Scalar product of complex valued square integrable functions

Question

So I was told that if $p,q \in \mathbb{L}^2 _\mathbb{C} [a,b]$ which is the set of all square integrable functions on the interval $[a,b]$ then: $$(p,q) = \int^b_a{dx\ p^*(x)q(x)}$$ is a scalar product. However, for this to be the case $(p,p) = 0 \leftrightarrow p =0$ should be satisfied. $(0,0) = 0$ is obvious but on $\mathbb{L}^2 _\mathbb{C}$ there can be other functions that can yield 0 upon integration (e.g. a function that takes a finite value at one point but is zero everywhere else), so how can this be true?

Answer 1

You'll need some measure theory for this. $L^2$ actually isn't a space of functions but a space of equivalence classes of functions. Two square-integral functions $f$, $g$ are defined to be equivalent if they differ only on a set of measure zero (If you don't know measure theory, think of this as a set with zero length/volume/area).

The function in your example differs from zero in one point only and since the measure of a one-point set is zero, your function is actually identical to zero in $L^2$.

More generally, if a function $p$ is nonzero on a set $A$ of positive measure, then $f = |p|^2 = p^* p$ is also nonzero on $A$. If you divide $A$ as to $A = \bigcup_{n \geq 1} A_n = \bigcup_{n \geq 1} \{x \in A | f(x) > \frac1n\}$ you get that one of the $A_n$ must have nonzero measure and therefore $\int_a^b f(x) \,\mathrm{d}x > \frac1n \mu(A_n) > 0$. (here $\mu$ denotes the measure). But this means that a function $p$ which is not equivalent to $0$ in $L^2$ also has a positive "scalar square" $(p, p)$, proving definiteness of the scalar product on $L^2$.

Answer 2

The set $\mathbb{L}^2[a, b]$ is defined as the quotient space $\mathcal{L}^2[a, b]$ / $\mathcal{N}$. Thus $\mathbb{L}^2$ is actually a set of equivalence classes. Here $\mathcal{N}$ denotes the set of all functions that are non-zero on a set with Lebesgue measure zero. $\mathcal{L}^2[a, b]$ denotes the set of all square-integrable functions (i.e. all measurable functions f such that $||f||_2^2 = \langle f, f\rangle = \int_a^b |f|^2(x) dx < \infty$.) So $\mathbb{L}^2$ is the set of all equivalence classes of functions that are equal almost everywhere (i.e. they differ only on a set with measure $0$).

Note that the above 2-norm on $\mathcal{L}^2[a, b]$ isn't an actual norm, as $\|f\|_2^2 = 0 \iff f \equiv 0$ is obviously not satisfied, similarly as how $\langle f, f \rangle = 0 \iff f \equiv 0$ is not satisfied on $\mathcal{L}^2[a, b]$. However, writing $f = 0$ for an $f \in \mathbb{L}^2$ means that $f$ is the equivalence class of the functions $f \in \mathcal{L}^2$ that are $0$ almost everywhere.

Note that, $\|f\|_2^2 = 0 \iff f = 0$ and $\langle f, f \rangle = 0 \iff f = 0$ are satisfied on $\mathbb{L}^2$ (contrary to $\mathcal{L}^2$), since the Integral of $|f|^2$ is $0$ if and only if $f$ is zero almost everywhere (i.e. if $f$ is the equivalence class of functions that are zero almost everywhere).

EDIT: Some further reading (also for more general $\mathbb{L}^2$ spaces than just $\mathbb{L}^2[a,b]$): https://en.wikipedia.org/wiki/Lp_space#Lp_spaces