The proof that L_{2}([a,b]) is an inner product space.

Question

In the following picture :

1-why we need the property that $|f|^2$ is Lebesgue integrable for the given space to be a vector space, could anyone explain this for me please?

2-And why we identify functions which are equal almost everywhere, could anyone explain this for me please?

Answer 1

You don't need these conditions for the set of complex-valued functions on $[a,b]$ to be a vector space. You can add and scale any such functions without a problem.

You need these conditions for the $L^2$-norm of a function $$ \Vert f \Vert = \sqrt{\left<f,f\right>} = \left(\int_a^b |f|^2 \,dx\right)^{1/2} $$ to be an actual norm. In particular, $\Vert f \Vert$ must be a nonnegative real number for any $f$, and $\Vert f \Vert = 0 \iff f = 0$.

1. If $|f|^2$ isn't integrable, the right-hand side of the above is undefined.
2. Unless we count functions that agree almost everywhere as identical, there are plenty of nonzero functions with a norm of zero. For instance, a function which is zero except at one point will have an integral of zero.

Answer 2

If $f$ and $g$, both measurable, have the property that $|f|^2$ and $|g|^2$ are L- integrable, then (Hölder !):

$f \overline{g} \in L_1$.

Hence we can define an inner product on $L_2$ as above.

Next we want is that

$\Vert f \Vert = \sqrt{\left<f,f\right>} = \left(\int_a^b |f|^2 \,dx\right)^{1/2}$

defines a norm on $L_2$.

But $\Vert f \Vert =0$ "only " gives $f=0$ a.e. If we identify functions which are equal a.e. , then we have

$\Vert f-g \Vert =0 \iff f=g$