Advanced Analysis

The Plancherel Formula

Theorem (The Plancherel Formula)
Suppose \(f\) and \(\hat{f}\) are both continuous and absolutely improperly Riemann integrable on \({\mathbb R}^n\). Then
\[ \int |f(x)|^2 dx = \int |\hat{f}(\xi)|^2 d \xi. \]
Note
As before, the assumption of continuity actually follows by Riemann-Lebesgue; it's merely convenient to emphasize that continuity holds and will be used.
Lemma (Multiplication Formula)
Suppose \(f\) and \(g\) are improperly Riemann integrable on \({\mathbb R}^n\). Then
\[ \int \hat{f}(x) g(x) dx = \int f(y) \hat{g}(y) dy. \]
In particular, both integrals exist in the improper Riemann sense.
Proof
Meta (Main Idea)
We've already seen one sketch of this result. Here's a second one using a somewhat different approach. For convenience, we restrict attention to the one-dimensional case.

Improper integrability of \(|f|\) and \(|g|\) imply that \(\hat{f}\) and \(\hat{g}\) are bounded functions, which gives existence of the two sides of the multiplication formula using what are now for us standard techniques.

Consider the left-hand side. For any \(\epsilon\), for all \(R\) sufficiently large, the quantity
\[{}\int_{-\infty}^\infty \hat{f}(x) g(x) dx{}\]
\[{}- \int_{-R}^R \hat{f}(x) g(x) dx{}\]
has magnitude at most \(\epsilon / 2\). Likewise, for all \(R\) sufficiently large,
\[ \hat{f}(\xi) - \int_{-R}^R e^{-2 \pi i y \xi} f(y) dy \]
has magnitude at most \(\epsilon/(2 + 2 \int |g|)\). Combining observations means that
\[{}\int_{-\infty}^\infty \hat{f}(x) g(x) dx{}\]
\[{}- \int_{-R}^R \left[ \int_{-R}^R e^{- 2 \pi i y \xi} f(y) dy\right] g(x) dx{}\]
has magnitude at most \(\epsilon\). Now use Fubini's theorem to switch the order of the integrals and use symmetry to see that both sides of the multiplication formula differ by at most \(2 \epsilon\) for some arbitrary \(\epsilon > 0\).
Proof (Proof of Plancherel)
Observe that if \(g(x) = \overline{\hat{f}(x)}\), then
\[{}\hat{g}(y){}\]
\[{}= \int e^{-2 \pi i y \cdot x} g(x) dx{}\]
\[{}= \int e^{-2 \pi i y \cdot x} \overline{\hat{f}(x)} dx{}\]
\[{}= \overline{\int e^{2 \pi i y \cdot x} \hat{f}(x) dx}{}\]
\[{}= \overline{f(y)}{}\]
thanks to the Fourier Inversion Formula applied to \(f\) (which is how the last line is derived from the line above it). Substituting this identity into the Multiplication Formula applied to \(f\) and \(g\) gives Plancherel.
Meta
The Plancherel formula is the continuous analogue of the Parseval identity for
Fourier series. It is one of the most important identities in analysis.
Corollary
Suppose \(f,g\) and \(\hat{f},\hat{g}\) are all continuous and absolutely improperly Riemann integrable on \({\mathbb R}^n\). Then
\[ \int f(x) \overline{g(x)} dx = \int \hat{f}(\xi) \overline{\hat{g}(\xi)} d \xi. \]
In other words, there is a natural inner product space such that the Fourier transform is well-defined on it and preserves orthogonality of functions.
Proof
Meta (Main Idea)
Use the polarization identity:
\[ f \overline{g} = \frac{1}{4} \sum_{k=1}^4 i^k |f + i^k g|^2\]
and apply Plancherel to the functions \(f + i^k g\).
Corollary (Uncertainty Principle)
If \(f\) is a Schwartz function on the real line,
\[{}\left( \int_{-\infty}^\infty x^2 |f(x)|^2 dx \right) \left( \int_{-\infty}^\infty \xi^2 |\hat{f}(\xi)|^2 d \xi \right){}\]
\[{}\geq \frac{\left( \int_{-\infty}^\infty |f(x)|^2 dx \right)^2}{16 \pi^2}. {}\]
Equality is attained for any Gaussian function \(f\) centered at the origin.
Meta (Informal Interpretation)
Informally, the inequality means that \(f\) and \(\hat{f}\) cannot both be concentrated arbitrarily close to the origin. To see this, think of \(\int |f|^2\) as fixed; the integral \(\int x^2 |f(x)|^2\) will be large if much of the “mass” of the integral of \(|f|^2\) lies far from the origin. Conversely, if the mass is concentrated near the origin, the integral of \(\int x^2 |f(x)|^2\) can be small even if \(\int |f(x)|^2\) retains its fixed value.
Proof (Uncertainty Principle)

We use the identity
\[ f = (xf)' - x f' \]
to compute the integral of \(|f|^2\):
\[{}\int |f|^2{}\]
\[{}= \int f \overline{f}{}\]
\[{}= \int \left[ (xf)' - xf' \right] \overline{f}{}\]
\[{}= \int \left[ (xf)'\overline{f} - xf' \overline{f} \right]{}\]
\[{}= \int \left[ - xf \overline{f'} - \overline{xf} f' \right]{}\]
By the Cauchy-Schwarz inequality (which always holds in any inner product space), each term on the right-hand side has magnitude at most
\[ \left(\int |xf|^2 \right)^{1/2} \left( \int |f'|^2 \right)^{1/2}. \]
Thus
\[{}\left(\int |f|^2 \right)^2{}\]
\[{} \leq 4 \int |xf(x)|^2 dx \int |f'(x)|^2 dx.{}\]
Now the Fourier transform of \(f'\) is equal to
\[{}\int e^{-2 \pi i \xi y} f'(y) dy{}\]
\[{}= 2 \pi i \xi \int e^{-2 \pi i \xi y} f(y) dy{}\]
thanks to the integration by parts formula, so for any Schwartz function \(f\),
\[ \int |f'(x)|^2 dx = \int |2 \pi i \xi \hat{f}(\xi)|^2 d \xi. \]
We conclude that
\[{}\left( \int |f|^2 \right)^2{}\]
\[{}\leq 16 \pi^2 \int |xf(x)|^2 dx \int |\xi \hat{f}(\xi)|^2 d \xi{}\]
which is the desired inequality. Gaussians give equality simply because there is only one step of the proof which isn't an equality, namely, the Cauchy-Schwarz step. For Gaussian functions, \(xf\) and \(f'\) are both real and constant multiples of one another, so Cauchy-Schwarz is an equality in this case.