The standard inner product for $f(x,t),g(x,t) \in L^2([a,b]\times[c,d])$

Question

I know that the standard inner product for $f,g \in L^2([a,b],\mathbb{R})$ is $$\langle f,g\rangle=\int_a^bf(x)g(x)\,dx.$$

Question: All right,

For $f(x,t),g(x,t) \in L^2([a,b]\times[c,d],\mathbb{R})$

Is the standard inner product as follows $$\langle f,g\rangle=\int_c^d \int_a^b f(x,t)g(x,t)\,dx\,dt$$

Answer 1

That is correct, except that (1) What's that comma doing there?, and (2) the use of the language of measure theory, as in $\text{“}L^2(\cdots\cdots)\text{''}$, incites some comment about fastidious distinctions: \begin{align} & \int_c^d \left( \int_a^b f(x,t)g(x,t)\,dx\right)\,dt & & \text{an interated integral} \\[8pt] = {} & \iint\limits_{[a,b]\times[c,d]} f(x,t)g(x,t)\, d(x,t) & & \text{a double integral} \\[8pt] = {} & \int_a^b \left( \int_c^d f(x,t)g(x,t)\,dt\right)\,dx & & \text{an iterated integral} \end{align} Are these the same? There are functions of $(x,t)$ for which the two iterated integrals have different values and the double integral is undefined (as a Lebesgue integral). For example $(x,t)\mapsto (x^2-t^2)/((x^2+t^2)^2$ on $[0,1]^2.$

But Fubini's theorem says these three are equal if $(x,t)\mapsto f(x,t)g(x,t)$ is in $L^1([a,b]\times[c,d]).$ So is it?

Since the measure of $[a,b]\times[c,d]$ is finite, we have $L^2([a,b]\times[c,d]) \subseteq L^1([a,b]\times[c,d]).$ So is this function in $L^2([a,b]\times[c,d])\text{?}$

Here the Cauchy–Schwarz inequality is what tells you that the pointwise product of two functions in that space is in that space.

Answer 2

That is correct. (There is an extra comma in your display.)

For a general measure space $(X,\mu)$, the inner product for $L^2(X,\mathbb{C})$ is defined as $$ \langle f,g\rangle=\int_Xf\ \overline{g}\,d\mu \tag{1}\label{eq1} $$ where $\overline{g}$ means the conjugate of $g$.

For $L^2(X,\mathbb{R})$, where the functions taking real values, one can drop the bar in (\ref{eq1}).