# Equality of absolute values of complex integrals

It was pretty hard finding a short and precise title. Here is my problem:

The equation $$\bigg|\int_\gamma f(z)\text{d}z\bigg|\le\int_\gamma\big|f(z)||\text{d}z|$$ holds if $f$ is integrable (where $\gamma$ is a differentiable path $[0,1]\rightarrow\mathbb{C}$). However, in what cases does "$=$" apply? Some cases are clear; for example if $f(z)=$const, $\forall z\in {\rm Img}(\gamma)$.

Our second idea was that both sides are equal if (or even iff?) $f(z)$ stays within one quadrant of the complex plane for all points in ${\rm Img}(\gamma)$.

So basically, the question is, when does "$=$" apply and is our second idea correct? Use "if" or "iff" (which is if and only if). Is there a nice proof to whatever answer there is? (No need to print the full proof; the basic idea suffices)

EDIT: instead of using the path definition of the integral, here is the corrected version: $$\bigg|\int_0^1f(\gamma(t))\gamma'(t)\text{d}t\bigg|\le\int_0^1|f(\gamma(t))||\gamma'(t)|dt.$$

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I think on the right hand side of your inequality $dz$ should be replaced by $|dz|$. –  23rd Dec 14 '12 at 17:55
this was the inequation we proved in our analysis reading; how id $|\text{d}z|$ defined? I will edit my question. –  CBenni Dec 14 '12 at 17:58
$|dz|$ denotes the infinitesimal arc length, i.e. if $\gamma$ is parameterized as $z=\gamma(t)$, then $dz=\gamma'(t)dt$ and $|dz|=|\gamma'(t)|dt$ –  23rd Dec 14 '12 at 18:07
@richard ok, that makes sense to me ;) Back to the question, any idea? –  CBenni Dec 14 '12 at 18:08
I still need you to clarify in what sense $f$ is integrable? –  23rd Dec 14 '12 at 18:28
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It suffices to consider the following reduced question: if $g:[0,1]\to\mathbb{C}$ is Riemann/Lebesgue integrable, in the following inequality $$\left|\int_0^1g(t)dt\right|\le \int_0^1|g(t)|dt,\tag{1}$$ when does "$=$" holds?

To answer this question, let us recall how to deduce $(1)$. Since if $g$ is Riemann integrable on $[0,1]$, it is automatically Lebesgue integrable, so let us focus on the Lebesgue integrable case. Denote $I=\int_0^1g(t)dt$. If $I=0$, then $(1)$ is trivially true, and "$=$" holds if and only if $g=0$ almost everywhere. If $I\ne 0$, then there exists a unique $\alpha\in\mathbb{C}$ with $|\alpha|=1$, such that $I=|I|\alpha$. Therefore, $$|I|=\overline{\alpha}I=\mathrm{Re}(\overline{\alpha}I)=\mathrm{Re}\int_0^1\overline{\alpha}g(t)dt=\int_0^1\mathrm{Re}(\overline{\alpha}g(t))dt\le\int_0^1|g(t)|dt.$$ Note that for the last inequaltiy, "$=$" holds if and only if $\mathrm{Re}(\overline{\alpha}g(t))=|g(t)|$ for almost every $t\in[0,1]$, i.e. $g(t)=\alpha|g(t)|$ for almost every $t\in[0,1]$. To sum up, "$=$" in $(1)$ holds if and only if there exists $\alpha\in\mathbb{C}$ with $|\alpha|=1$, such that $g(t)=\alpha|g(t)|$ for almost every $t\in[0,1]$. To go back to your original question, simply let $g(t)=f(\gamma(t))\gamma'(t)$.

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I finally understood your proof ^.^ Heres the accept. –  CBenni Dec 16 '12 at 9:13
@CBenni: Thank you! Actually this is a classical proof which can be found in standard textbooks of complex analysis. For example, I just checked that it appeared in section 1.1, chapter 4 of Ahlfors' complex analysis. –  23rd Dec 16 '12 at 13:07
Well that amazes me to be honest. We asked both our professor as well as our assistant and even dozens of fellow students, noone knew an answer right away... –  CBenni Dec 16 '12 at 13:16
@CBenni: Really? Have you heard Minkowski inequality? In fact the inequality in your question is a very special case of the Minkowski's integral inequality. –  23rd Dec 16 '12 at 13:26
I havent (Until I read wisefools answer), but thank you nonetheless ;) –  CBenni Dec 16 '12 at 13:29

If you write $$\bigg|\int_0^1f(\gamma(t))\gamma'(t)\text{d}t\bigg|\le\int_0^1|f(\gamma(t))||\gamma'(t)|dt$$ then it should be clear that you are asking this other question: take $a,b:[0,1]\to\mathbb{R}$ and write $$A=\int_0^1a(t)dt\qquad B=\int_{0}^1b(t)dt\;,$$ when does one have $$\sqrt{A^2+B^2}=\int_0^1(a^2(t)+b^2(t))^{1/2}dt$$ ? By squaring you get $$A^2+B^2= \left(\int_{0}^{1}(a^2(t)+b^2(t))^{1/2}dt\right)^2=\|a^2+b^2\|_{1/2}$$ and then $$A^2+B^2=\left(\int_0^1 a(t)dt\right)^2+\left(\int_{0}^1b(t)dt\right)^2=\left(\int_0^1 \sqrt{a^2(t)}dt\right)^2+\left(\int_0^1 \sqrt{b^2(t)}dt\right)^2=\left(\int_{0}^{1}(a^2(t)+b^2(t))^{1/2}dt\right)^2$$ because $$A^2+B^2\le\|a^2\|_{1/2}+\|b^2\|_{1/2}\le\|a^2+b^2\|_{1/2}$$ which is the reverse Minkowski inequality for $p=1/2$ (Minkowski and Holder hold for $0<p<1$ with the opposite inequality). The only way to have equality in the first one is $a=\pm|a|$, $b=\pm|b|$ to hold for a.e. $t$;for the second, like in the usual Minkowsky inequality, is to have $a^2=\lambda^2 b^2$ for a fixed real $\lambda$, that is $a+ib=(\lambda + i)b$.

Now, in your setting this means that $f(\gamma(t))\gamma'(t)$ has to be of the form $e^{i\alpha}\phi(t)$, where $\alpha\in\mathbb{R}$ is fixed and $\phi:[0,1]\to\mathbb{R^+}$, i.e. $$f(\gamma(t))=\frac{e^{i\alpha}\phi(t)}{\gamma'(t)}\;.$$

If, wlog, you take $\gamma$ a $\mathcal{C}^1$ curve, parametrized with respect to arc length, then $|\gamma'(t)|=1$ and $\gamma'(t)=e^{i\beta(t)}$; then you have $$f(\gamma(t))=e^{i\alpha-i\beta(t)}\phi(t)$$ which, upon rescaling $f$ by a constant factor, can be reduced to $$f(\gamma(t))=e^{-i\beta(t)}\phi(t)$$ which is not as neat as before (errors always improve results!) but yields nonetheless some link with curvature: $$\frac{df}{dz}(\gamma(t))e^{i\beta(t)}=f'(\gamma(t))\gamma'(t)=\frac{d}{dt}f(\gamma(t))=-i\beta'(t)e^{-i\beta(t)}\phi(t)+e^{-i\beta(t)}\phi'(t)$$ that is $$\frac{df}{dz}(\gamma(t))=e^{-2i\beta(t)}(-i\beta'(t)\phi(t)+\phi'(t))$$ and $\beta'(t)$ is related to the curvature of $\gamma$.

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@CBenni: The correct form of the sixth line should be $\sqrt{A^2+B^2}=\int_0^1\sqrt{a^2(t)+b^2(t)}dt$. –  23rd Dec 14 '12 at 21:12
Whops, you are right, I'll correct it (and what follows, obviously). –  wisefool Dec 14 '12 at 21:34
For simplicity's sake define $$g(t)=f(\gamma(t))\gamma'(t)\tag{1}$$ Note that $$\int_0^1|g(t)|\,\mathrm{d}t =\sup_{|u(z)|=1}\mathrm{Re}\left(\int_0^1g(t)u(t)\,\mathrm{d}t\right)\tag{2}$$ and $$\left|\,\int_0^1g(t)\,\mathrm{d}t\,\right| =\sup_{\substack{|u(z)|=1\\u\text{ constant}}}\mathrm{Re}\left(\int_0^1g(t)u(t)\,\mathrm{d}t\right)\tag{3}$$ By the principle that the sup over a larger set is larger, $(2)\ge(3)$.
Suppose that $(3)\ge(2)$. Then, by the compactness of the unit circle, there is a $u_0\in\partial D$ so that for any function $u:[0,1]\mapsto\partial D$, we have $$\mathrm{Re}\left(\int_0^1g(t)u(t)\,\mathrm{d}t\right)\le\mathrm{Re}\left(\int_0^1g(t)u_0\,\mathrm{d}t\right)\tag{4}$$ Set $$u(t)=\left\{\begin{array}{} \frac{|g(t)|}{g(t)}&\text{when }g(t)\ne0\\ 1&\text{when }g(t)=0 \end{array}\right.\tag{5}$$ Then we get from $(4)$ that \begin{align} \int_0^1|g(t)|\,\mathrm{Re}\left(1-\frac1{u(t)}u_0\,\right)\,\mathrm{d}t &=\int_0^1\,\mathrm{Re}\left(\,g(t)u(t)-g(t)u_0\,\right)\,\mathrm{d}t\\ &\le0\tag{6} \end{align} However, since $\left|\frac1{u(t)}\right|=1$ and $|u_0|=1$, we have $\mathrm{Re}\left(1-\frac1{u(t)}u_0\,\right)\ge0$. Since $(6)$ represents the integral of two non-negative functions, yet is non-positive, we get that for almost all $t\in[0,1]$, either $g(t)=0$ or $u(t)=u_0$. Thus, $$g(t)=\frac{|g(t)|}{u_0}\tag{7}$$ Therefore, equality can hold only when $g(t)$ has the same argument for all $t\in[0,1]$.