Subharmonic function bounded by linear functions The Next CEO of Stack OverflowSubharmonic/Superharmonic Inequality in Gilbarg/Trudinger [Section 2.8]Maximum of a subharmonic function on it's boundary.Maximum principle for subharmonic functionsGiven $u$ subharmonic, show that $u^p$ subharmonic for $pgeq 1$Show that the “Hartogs Regularity Radius” $R(z)$ is subharmonicMean-value theorem for subharmonic functions:Squared gradient of a harmonic function is subharmonicEquivalent definition of subharmonic functions.Harmonic lifts of subharmonic functions are subharmonic.Poisson problem, minimun value of subharmonic function in the interior.
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Subharmonic function bounded by linear functions
The Next CEO of Stack OverflowSubharmonic/Superharmonic Inequality in Gilbarg/Trudinger [Section 2.8]Maximum of a subharmonic function on it's boundary.Maximum principle for subharmonic functionsGiven $u$ subharmonic, show that $u^p$ subharmonic for $pgeq 1$Show that the “Hartogs Regularity Radius” $R(z)$ is subharmonicMean-value theorem for subharmonic functions:Squared gradient of a harmonic function is subharmonicEquivalent definition of subharmonic functions.Harmonic lifts of subharmonic functions are subharmonic.Poisson problem, minimun value of subharmonic function in the interior.
$begingroup$
Problem: Let $Omega = (x,y) in mathbbR^2 : xy > 0$ and suppose $u colon overlineOmega to mathbbR$ is continuous and subharmonic on $Omega$ and satisfies
$$u(x,y) leq sqrtx^2 + y^2$$
for $sqrtx^2 + y^2$ sufficiently large and for some $a,b > 0$ we have
beginequation*
beginaligned
u(x,0) &leq ax \
u(0,y) & leq by
endaligned
endequation*
for all $x,y geq 0$. Show that
$$u(x,y) leq ax + by$$
for all $x,y geq 0$.
If $a,b geq 1$, then we have $sqrtx^2 + y^2 leq ax + by$, so by taking a square $S = [R,0] times [0,R]$ with $R > 0$ sufficiently large, we see that
$$u(x,y) leq ax + by$$
on the boundary of $S$. Since $ax + by$ is a harmonic function and $u$ is subharmonic, the above inequality holds throughout $S$, hence for all $x,y geq 0$ since it holds for all $R$ sufficiently large. On the other hand, without this extra hypothesis, I'm not sure how to proceed. Since squares are simply connected I believe I can reduce to the case where $u$ is the real part of a holomorphic function, but I'm not sure how to use this extra information. Any hints?
analysis harmonic-functions
asked Mar 20 at 9:09
Ethan AlwaiseEthan Alwaise
6,471717
$endgroup$
add a comment |
$begingroup$
Problem: Let $Omega = (x,y) in mathbbR^2 : xy > 0$ and suppose $u colon overlineOmega to mathbbR$ is continuous and subharmonic on $Omega$ and satisfies
$$u(x,y) leq sqrtx^2 + y^2$$
for $sqrtx^2 + y^2$ sufficiently large and for some $a,b > 0$ we have
beginequation*
beginaligned
u(x,0) &leq ax \
u(0,y) & leq by
endaligned
endequation*
for all $x,y geq 0$. Show that
$$u(x,y) leq ax + by$$
for all $x,y geq 0$.
If $a,b geq 1$, then we have $sqrtx^2 + y^2 leq ax + by$, so by taking a square $S = [R,0] times [0,R]$ with $R > 0$ sufficiently large, we see that
$$u(x,y) leq ax + by$$
on the boundary of $S$. Since $ax + by$ is a harmonic function and $u$ is subharmonic, the above inequality holds throughout $S$, hence for all $x,y geq 0$ since it holds for all $R$ sufficiently large. On the other hand, without this extra hypothesis, I'm not sure how to proceed. Since squares are simply connected I believe I can reduce to the case where $u$ is the real part of a holomorphic function, but I'm not sure how to use this extra information. Any hints?
analysis harmonic-functions
asked Mar 20 at 9:09
Ethan AlwaiseEthan Alwaise
6,471717
$endgroup$
add a comment |
$begingroup$
Problem: Let $Omega = (x,y) in mathbbR^2 : xy > 0$ and suppose $u colon overlineOmega to mathbbR$ is continuous and subharmonic on $Omega$ and satisfies
$$u(x,y) leq sqrtx^2 + y^2$$
for $sqrtx^2 + y^2$ sufficiently large and for some $a,b > 0$ we have
beginequation*
beginaligned
u(x,0) &leq ax \
u(0,y) & leq by
endaligned
endequation*
for all $x,y geq 0$. Show that
$$u(x,y) leq ax + by$$
for all $x,y geq 0$.
If $a,b geq 1$, then we have $sqrtx^2 + y^2 leq ax + by$, so by taking a square $S = [R,0] times [0,R]$ with $R > 0$ sufficiently large, we see that
$$u(x,y) leq ax + by$$
on the boundary of $S$. Since $ax + by$ is a harmonic function and $u$ is subharmonic, the above inequality holds throughout $S$, hence for all $x,y geq 0$ since it holds for all $R$ sufficiently large. On the other hand, without this extra hypothesis, I'm not sure how to proceed. Since squares are simply connected I believe I can reduce to the case where $u$ is the real part of a holomorphic function, but I'm not sure how to use this extra information. Any hints?
analysis harmonic-functions
asked Mar 20 at 9:09
Ethan AlwaiseEthan Alwaise
6,471717
$endgroup$
Problem: Let $Omega = (x,y) in mathbbR^2 : xy > 0$ and suppose $u colon overlineOmega to mathbbR$ is continuous and subharmonic on $Omega$ and satisfies
$$u(x,y) leq sqrtx^2 + y^2$$
for $sqrtx^2 + y^2$ sufficiently large and for some $a,b > 0$ we have
beginequation*
beginaligned
u(x,0) &leq ax \
u(0,y) & leq by
endaligned
endequation*
for all $x,y geq 0$. Show that
$$u(x,y) leq ax + by$$
for all $x,y geq 0$.
If $a,b geq 1$, then we have $sqrtx^2 + y^2 leq ax + by$, so by taking a square $S = [R,0] times [0,R]$ with $R > 0$ sufficiently large, we see that
$$u(x,y) leq ax + by$$
on the boundary of $S$. Since $ax + by$ is a harmonic function and $u$ is subharmonic, the above inequality holds throughout $S$, hence for all $x,y geq 0$ since it holds for all $R$ sufficiently large. On the other hand, without this extra hypothesis, I'm not sure how to proceed. Since squares are simply connected I believe I can reduce to the case where $u$ is the real part of a holomorphic function, but I'm not sure how to use this extra information. Any hints?
analysis harmonic-functions
analysis harmonic-functions
asked Mar 20 at 9:09
Ethan AlwaiseEthan Alwaise
6,471717
asked Mar 20 at 9:09
Ethan AlwaiseEthan Alwaise
6,471717
asked Mar 20 at 9:09
Ethan AlwaiseEthan Alwaise
6,471717
asked Mar 20 at 9:09
Ethan AlwaiseEthan Alwaise
6,471717
asked Mar 20 at 9:09
Ethan AlwaiseEthan Alwaise
6,471717
6,471717
add a comment |
add a comment |
0
active
oldest
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