I Norm of integral less than or equal to integral of norm of function

psie
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I have a question concerning the inequality $$\left|\int f\,\mathrm{d}\mu\right|\leq\int |f|\,\mathrm{d}\mu,$$where ##f## is complex-valued, measurable and integrable.
Let ##(E,\mathcal A)## be a measurable space equipped with a measure ##\mu##. If ##f:E\to\mathbb R## is integrable, then we have ##\left|\int f\,\mathrm{d}\mu\right|\leq\int |f|\,\mathrm{d}\mu##. If ##f:E\to\mathbb C## is integrable, Le Gall in his book Measure Theory, Probability and Stochastic Processes argues that (on page 29, bottom) the easiest way to obtain the inequality is by noticing $$\left|\int f\,\mathrm{d}\mu\right|=\sup_{a\in\mathbb C,|a|=1}a\cdot \int f\,\mathrm{d}\mu=\sup_{a\in\mathbb C,|a|=1}\int a\cdot f\,\mathrm{d}\mu,\tag1$$where ##a\cdot z## denotes the Euclidean scalar product on ##\mathbb C## identified with ##\mathbb R^2##. I wonder
  1. What is the author using in the first equality? Why does the second equality hold?
  2. How does one obtain the inequality from ##(1)##?
 
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Note the identity z_1 \cdot z_2 \equiv |z_1||z_2|\cos(\arg z_1 - \arg z_2).
1. The first equality follows from applying the above identity to z_1 = a = e^{i\psi} and z_2 = \int f\,d\mu. The second equality follows because for constant a, <br /> a \cdot \int f\,d\mu = \int a \cdot f\,d\mu as can be verified by writing both sides in terms of real and imaginary parts.

2. I think the idea is to show that <br /> \left| \int f\,d\mu \right| = \sup_{\psi\,\mathrm{constant}} \int e^{i\psi} \cdot f\,d\mu \leq \sup_{\psi\,\mathrm{variable}} \int e^{i\psi} \cdot f\,d\mu = \int |f|\,d\mu. (Naturally one must constrain \psi so that e^{i\psi} \cdot f is integrable.)
 
Good question. I don't know how to interpret the sup of a set of complex numbers or vectors in ##\mathbb {R}^2## (which are not ordered). Does '.' mean the dot product or complex multiplication?
But here is my best guess at what he means.
1) The first one is just rotating the integral around until it is positive real and equal to its norm. The second one is the linearity of integration.
2) Bringing sup inside the integral can only give larger values. And the sup equals the norm.
 
I hope you realize that this inequality is intuitively clear. There can be a lot of cancellation (full or partial) in the integral on the left side but no cancellation on the right side.
 
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Likes pasmith and psie
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.

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