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,

$⟨f,g⟩=\underset{E}{\int }fg$

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 $\overrightarrow{0}$ 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 $\overrightarrow{v}\in V$, we have $\Vert \overrightarrow{v}{\Vert }_2\ge 0$.

Also show that $\Vert \overrightarrow{0}{\Vert }_2=0$.

Now consider proving that if $\Vert \overrightarrow{v}\Vert =0$ then $\overrightarrow{v}=\overrightarrow{0}$. First show that, in fact, there is a function $f\in {ℒ}^2(E)$ for which $\underset{E}{\int }{f}^2=0$ 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 $\Vert \overrightarrow{v}{\Vert }_2=0$ then as an equality of equivalence classes, it follows that $\overrightarrow{v}=\overrightarrow{0}$.

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 $\Vert \overrightarrow{v}{\Vert }_2=⟨\overrightarrow{v},\overrightarrow{v}⟩$.

Metric Space

Prove that every normed space is a metric space, and infer that $L^2(E)$ is a metric space with metric $d(\overrightarrow{v},\overrightarrow{w})=\Vert \overrightarrow{v}-\overrightarrow{w}{\Vert }_2$.