Proof of a weakening of Moreras's Theorem for square archs
$begingroup$
Suppose that $f(z)$ is a continuous complex valued function on a disc such that
the integral $ int_C f dz = 0 $ for every contour C that is the boundary
of a square in the disc. Prove that $f$ must be analytic.
I am trying to adapt other proofs of Morera's theorem to this case of square paths, but without much success. The general strategy I believe should be a construct an antidervative by using the fact that we can integrate along squares, but I cannot seem to get started.
complex-analysis
edited Jan 1 at 22:37
A.Γ.
22.7k32656
asked Jan 1 at 22:33
David WarrenDavid Warren
586313
$endgroup$
add a comment |
$begingroup$
Suppose that $f(z)$ is a continuous complex valued function on a disc such that
the integral $ int_C f dz = 0 $ for every contour C that is the boundary
of a square in the disc. Prove that $f$ must be analytic.
I am trying to adapt other proofs of Morera's theorem to this case of square paths, but without much success. The general strategy I believe should be a construct an antidervative by using the fact that we can integrate along squares, but I cannot seem to get started.
complex-analysis
edited Jan 1 at 22:37
A.Γ.
22.7k32656
asked Jan 1 at 22:33
David WarrenDavid Warren
586313
$endgroup$
$begingroup$
see here: math.stackexchange.com/questions/42098/morera-type-theorems
$endgroup$
– user8268
Jan 1 at 23:27
$begingroup$
I am trying to answer it by constructing an antiderivative if possible
$endgroup$
– David Warren
Jan 2 at 3:51
add a comment |
$begingroup$
Suppose that $f(z)$ is a continuous complex valued function on a disc such that
the integral $ int_C f dz = 0 $ for every contour C that is the boundary
of a square in the disc. Prove that $f$ must be analytic.
I am trying to adapt other proofs of Morera's theorem to this case of square paths, but without much success. The general strategy I believe should be a construct an antidervative by using the fact that we can integrate along squares, but I cannot seem to get started.
complex-analysis
edited Jan 1 at 22:37
A.Γ.
22.7k32656
asked Jan 1 at 22:33
David WarrenDavid Warren
586313
$endgroup$
Suppose that $f(z)$ is a continuous complex valued function on a disc such that
the integral $ int_C f dz = 0 $ for every contour C that is the boundary
of a square in the disc. Prove that $f$ must be analytic.
I am trying to adapt other proofs of Morera's theorem to this case of square paths, but without much success. The general strategy I believe should be a construct an antidervative by using the fact that we can integrate along squares, but I cannot seem to get started.
complex-analysis
complex-analysis
edited Jan 1 at 22:37
A.Γ.
22.7k32656
asked Jan 1 at 22:33
David WarrenDavid Warren
586313
edited Jan 1 at 22:37
A.Γ.
22.7k32656
asked Jan 1 at 22:33
David WarrenDavid Warren
586313
edited Jan 1 at 22:37
A.Γ.
22.7k32656
edited Jan 1 at 22:37
A.Γ.
22.7k32656
edited Jan 1 at 22:37
A.Γ.
22.7k32656
22.7k32656
asked Jan 1 at 22:33
David WarrenDavid Warren
586313
asked Jan 1 at 22:33
David WarrenDavid Warren
586313
asked Jan 1 at 22:33
David WarrenDavid Warren
586313
586313
$begingroup$
see here: math.stackexchange.com/questions/42098/morera-type-theorems
$endgroup$
– user8268
Jan 1 at 23:27
$begingroup$
I am trying to answer it by constructing an antiderivative if possible
$endgroup$
– David Warren
Jan 2 at 3:51
add a comment |
$begingroup$
see here: math.stackexchange.com/questions/42098/morera-type-theorems
$endgroup$
– user8268
Jan 1 at 23:27
$begingroup$
I am trying to answer it by constructing an antiderivative if possible
$endgroup$
– David Warren
Jan 2 at 3:51
$begingroup$
see here: math.stackexchange.com/questions/42098/morera-type-theorems
$endgroup$
– user8268
Jan 1 at 23:27
$begingroup$
see here: math.stackexchange.com/questions/42098/morera-type-theorems
$endgroup$
– user8268
Jan 1 at 23:27
$begingroup$
I am trying to answer it by constructing an antiderivative if possible
$endgroup$
– David Warren
Jan 2 at 3:51
$begingroup$
I am trying to answer it by constructing an antiderivative if possible
$endgroup$
– David Warren
Jan 2 at 3:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can patch 2 adjacent square paths to get a $1:2$ rectangle path. In the same manner, we can get $1:n$ and $m:n$ rectangle path on which $int_C f(z)dz = 0$. Then, by approximation, we can get any rectangular path of real ratio from those of rational ratios, and by continuity of $f$, it holds $$int_{partial R}f(z)dz = 0$$ for any rectangle $Rsubset mathbb{D}={z;|;|z|<1}$. Using this, define anti-derivative $F$ as
$$
F(x+iy) = int_{gamma_1} f(z)dz + int_{gamma_2} f(z)dz
$$ where $gamma_1(t) = tx, 0le tle 1$ and$gamma_2(t) = x+ity, 0le tle 1$. Then we can see that $F$ satisfies the Cauchy-Riemann equation, and thus $F$ is complex differentiable and $F'=f$.
answered Jan 2 at 5:47
SongSong
9,869627
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We can patch 2 adjacent square paths to get a $1:2$ rectangle path. In the same manner, we can get $1:n$ and $m:n$ rectangle path on which $int_C f(z)dz = 0$. Then, by approximation, we can get any rectangular path of real ratio from those of rational ratios, and by continuity of $f$, it holds $$int_{partial R}f(z)dz = 0$$ for any rectangle $Rsubset mathbb{D}={z;|;|z|<1}$. Using this, define anti-derivative $F$ as
$$
F(x+iy) = int_{gamma_1} f(z)dz + int_{gamma_2} f(z)dz
$$ where $gamma_1(t) = tx, 0le tle 1$ and$gamma_2(t) = x+ity, 0le tle 1$. Then we can see that $F$ satisfies the Cauchy-Riemann equation, and thus $F$ is complex differentiable and $F'=f$.
answered Jan 2 at 5:47
SongSong
9,869627
$endgroup$
add a comment |
$begingroup$
We can patch 2 adjacent square paths to get a $1:2$ rectangle path. In the same manner, we can get $1:n$ and $m:n$ rectangle path on which $int_C f(z)dz = 0$. Then, by approximation, we can get any rectangular path of real ratio from those of rational ratios, and by continuity of $f$, it holds $$int_{partial R}f(z)dz = 0$$ for any rectangle $Rsubset mathbb{D}={z;|;|z|<1}$. Using this, define anti-derivative $F$ as
$$
F(x+iy) = int_{gamma_1} f(z)dz + int_{gamma_2} f(z)dz
$$ where $gamma_1(t) = tx, 0le tle 1$ and$gamma_2(t) = x+ity, 0le tle 1$. Then we can see that $F$ satisfies the Cauchy-Riemann equation, and thus $F$ is complex differentiable and $F'=f$.
answered Jan 2 at 5:47
SongSong
9,869627
$endgroup$
add a comment |
$begingroup$
We can patch 2 adjacent square paths to get a $1:2$ rectangle path. In the same manner, we can get $1:n$ and $m:n$ rectangle path on which $int_C f(z)dz = 0$. Then, by approximation, we can get any rectangular path of real ratio from those of rational ratios, and by continuity of $f$, it holds $$int_{partial R}f(z)dz = 0$$ for any rectangle $Rsubset mathbb{D}={z;|;|z|<1}$. Using this, define anti-derivative $F$ as
$$
F(x+iy) = int_{gamma_1} f(z)dz + int_{gamma_2} f(z)dz
$$ where $gamma_1(t) = tx, 0le tle 1$ and$gamma_2(t) = x+ity, 0le tle 1$. Then we can see that $F$ satisfies the Cauchy-Riemann equation, and thus $F$ is complex differentiable and $F'=f$.
answered Jan 2 at 5:47
SongSong
9,869627
$endgroup$
We can patch 2 adjacent square paths to get a $1:2$ rectangle path. In the same manner, we can get $1:n$ and $m:n$ rectangle path on which $int_C f(z)dz = 0$. Then, by approximation, we can get any rectangular path of real ratio from those of rational ratios, and by continuity of $f$, it holds $$int_{partial R}f(z)dz = 0$$ for any rectangle $Rsubset mathbb{D}={z;|;|z|<1}$. Using this, define anti-derivative $F$ as
$$
F(x+iy) = int_{gamma_1} f(z)dz + int_{gamma_2} f(z)dz
$$ where $gamma_1(t) = tx, 0le tle 1$ and$gamma_2(t) = x+ity, 0le tle 1$. Then we can see that $F$ satisfies the Cauchy-Riemann equation, and thus $F$ is complex differentiable and $F'=f$.
answered Jan 2 at 5:47
SongSong
9,869627
answered Jan 2 at 5:47
SongSong
9,869627
answered Jan 2 at 5:47
SongSong
9,869627
answered Jan 2 at 5:47
SongSong
9,869627
9,869627
add a comment |
add a comment |
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$begingroup$
see here: math.stackexchange.com/questions/42098/morera-type-theorems
$endgroup$
– user8268
Jan 1 at 23:27
$begingroup$
I am trying to answer it by constructing an antiderivative if possible
$endgroup$
– David Warren
Jan 2 at 3:51