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Goursat's theorem: Let $f : U \to \mathbb{C}$ be a function that is holomorphic on the open set $U$. If $T$ is a triangle in $U$ and $\gamma$ is some smooth parametrization of that triangle, then $\int_{\gamma} f = 0$.

My question: Is this no longer true if I replace holomorphic with differentiable?

The proof I am reading proceeds by dividing $T$ into quarter triangles, and arguing that the integral over $T$ must be bounded above by 4 times the integral of one of the quarter triangles. The argument proceeds by induction to show that this bound forces the integral over $T$ to vanish.

There seems to be only one place where the assumption that $f$ is holomorphic is used - namely, we use it to write $f(z) = f(z_0) + f'(z_0)h + h \phi(h)$, with $h = (z - z_0)$ $\phi(h)$ converging to zero as $h \to 0$. Let $T^n$ be the $nth$ triangle chosen in the typical method of proof. $p^n$ denotes its perimeter, and $d^n$ the diameter.

Supposing that $f$ was merely $R^2$ differentiable, I can create a similar expression: $f(z) = f(z_0) + f'(z_0)(h) + |h| \phi(h)$, and conclude that $\int_{T^n} f(z) = \int_{T^n} (f(z_0) + f'(z_0)(h)) + \int_{T^n} (|h| \phi(h))$.

I want to claim that $(f(z_0) + f'(z_0)(h))$ has a primitive since it is the sum of a constant function and a linear function (however, I think that the constant part has no primitive, since it is a vector and not scalar, so it is not the derivitive of $"f(z_0) x"$ - this is wrong, see below), so that we are left with $\int_{T^n} f(z) = \int_{T^n} (|h| \phi(h)) \leq \sup (|h| \phi(h)) p^{(n)}$ ($p^n$ is the perimeter of $T^n$).

Since $\sup |h| \phi(h) \leq d^n \sup \phi(h)$, it seems like the same exact proof goes through as in the holomorphic case.

Can someone help me find my mistake?

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I wonder, is the primitive really a primitive as you claim? Would be nice to see more detail there. – James S. Cook Jun 14 '14 at 3:18
Hmm, my comment regarding typos didn't get entered. The most serious notational issue is the $f'(h)$. The proof I address below. – Ted Shifrin Jun 14 '14 at 3:30
@JamesS.Cook I think the constant part has no primitive, since the function $f'(z_0) x$ is not in general defined as $f'(z_0)$ is a vector and not a scalar, and $x$ takes vector values. I'm not sure how to prove that it has no primitive, but at least my naive assumption is wrong. – AreaMan Jun 14 '14 at 3:41
No problem with constants. It's the linear part that's the issue. See my computation. – Ted Shifrin Jun 14 '14 at 3:52
up vote 4 down vote accepted

When you have a statement you know is incorrect, but you think you have a proof, it is always a good idea to see what happens when you apply your proof to an explicit counterexample. In this way, you should be able to find the mistake or misconception.

In this case, one of the easiest counterexamples is $f(z) = \overline{z}$ (the complex conjugate). Take the triangle to be the real axis from $0$ to $1$, the diagonal line from $1$ to $i$, and then the imaginary axis from $i$ to $0$. The integral should have value equal to $i$ (if I haven't miscalculated), rather than $0$. Now try to see where your argument goes wrong!

Added: for this kind of question, it is helpful to think of the domain copy of $\mathbb C$ as being $\mathbb R^2$ (so depending on two parameters), but the target copy of $\mathbb C$ as just being $\mathbb C$, a field of scalars.

In other words, we think of complex valued functions of two real variables.

Now we usually work in terms of $\partial_x$ and $\partial_y$ when studying functions on $\mathbb R^2$, but when the functions are complex valued, we can instead work in terms of $\partial_z := \dfrac{1}{2}(\partial_x - i \partial_y)$ and $\partial_{\overline{z}} := \dfrac{1}{2}(\partial_x + i \partial_y).$

If $F$ is any (sufficiently differentiable) function, and $\gamma$ is a path between two complex numbers $a$ and $b$, we then find that $$\int_{\gamma} \partial_z F(z) d z + \int_{\gamma} \partial_{\overline{z}} F(z) d\overline{z} = F(b) - F(a).$$ You can check this for yourself; it is just a rewriting of the (probably) more familiar formula $$ \int_{\gamma} \partial_x F(z) dx + \int_{\gamma} \partial_y F(z) dy = F(b) - F(a).$$

When you are looking for a "primitive" to compute a line integral $\int_{\gamma} f(z) dz,$ then, what you are trying to do is to write $f(z)$ as $\partial_z F(z)$ for some $F(z)$ which satisfies $\partial_{\overline{z}} F = 0$ identically, and then apply the above formula.

This isn't always possible, anymore then it is possible to always write a function of two variables as $\partial_x$ of some function whose $\partial_y$ vanishes.

Indeed, the Cauchy--Riemann equations can be expressed as saying that $\partial_{\overline{z}} F = 0$ at every point. You can further check that, for a holomorphic functions, $\partial_z$ is just the usual complex derivative.

So if you can find a "primitive" $F$for $f$ as above, then $F$ is holomorphic, and hence so is $f$ (being the derivative of a holomorphic function). So you can find a primitive, and hence get integral zero, precisely in the holomorphic case.

The discussion is perhaps easier when expressed in terms of differential forms, if you know that language. Then we have the formula, for smooth functions on $\mathbb R^2$:
$$\int_{\gamma} dF = F(b) - F(a)$$ (the one-dimensional form of Stokes's theorem, which I guess classically is just the fundamental theorem of calculus), and we can write $dF$ either as $\partial_x F dx + \partial_y F dy,$ or instead as $\partial_z F dz + \partial_{\overline{z}}F d\overline{z}$. (It is just a question of choosing $dz$ and $d\overline{z}$ as a basis for the complexified one-forms, rather than $dx$ and $dy$.)

We also have the formula, for a closed curve $\gamma$ bounding a region $\Delta$, that $$\int_{\gamma} f dz = \int_{\Delta} d(f dz) = \int_{\Delta} \partial_{\overline{z}} f d\overline{z} \wedge dz$$ (the two-dimensional form of Stokes's theorem, I guess classically known as Green's theorem in the plane). Now one computes $d \overline{z} \wedge dz = 2i dx \wedge dy,$ so we find that $$\int_{\gamma} f dz = 2i \int_{\Delta} \partial_{\overline{z}} f dx \wedge dy,$$ where $dx \wedge dy$ is the usual area form.

If $f$ is holomorphic, then $\partial_{\overline{z}} f = 0,$ and so this integral vanishes. But if e.g. $f = \overline{z}$, then this partial derivative equals $1$, so we are just computing $2i$ times the area of $\Delta$. (This is one way to get the answer of $i$ for the line integral around a triangle I described above.)

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Thank you so much for taking the time to write this out. This is very helpful. – AreaMan Jun 14 '14 at 9:55

We can write $f(z) = f(z_0) + f'(z_0)h + h \phi(h)$, with $h = (z - z_0)$ and $\phi(h) \to 0$ as $h \to 0$ exactly when $f$ is holomorphic (that is, complex differentiable).

When $f$ is just $\mathbb R^2$ differentiable you get $f(z) = f(z_0) + \nabla f(z_0) \cdot (h_x,h_y) + \cdots$, which is not the same thing.

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When $f$ is $R^2$ differentiable you can write $f(z) = f(z_0) + f'(h) + |h| \phi(h)$, with $\phi(h) \to 0$ as $|h| \to 0$. This is almost the same thing (h is replaced by $|h|$, since vector addition is not defined), and I don't think that it affects the estimation of the integral. So I don't think that this is were the issue is. – AreaMan Jun 14 '14 at 2:49
Again, there are notational issues here. But, yes, the definition of differentiability for a map $\Bbb R^2\to\Bbb R^2$ is that $f(z) = f(z_0) + Df(z_0)(z-z_0) + \varepsilon (z)$, where $\varepsilon(z)/|z-z_0|\to 0$ as $z\to z_0$. – Ted Shifrin Jun 14 '14 at 3:05
@TedShifrin: Dear Ted, If you were going to write a smooth function in terms of complex coordinates, I would have expected to see a $\overline{z}$-term appearing in the linear approximation as well, so I think I'm a bit confused by your comment. Best wishes, – Matt E Jun 14 '14 at 3:15
@user54092: Dear user, Your description of differentiability is not correct. You are looking at a function from $\mathbb R^2$ to $\mathbb C$, and so the linear term will involve both the $x$ and $y$-coord. of $h$ (as lhf indicates). Regards, – Matt E Jun 14 '14 at 3:18
@Matt: I admit I was being lazy and writing $z$ for $(x,y)$. These are all vectors and $Df(z_0)$ is a $2\times 2$ matrix. – Ted Shifrin Jun 14 '14 at 3:27

Remember that $f=u+iv$ is a $\Bbb C$-valued function and we are considering the line integral $$\int_\gamma f(z)\,dz = \int_\gamma (u+iv)(dx+i\,dy)= \int_\gamma (u\,dx - v\,dy)+i(v\,dx+u\,dy).$$ When you get down to the linear approximation of a differentiable function $(u,v)\colon\Bbb R^2\to\Bbb R^2$, you will have something resembling $$\int_\gamma \big((Ax + By) + i(Cx+Dy)\big)\, dx + i\big((Ax + By) + i(Cx+Dy)\big)\, dy\,,$$ which won't have a primitive (and hence won't have integral $0$) unless $B+iD=iA-C$. This looks a lot like the Cauchy-Riemann equations.

ADDED: Where did $B+iD=iA-C$ come from? $\displaystyle\int_\gamma ax\,dx + by\,dy = 0$ for all $a,b$ and closed curves $\gamma$. This follows from Green's Theorem or from noticing that $F(x,y)=\frac12(ax^2+by^2)$ is a primitive. But what about $\displaystyle\int_\gamma ay\,dx + bx\,dy$? Again, Green's Theorem tells us that when $\gamma$ bounds a region $R$, this integral is $\displaystyle\iint_R (b-a)\,dxdy$. Thus, if we want this integral to be $0$ for all such $\gamma$, we need $a=b$. Alternatively, we can see that $a=b$ is necessary and sufficient for us to have a (smooth) function $F$ with $\displaystyle\frac{\partial F}{\partial x} = ay$ and $\displaystyle\frac{\partial F}{\partial y} = by$, inasmuch as $b=\displaystyle\frac{\partial^2 F}{\partial x\partial y} = \displaystyle\frac{\partial^2 F}{\partial y\partial x}=a$.

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I see why there is an answer here, but I am absolutely confused about how to evaluate that second line integral in order to get the result $B + iD = iA - C$. I'm not sure what to do with $\int_{\gamma} x dx$ or $\int_{\gamma} y dx$... I'm not sure how to interpret these integrals, really. Is the first a path integral on x axis with values just the x coordinate of $\gamma$? That would give \int_0^1 x(\gamma(t)) \gamma'(t) dt = \int d/dt (x(\gamma(t)) dt = x\gamma(1) - x\gamma(0) = 0... the other would be the same, giving a (useless) total of 0. So I'm certainly misinterpreting something... – AreaMan Jun 14 '14 at 4:40
(Thank you for your help, by the way.) – AreaMan Jun 14 '14 at 4:40
I've edited, but it does sound like you need to learn something about line integrals and Green's Theorem. :) And, no, $\int_\gamma y\,dx$ is never $0$ for a closed curve $\gamma$, unless the curve is just a point. – Ted Shifrin Jun 14 '14 at 11:29
I really do - is there a resource you can recommend? I know enough to make sense of your manipulations, at least formally, so I can see how you arrive at that equation. – AreaMan Jun 14 '14 at 18:28
A standard multivariable calculus book should do. If you want more rigor, my profile has a link to my course from this spring with YouTube videos (this includes integration, line integrals, Green's Theorem, etc. — the first half of the course should get videoed this fall). – Ted Shifrin Jun 14 '14 at 18:40

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