Question on Parseval's Theorem and Plancherel’s formula

Question

I've come across Parseval's theorem and Plancherel’s formula several times on this forum. Each time they're referenced they're mentioned in regards to inner products in general. However, every proof I find of them deals specifically with inner products of the form $$\langle f,g\rangle = \int_{- \infty}^{\infty} f(x)g(x)dx$$

I'm attempting to apply this theorem to a situation such as this: If $\bar{u}(x,y,z)$ is a field of 3-D vectors over $\mathbb{R}^3$ (by this I mean that at for every coordinate $(a,b,c) \in \mathbb{R}^3$, $\bar{u}(a,b,c)$ is a three dimensional vector in the Euclidian sense) and we define the inner product of two such fields, $\bar{u}(x,y,z)$ and $\bar{w}(x,y,z)$, to be $$<\bar{u}(x,y,z),\bar{w}(x,y,z)> = \phi(x,y,z)$$ Where $\phi(x,y,z)$ is a scalar field such that $$\phi(a,b,c) = \bar{u}(a,b,c) \cdot \bar{w}(a,b,c)$$ then can we apply Parseval's theorem to this inner product to say that $$<\bar{u}(x,y,z),\bar{w}(x,y,z)> = <\hat{u}(k_x,k_y,k_z),\hat{w}(k_x,k_y,k_z)>$$ where the hats indicate a three dimensional Fourier transform, and $k_x,k_y,k_z$ signify the Fourier units for $x,y,z$ respectively?

I feel like the answer is no since this isn't exactly an inner product due to the fact that it produces a scalar field rather than a scalar.

Furthermore, I think this serves as a counter example: Let $\nabla \times \bar{u}(x,y,z)$ be the curl of $\bar{u}(x,y,z)$. Then if this was true we could express the inner product of these two vector fields as $$<\bar{u},\nabla \times \bar{u}>=<\hat{u},i\bar{k}\times \hat{u}>$$ where i is the imaginary unit and $\bar{k}$ is the vector $(k_x,k_y,k_z)$ and $\times$ indicates a cross product. This inner product would then evaluate to $0$, even though the original expression is not necessarily zero.

Does anyone have any thoughts or insights on this though? I'm still not too sure and would love to hear some other opinions.If Parseval's theorem doesn't apply here, does anyone have any ideas of other theorems or things we could say about this?

Answer 1

Let $v = (v_0, v_1, v_2)$ and $w = (w_0, w_1, w_2)$. Consider the functions $$f_v(x) = \begin{cases} v_0 & \mbox{if } x \in [0,1) \\ v_1 & \mbox{if } x \in [1,2) \\ v_2 & \mbox{if } x \in [2,3) \\ 0 & \mbox{otherwise} \end{cases} $$ and $$f_w(x) = \begin{cases} w_0 & \mbox{if } x \in [0,1) \\ w_1 & \mbox{if } x \in [1,2) \\ w_2 & \mbox{if } x \in [2,3) \\ 0 & \mbox{otherwise} \end{cases} $$ Then $$\langle f_v, f_w \rangle = \int_{-\infty}^\infty f_v(x)f_w(x)\,dx = v_0w_0 + v_1w_1 + v_2 w_2 = \langle v,w \rangle.$$ In this sense, the inner product of vectors is a special case of the inner product of functions. (It is the special case where the functions are of the form $f_v$ and $f_w$.)

Answer 2

Ok, I think I see what you're saying. You have two vector fields $\vec{u}(\vec{x})$ and $\vec{w}(\vec{x})$, and you want to ask whether there is a relationship between $\vec{u}(\vec{x}) \cdot \vec{w}(\vec{x})$ and $\hat{u}(\vec{k}) \cdot \hat{w}(\vec{k})$, where the hats indicate Fourier transforms. The question makes sense to me now.

I think that the answer to this is almost certainly "no," i.e., there is no equality that can relate these two quantities. The first problem to discuss here is: Suppose that I want to find an equality relating $\vec{u}(\vec{x}) \cdot \vec{w}(\vec{x})$, for some particular value of $\vec{x}$, to $\hat{u}(\vec{k}) \cdot \hat{w}(\vec{k})$, for some particular value of $\vec{k}$. Given $\vec{x}$, which value of $\vec{k}$ should I use? In the inner product used in Plancherel's theorem, this isn't a problem since we integrate over $\vec{x}$ and $\vec{k}$. But in your case, you have to pick a particular value of $\vec{x}$ and $\vec{k}$, but there is no obvious way to pick a value.

Suppose that you come up with a resolution to this problem. For example, you could say that $\vec{x}$ and $\vec{k}$ are both in $\mathbb{R}^3$, so you just choose $\vec{k}=\vec{x}$. Then you will run into the following problem.

The value of the Fourier transform depends on very nonlocal properties of the function. The value of $\hat{u}(\vec{k})$ for any particular $\vec{k}$ depends on the value of $\vec{u}(\vec{x})$ for every single value of $\vec{x}$. Any equality that you come up with between a particular $\vec{x}$ and a particular $\vec{k}$ could be ruined by changing the values of the functions for different values of $\vec{x}$ or $\vec{k}$ that do not appear in your equality.

Another way of saying this is: If you give me a function where $\vec{u}(\vec{x})$ is specified at a given point $\vec{x}$, and you ask me what the value of $\hat{u}(\vec{k})$ is at a given point $\vec{k}$, the answer is: It could be anything. I can find some function $\vec{u}$ that will give me whatever answer I want.

While none of this amounts to a rigorous proof that such a relationship does not exist, it is a reasonably strong argument that you won't find anything useful along these lines.