maximum-principle
complex-analysis holomorphic-functions maximum-principle
asked Dec 10 at 9:54
Renato Faraone
2,32911627
asked Dec 10 at 9:54
Renato Faraone
2,32911627
edited Dec 10 at 10:03
asked Dec 10 at 9:54
Renato Faraone
2,32911627
asked Dec 10 at 9:54
Renato Faraone
2,32911627
asked Dec 10 at 9:54
Renato Faraone
2,32911627
2,32911627
out of curiosity, is the bound tight?
– AccidentalFourierTransform
Dec 10 at 18:11
@AccidentalFourierTransform As far as the text of my exercises says no
– Renato Faraone
Dec 11 at 10:04
add a comment |
out of curiosity, is the bound tight?
– AccidentalFourierTransform
Dec 10 at 18:11
@AccidentalFourierTransform As far as the text of my exercises says no
– Renato Faraone
Dec 11 at 10:04
out of curiosity, is the bound tight?
– AccidentalFourierTransform
Dec 10 at 18:11
out of curiosity, is the bound tight?
– AccidentalFourierTransform
Dec 10 at 18:11
@AccidentalFourierTransform As far as the text of my exercises says no
– Renato Faraone
Dec 11 at 10:04
@AccidentalFourierTransform As far as the text of my exercises says no
– Renato Faraone
Dec 11 at 10:04
add a comment |
2 Answers
2
active
oldest
votes
First try. By Cauchy integral
$$f(0) = frac{1}{2pi i} int_{lvert zrvert = 1} frac{f(z)}{z},dz
=frac{1}{2pi} int_0^{2pi} f(e^{ivarphi}),dvarphi.$$
Hence
$$|f(0)|leq frac{1}{2pi} int_0^{2pi} |f(e^{ivarphi})|,dvarphileqfrac{2pi+1pi}{2pi}=frac{3}{2}.$$
But unfortunately $sqrt{2}<3/2$.
Second try. Consider the function $F(z)=f(z)f(−z)$ which is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Then, $text{Re}(z)leq 0$ iff $text{Re}(-z)geq 0$ and therefore, for $|z|=1$ we have that
$$|F(z)|leq |f(z)||f(−z)|leq 2cdot 1.$$
Now apply the Cauchy integral to $F$:
$$|f(0)|^2=|F(0)|leq frac{1}{2pi} int_0^{2pi} |F(e^{ivarphi})|,dvarphileq 2implies |f(0)|leq sqrt{2}.$$
answered Dec 10 at 10:00
Robert Z
93.2k1061132
add a comment |
For a slightly different proof than the one RobertZ gave, note that $loglvert frvertcolonoverline{D}tobar{mathbb{R}}$ is subharmonic (it is actually harmonic with poles), since it is $Relog f$ away from the zeros of $f$, and if $f(z)=0$ then $loglvert f(z)rvert=-infty$. Now the mean value property of harmonic function gives $loglvert f(0)rvert$ is at most the average value of $loglvert frvert$ on the unit circle, and the latter is bounded by $frac12log 2$.
answered Dec 10 at 10:44
user10354138
7,3942824
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
First try. By Cauchy integral
$$f(0) = frac{1}{2pi i} int_{lvert zrvert = 1} frac{f(z)}{z},dz
=frac{1}{2pi} int_0^{2pi} f(e^{ivarphi}),dvarphi.$$
Hence
$$|f(0)|leq frac{1}{2pi} int_0^{2pi} |f(e^{ivarphi})|,dvarphileqfrac{2pi+1pi}{2pi}=frac{3}{2}.$$
But unfortunately $sqrt{2}<3/2$.
Second try. Consider the function $F(z)=f(z)f(−z)$ which is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Then, $text{Re}(z)leq 0$ iff $text{Re}(-z)geq 0$ and therefore, for $|z|=1$ we have that
$$|F(z)|leq |f(z)||f(−z)|leq 2cdot 1.$$
Now apply the Cauchy integral to $F$:
$$|f(0)|^2=|F(0)|leq frac{1}{2pi} int_0^{2pi} |F(e^{ivarphi})|,dvarphileq 2implies |f(0)|leq sqrt{2}.$$
answered Dec 10 at 10:00
Robert Z
93.2k1061132
add a comment |
First try. By Cauchy integral
$$f(0) = frac{1}{2pi i} int_{lvert zrvert = 1} frac{f(z)}{z},dz
=frac{1}{2pi} int_0^{2pi} f(e^{ivarphi}),dvarphi.$$
Hence
$$|f(0)|leq frac{1}{2pi} int_0^{2pi} |f(e^{ivarphi})|,dvarphileqfrac{2pi+1pi}{2pi}=frac{3}{2}.$$
But unfortunately $sqrt{2}<3/2$.
Second try. Consider the function $F(z)=f(z)f(−z)$ which is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Then, $text{Re}(z)leq 0$ iff $text{Re}(-z)geq 0$ and therefore, for $|z|=1$ we have that
$$|F(z)|leq |f(z)||f(−z)|leq 2cdot 1.$$
Now apply the Cauchy integral to $F$:
$$|f(0)|^2=|F(0)|leq frac{1}{2pi} int_0^{2pi} |F(e^{ivarphi})|,dvarphileq 2implies |f(0)|leq sqrt{2}.$$
answered Dec 10 at 10:00
Robert Z
93.2k1061132
add a comment |
First try. By Cauchy integral
$$f(0) = frac{1}{2pi i} int_{lvert zrvert = 1} frac{f(z)}{z},dz
=frac{1}{2pi} int_0^{2pi} f(e^{ivarphi}),dvarphi.$$
Hence
$$|f(0)|leq frac{1}{2pi} int_0^{2pi} |f(e^{ivarphi})|,dvarphileqfrac{2pi+1pi}{2pi}=frac{3}{2}.$$
But unfortunately $sqrt{2}<3/2$.
Second try. Consider the function $F(z)=f(z)f(−z)$ which is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Then, $text{Re}(z)leq 0$ iff $text{Re}(-z)geq 0$ and therefore, for $|z|=1$ we have that
$$|F(z)|leq |f(z)||f(−z)|leq 2cdot 1.$$
Now apply the Cauchy integral to $F$:
$$|f(0)|^2=|F(0)|leq frac{1}{2pi} int_0^{2pi} |F(e^{ivarphi})|,dvarphileq 2implies |f(0)|leq sqrt{2}.$$
answered Dec 10 at 10:00
Robert Z
93.2k1061132
First try. By Cauchy integral
$$f(0) = frac{1}{2pi i} int_{lvert zrvert = 1} frac{f(z)}{z},dz
=frac{1}{2pi} int_0^{2pi} f(e^{ivarphi}),dvarphi.$$
Hence
$$|f(0)|leq frac{1}{2pi} int_0^{2pi} |f(e^{ivarphi})|,dvarphileqfrac{2pi+1pi}{2pi}=frac{3}{2}.$$
But unfortunately $sqrt{2}<3/2$.
Second try. Consider the function $F(z)=f(z)f(−z)$ which is continuous in the closed unit disk $bar{D}(0,1)$ and holomorphic over its interior $D(0,1)$. Then, $text{Re}(z)leq 0$ iff $text{Re}(-z)geq 0$ and therefore, for $|z|=1$ we have that
$$|F(z)|leq |f(z)||f(−z)|leq 2cdot 1.$$
Now apply the Cauchy integral to $F$:
$$|f(0)|^2=|F(0)|leq frac{1}{2pi} int_0^{2pi} |F(e^{ivarphi})|,dvarphileq 2implies |f(0)|leq sqrt{2}.$$
answered Dec 10 at 10:00
Robert Z
93.2k1061132
edited Dec 10 at 10:20
answered Dec 10 at 10:00
Robert Z
93.2k1061132
answered Dec 10 at 10:00
Robert Z
93.2k1061132
answered Dec 10 at 10:00
Robert Z
93.2k1061132
93.2k1061132
add a comment |
add a comment |
For a slightly different proof than the one RobertZ gave, note that $loglvert frvertcolonoverline{D}tobar{mathbb{R}}$ is subharmonic (it is actually harmonic with poles), since it is $Relog f$ away from the zeros of $f$, and if $f(z)=0$ then $loglvert f(z)rvert=-infty$. Now the mean value property of harmonic function gives $loglvert f(0)rvert$ is at most the average value of $loglvert frvert$ on the unit circle, and the latter is bounded by $frac12log 2$.
answered Dec 10 at 10:44
user10354138
7,3942824
add a comment |
For a slightly different proof than the one RobertZ gave, note that $loglvert frvertcolonoverline{D}tobar{mathbb{R}}$ is subharmonic (it is actually harmonic with poles), since it is $Relog f$ away from the zeros of $f$, and if $f(z)=0$ then $loglvert f(z)rvert=-infty$. Now the mean value property of harmonic function gives $loglvert f(0)rvert$ is at most the average value of $loglvert frvert$ on the unit circle, and the latter is bounded by $frac12log 2$.
answered Dec 10 at 10:44
user10354138
7,3942824
add a comment |
For a slightly different proof than the one RobertZ gave, note that $loglvert frvertcolonoverline{D}tobar{mathbb{R}}$ is subharmonic (it is actually harmonic with poles), since it is $Relog f$ away from the zeros of $f$, and if $f(z)=0$ then $loglvert f(z)rvert=-infty$. Now the mean value property of harmonic function gives $loglvert f(0)rvert$ is at most the average value of $loglvert frvert$ on the unit circle, and the latter is bounded by $frac12log 2$.
answered Dec 10 at 10:44
user10354138
7,3942824
For a slightly different proof than the one RobertZ gave, note that $loglvert frvertcolonoverline{D}tobar{mathbb{R}}$ is subharmonic (it is actually harmonic with poles), since it is $Relog f$ away from the zeros of $f$, and if $f(z)=0$ then $loglvert f(z)rvert=-infty$. Now the mean value property of harmonic function gives $loglvert f(0)rvert$ is at most the average value of $loglvert frvert$ on the unit circle, and the latter is bounded by $frac12log 2$.
answered Dec 10 at 10:44
user10354138
7,3942824
answered Dec 10 at 10:44
user10354138
7,3942824
answered Dec 10 at 10:44
user10354138
7,3942824
answered Dec 10 at 10:44
user10354138
7,3942824
7,3942824
add a comment |
add a comment |
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out of curiosity, is the bound tight?
– AccidentalFourierTransform
Dec 10 at 18:11
@AccidentalFourierTransform As far as the text of my exercises says no
– Renato Faraone
Dec 11 at 10:04