Measure Theory/L2 Inner Product Space

Inner Products
We have defined, in Lesson 0, the inner product for $$L^2(E)$$. However, when calling something by the name "inner product" we mean to communicate a few properties. We should ensure that our inner product,


 * $$\langle f,g\rangle=\int_E fg$$

deserves the name.

Besides justifying the name, out of a sense of principle, these properties will also be useful common manipulations in the proofs of the theorems that we care about -- the most important, for this section, being completeness.

Recall that you have already decided what $$\vec 0$$ is when the vector space in question is $$L^2(E)$$, in the previous lesson, Lesson 1.

Show that the $$L^2(E)$$ inner product is symmetric and linear. (This should be trivial.)

Also show that for every $$\vec v\in V$$, we have $$\|\vec v\|_2\ge 0$$.

Also show that $$\|\vec 0\|_2=0$$.

Now consider proving that if $$\|\vec v\|=0$$ then $$\vec v = \vec 0$$. First show that, in fact, there is a function $$f\in \mathcal L^2(E)$$ for which $$\int_E f^2 =0$$ even though $$f\ne 0$$.

Next recall that, technically, $$L^2(E)$$ is a set of equivalence classes. Recalling also a result from a previous section, prove that if $$\|\vec v\|_2 = 0$$ then as an equality of equivalence classes, it follows that $$\vec v=\vec 0$$.

Conclude that the $$L^2(E)$$ inner product, is an inner product on $$L^2(E)$$.

Normed Space
Show that every inner product space is a normed space, and infer that $$L^2(E)$$ is a normed space with norm $$\|\vec v\|_2=\langle \vec v,\vec v\rangle$$.

Metric Space
Prove that every normed space is a metric space, and infer that $$L^2(E)$$ is a metric space with metric $$d(\vec v,\vec w) = \|\vec v-\vec w\|_2$$.