# Measure Theory/Generalizing
## For a General p

Early after the start of the study of $L^{2}(E)$ spaces, people realized that we could generalize 2 to any real number $1\leq p$ without too much difficulty. This doesn't have any obvious value at this point, and as far as I can tell, mathematicians initially studied this generalization because it was relatively easy and because mathematicians like generality.
If that were the only motivation, then I would have omitted this subject from the course. However, later on it does turn out to have useful applications to other topics. In particular, when we generalize from length-measure to abstract measure spaces, it will be valuable to have a general $L^{p}(E)$ space in mind. This is not the only application of $L^{p}(E)$ spaces that people have discovered, but it's the only one that our course will eventually discuss.
You'll notice the following definitions strongly parallel the $L^{2}$ definitions, with a glaring exception. There is no $L^{p}$ inner-product.
One other, smaller exception is the use of the absolute value. In order for these objects to be like a measure of size, we require them to be nonnegative.
Definition: L$^p$ norm, space, distance
Let $f:E\to {\mathbb {R}}$ be a function on $E\subseteq {\mathbb {R}}$. Define the $L^{p}(E)$ norm by
    $\|f\|_{p}=\left(\int _{E}|f|^{p}\right)^{1/p}$
Define the $L^{p}(E)$ space of functions as the set of functions with finite $L^{p}(E)$ norm.
    $L^{p}(E)=\{f|\ \|f\|_{p}<\infty \}$
Define the $L^{p}(E)$ distance metric as the norm of the difference of two functions. That is to say, if $f,g\in L^{p}(E)$ then
    $d(f,g)=\|f-g\|_{p}$
* Exercise 1. L$^p$ Is a Vector Space
* Prove that $L^{p}(E)$ is a vector space.
* The proof should be nearly identical to the proof that $L^{2}(E)$ is a vector space.

## L$^p$(E) Is a Normed Vector Space

Most of the proof that $L^{p}(E)$ is a normed vector space is easy and just like the proof that $L^{2}(E)$ is, but with one very glaring and in fact very difficult exception: The triangle inequality for $\|\cdot \|_{p}$.
The way that we proved the triangle inequality for $L^{2}(E)$ simply does not apply here because we made essential use of the fact that $L^{2}(E)$ is an inner-product space, and all inner product spaces are normed spaces.
Therefore in this setting we need a more direct proof that $\|\cdot \|_{p}$ is a norm, and in particular that it satisfies the triangle inequality.
    $\|f+g\|_{p}\leq \|f\|_{p}+\|g\|_{p}$ for $f,g\in L^{p}(E)$
Let us focus for now on $\|f+g\|_{p}^{p}=\int _{E}|f+g|^{p}$ so that we ignore the fractional exponent, which we may account for later. In a somewhat desperate attempt to apply a triangle inequality-like move, we may split off a single factor of $|f+g|$.
    $\int _{E}|f+g||f+g|^{p-1}\leq \int _{E}(|f|+|g|)|f+g|^{p-1}=\int _{E}|f||f+g|^{p-1}+\int _{E}|g||f+g|^{p-1}$
From here we hope to find some kind of ability to integrate a product -- in particular, the products $|f||f+g|^{p-1}$ and $|g||f+g|^{p-1}$.
We may find it more general, natural, and simple to investigate the integral of any product $\int _{E}|fg|$, especially relating this to the $L^{p}(E)$ norm. When we find this relationship below, we will call it the "p norm product bound".
In order to obtain some such bound, it would be helpful if we could more directly bound $|fg|$ -- because the p norm has an exponent of p then we will be perfectly happy if the bound that we find also has an exponent of p.
When we obtain the bound on $|fg|$ we will call it the "generalized AM-GM".

### AM-GM

A famous fact, which one might discover from an entirely different set of mathematical interests and considerations, is the AM-GM inequality, which states that for any real numbers $a,b\in {\mathbb {R}}$, their geometric mean is never more than their arithmetic mean.
    ${\sqrt {ab}}\leq {\frac {a+b}{2}}$
There is a nice elementary proof from the formula for the square of a sum.
* Exercise 2. Prove the AM-GM
* Prove the AM-GM inequality.
* Hint: Any square is nonnegative.
* Exercise 3. Reformulate the AM-GM
* Use the AM-GM inequality to show that
* $ab\leq {\frac {a^{2}+b^{2}}{2}}$
* Hint: Reassign a to $a^{2}$ and so on.
One should suspect that there is a generalization or two which we could use to extend the AM-GM inequality. We could extend it to more terms, but that is not really our interest here. We're more interested in exponents of p rather than 2.
A generalization of the arithmetic mean is the weighted average. If we have numbers $a,b$ and a weighting $w_{1},w_{2}\in {\mathbb {R}}$ then this means that $0\leq w_{1},w_{2}$ and $w_{1}+w_{2}=1$. The weighted average is then defined by $w_{1}a+w_{2}b$.
A generalization of the geometric mean is $a^{w_{1}}b^{w_{2}}$ and then the natural conjecture for a generalized AM-GM is
    $a^{w_{1}}b^{w_{2}}\leq w_{1}a+w_{2}b$
The back-and-forth between the product and sum is evocative of the logarithm, which turns a product into a sum. The desired inequality is equivalent to
    $\ln(a^{w_{1}}b^{w_{2}})\leq \ln(w_{1}a+w_{2}b)$
which is the same as
    $w_{1}\ln a+w_{2}\ln b\leq \ln(w_{1}a+w_{2}b)$
This now suggests a very important realization. This looks exactly like the kind of thing that geometers say about the convexity of shapes.

## Summary of This Lesson

The plan for this lesson is to:
1. Investigate convexity in order to prove the weighted AM-GM inequality.
2. Use the weighted AM-GM inequality to obtain a bound on the product of any real numbers.
3. Use the bound on the product of real numbers to obtain the p norm product bound.
4. Use the p norm product bound to prove that $\|\cdot \|_{p}$ is a norm.
5. Prove that $L^{p}(E)$ is a complete metric space.