$L^2([0,1]^2, \mathbb{R}^2)$ inner product definition

Question

Given $L^2([0,1], \mathbb{R})$, we have

$\langle f,g\rangle = \int_0^1 f(x) g(x) dx$.

Now, given $L^2([0,1]^2, \mathbb{R}^2)$,

is

$\langle X,Y\rangle = \int_0^1 \int_0^1 X^1(x,y)Y^1(x,y) + X^2(x,y) Y^2(x,y) \; dx dy $

where $X = (X^1(x,y), X^2(x,y))$ and $Y = (Y^1(x,y), Y^2(x,y))$

the standard definition ?

Answer 1

Indeed, this is a special case of a Bochner space, where we take $\mathbb{R}^2$ with the standard Euclidean inner product $\langle (x_1, y_1), (x_2, y_2)\rangle=x_1x_2+y_1y_2.$ The generalization to $\mathbb{R}^n$ is precisely $$\langle f,g\rangle=\int\limits_{[0,1]^n}\sum\limits_{j=1}^nf_j(x)g_j(x)\, dx,\qquad f=(f_1,f_2,\cdots, f_n),\ g=(g_1,g_2,\cdots, g_n).$$ Note that the inner product within the integral is, once again, the standard Euclidean inner product (this time defined on $\mathbb{R}^n$). You can generate to the space $L^2(X,Y)$ when $X$ is a measure space and $Y$ is a Hilbert space. If the inner product on $Y$ is denoted $\langle \cdot,\cdot\rangle$ and the measure on $X$ is $\mu$, then we can form $L^2(X,Y)$ into a Hilbert space using the inner product $$(f,g)=\int \langle f(x),g(x)\rangle \, d\mu(x).$$