Uniqueness of the Plancherel transform

Question

The first sentence in Note says that (a) and (b) together determine the mapping uniquely.

My understanding is this. By (c), any mapping satisfying the theorem is a linear transformation from $L^2$ onto $L^2.$ Then, by (b), the operator norm of any such mapping is 1, so that it is bounded. Therefore, any such mapping is continuous on $L^2.$ Then, we already know from (a) that any two mappings satisfying the theorem coincide on a denset subset of $L^2, $ which together with continuity implies that any such mappings coincide on $L^2$ itself. Uniqueness therefore follows. But, here (c) is used, as opposed to the statement invoking only (a) and (b). But I am feeling that (c) must be used to deduce the statement. Any comment on my understanding will be appreciated. Thanks.

Answer 1

There is already a detailed answer posted to your question. There is essentially nothing new to add, but as you have not accepted the previous answer I thought it might be helpful to elaborate a little further.

I would read the theorem as follows. For any $f\in L^1,$ we have a formula defining $\hat{f},$ that is, $$\hat{f}(y):=\int f(x)e^{-ix\cdot y}dx.$$

So far so good. Now if we assume that $f$ is also in $L^2,$ then it is not hard to show that $\hat{f}\in L^2.$ But there is one little caveat, namely, the above formula does not make sense for an arbitrary function $f\in L^2.$ But, this suggests us to work with $L^1\cap L^2.$ The great thing here is that we have an explicit formula defining $\hat{f}$ and the mapping $f\mapsto \hat{f}$ satisfies that $||\hat{f}||_2=||f||_{2}.$

Note that the map $f\mapsto \hat{f}$ is defined only on $L^1\cap L^2.$ But we realize that $L^1\cap L^2$ is a dense subspace of $L^2.$ This allows us to extend the map $f\mapsto \hat{f}$ to $L^2.$ Why so? Take an $f\in L^2,$ and obtain a sequence $f_n\in L^1\cap L^2$ such that $f_n \to f$ in $L^2.$ Using the fact that $||\hat{f_n}||_2=||f_n||_2,$ we obtain that $\hat{f_n}$ is Cauchy in $L^2$ and therefore has a limit, say $g.$ We set $\hat{f}:=g.$ One needs to pause here for a moment and think carefully why this assignment defines a mapping. In other words, if we take a difrent approximating sequence $h_n\to f$ in $L_2,$ we need to argue that $||\hat{h}_n-\hat{f_n}||_2\to 0.$

This is more-or-less the content of $a)$ and $b).$ There is one more question which $b)$ answers. On $L^1\cap L^2,$ we have a "god-given" definition of what $\hat{f}$ mean. Now, we extended this definition to $L^2.$ Is it possible that there is some way to extend this map to whole $L^2$ in some other way? The answer is, No! This is what the uniqueness means. As long as you want a continuous extension, this is the only way to do it. (But this fact has nothing to do with Fourier analysis. It is a very down-to-earth fact that is true in any metric space setting. Two continuous functions that agree on a dense set must agree everywhere.)

This completes $a)$ and $b).$ To answer one of your questions, you do not need $c)$ to deduce $a)$ and $b).$

Once we have obtained a Fourier transform $\mathcal{F}: L^2\to L^2,$ it is natural to ask what more we can say about it? From part $b)$ we already know that this map is an isometry, that is, $||f||_2=||\hat{f}||_2.$ The $c)$ part of the theorem says that it is an isometric isomorphic. Note that isometries are always injective. But an isometry may fail to be surjective. But the Fourier transform is not one such. It is an isomorphism. That is every $L^2$ function can be obtained as the Fourier transform of some $L^2$ function. (It is so good to know this, isn't it?)

Once you know from part $c)$ that the Fourier transform map can be inverted, the next (and I would say a little greedy) question to ask is the following:

Can we write a "formula" for the inverse of the Fourier transform? Note that we do have a formula for the Fourier transform (not for all $f\in L^2$) for at least a dense class of functions. So this is a reasonable ambition to have. Part $d)$ of the theorem essentially answers this question in (almost) an affirmative tone.

I hope this explanation helps.

Answer 2

I'm not exactly sure what the question is, but it seems like you're asking "why is this map unique?" You might also be asking "why does this map exist?" which is related, and you basically do the same proof but you need to do a little bit more work.

The general statement is this: Suppose that $X$, $Z$ are metric spaces, and that $Y \subset X$ and $Y$ is dense. Then, if $F,G: X \to Z$ are continuous and agree on $Y$, they agree everywhere.

Why? Suppose $x \in X$. Then, take a sequence $y_n \in Y$ with $y_n \to x$ (which we can do by the density of $Y$).

By continuity we have $F(x) = \lim_{n \to \infty} F(y_n)$ and $G(x) = \lim_{x \to \infty} G(y_n)$. But, $F(y_n) = G(y_n)$, hence $\lim_{n \to \infty} F(y_n) = \lim_{n \to \infty} G(y_n)$. So, $F(x) = G(x)$ for all $x \in X$.

This is basically the proof you have above. You just need to employ the fact that $\Lambda_1$ and $\Lambda_2$ are assumed continuous.