The Inner Product Space of Square Lebesgue Integrable Functions

The Inner Product Space of Square Lebesgue Integrable Functions

Recall from the Inner Product Spaces Over the Field of Real Numbers page that if $V$ is a vector space over the field $\mathbb{R}$ then a function $(\cdot, \cdot) : V \times V \to \mathbb{R}$ is said to be an inner product on $V$ if the following properties are satisfied:

• (1) $(x, x) \geq 0$ for all $x \in V$ and $(x, x) = 0$ if and only if $x = 0$.
• (2) $(x + y, z) = (x, z) + (y, z)$ for all $x, y, z \in V$.
• (3) $(\alpha x, y) = \alpha (x, y)$ for all $x, y \in V$ and for all $\alpha \in \mathbb{R}$.
• (4) $(x, y) = (y, x)$ for all $x, y \in V$.

An vector space with an inner product defined on it is called an inner product space.

The main inner product space that we will be interested in will be described as follows. Let $I$ be any interval. Then $L^2(I)$ is the set of all square Lebesgue integrable functions on $I$, and $L^2(I)$ is indeed a vector space under the operations of function addition and scalar multiplication. This is because the set of all real-valued functions is a vector space, so $L^2(I)$ is a subset of this vector space, and it suffices to show that $L^2(I)$ is closed under addition, closed under scalar multiplication, and contains the $0$ element.

We have already seen that the sum of two square Lebesgue integrable functions on an interval $I$ is square Lebesgue integrable on $I$. Moreover, we also proved that any scalar multiple of a square Lebesgue integrable function on $I$ is square Lebesgue integrable on $I$. Moreover, it’s not hard to verify that the identity function $0$ is square Lebesgue integrable on $I$. Thus, $L^2(I)$ is a vector subspace of the set of all real-valued functions and is hence a vector space itself.

We now define an inner product on $L^2(I)$. For each $f, g \in L^2(I)$ let:

(1)
\begin{align} \quad (f, g) = \int_I f(x)g(x) \: dx \end{align}

We regard the functions $f$ and $g$ to be equivalently equal if $f = g$ almost everywhere on $I$. Let’s verify that this is indeed an inner product.

Let $f \in L^2(I)$. Then $f^2 \in L(I)$ (by definition), and $f^2(x) \geq 0$ for all $x \in I$. Therefore we see that:

(2)
\begin{align} (f, f) = \int_I f(x)f(x) \: dx = \int_I f^2(x) \: dx \geq 0 \end{align}

Moreover, we see that if $f = 0$ (or if $f(x) = 0$ almost everywhere on $I$ which we regard to be the same function as the zero function) then $(f, f) = 0$, so property (1) is satisfied.

Now let $f, g, h \in L^2(I)$. Then from the results on the Linearity of Sums and Scalar Multiples of Square Lebesgue Integrable Functions page, it is clear that:

(3)
\begin{align} \quad (f + g, h) = \int_I [f(x) + g(x)]h(x) \: dx = \int_I [f(x)h(x) + g(x)h(x)] \: dx = \int_I f(x)h(x) \: dx + \int_I g(x)h(x) \: dx = (f, h) + (g, h) \end{align}

So property (2) is satisfied. Let $f, g \in L^2(I)$ and let $\alpha \in \mathbb{R}$. Then clearly:

(4)
\begin{align} \quad (\alpha f, g) = \int_I \alpha f(x)g(x) \: dx = \alpha\int_I f(x)g(x) \: dx = \alpha(f, g) \end{align}

So property (3) is satisfied. Lastly, for all $f, g \in L^2(I)$ we have that:

(5)
\begin{align} \quad (f, g) = \int_I f(x)g(x) \: dx = \int_I g(x)f(x) \: dx = (g, f) \end{align}

So property (4) is satisfied. So indeed, $L^2(I)$ is an inner product space with the inner product defined above.