Course 4 Book 4A: Function Spaces and Sobolev SpacesChapter 5

The Fourier Transform

Plancherel, Gaussians, uncertainty, and solving the heat equation by frequencies.

26 min read · Updated Oct 2, 2026

Neither Kreyszig nor Brezis covers the Fourier transform. Read Stein and Shakarchi, Fourier Analysis: An Introduction, chapters 5–6 (the Fourier transform on $\mathbb{R}$ and on $\mathbb{R}^d$ ), which uses the same normalisation as this guidebook; or Evans, Partial Differential Equations, §4.3.1, which uses a different one (see the note below).

In this chapter · 9 sections

5.1An image measured in frequency
5.2The Fourier transform on L1L^1L1
5.3The Gaussian
5.4Inversion and Plancherel
5.4.1Poisson summation
5.5The uncertainty principle
5.6Smoothness is decay
5.7The heat equation by Fourier transform
5.8History
5.9Exercises

2B.7 Fourier Series and the First Heat Equation broke periodic functions into frequencies $e^{2\pi ikx}$ , $k \in \mathbb{Z}$ , and solved the heat equation on a ring frequency by frequency. On $\mathbb{R}^n$ there is no period, so every real frequency $\xi \in \mathbb{R}^n$ is needed, and the Fourier series becomes the Fourier transform

\hat f(\xi) = \int_{\mathbb{R}^n}f(x)\,e^{-2\pi ix\cdot\xi}\,dx.

It turns differentiation into multiplication by $2\pi i\xi$ , so constant-coefficient differential equations become algebra; it turns convolution into multiplication; it is an isometry of $L^2$ (Plancherel); and it maps the Gaussian to itself. With it, the heat equation on $\mathbb{R}^n$ is solved in three lines, and "having derivatives in $L^2$ " becomes "having a Fourier transform that decays", the idea behind the Sobolev spaces of 4A.9 Sobolev Spaces.

Note

Normalisation. This guidebook uses $\hat f(\xi) = \int f(x)e^{-2\pi ix\cdot\xi}dx$ , as Stein and Shakarchi do (fixed in the front matter, P.5). With it, Plancherel and the inversion formula have no constants, and $e^{-\pi|x|^2}$ is its own transform. Evans uses $\hat u(y) = (2\pi)^{-n/2}\int u(x)e^{-ix\cdot y}dx$ ; to convert, set $y = 2\pi\xi$ , so his $\hat u(y)$ equals $(2\pi)^{-n/2}$ times ours at $\xi = y/2\pi$ . Physicists often omit the $2\pi$ in the exponent and put a $\frac{1}{2\pi}$ in front of the inverse transform.

By the end of this chapter you will be able to:

compute Fourier transforms, including the Gaussian's, and use the dictionary between operations on $f$ and on $\hat f$ ;
prove the inversion formula and Plancherel's theorem, and the Poisson summation formula;
state and prove the uncertainty principle;
measure smoothness by the decay of $\hat f$ , and prove that $H^s(\mathbb{R}^n) \subseteq C(\mathbb{R}^n)$ when $s > n/2$ ;
solve the heat equation on $\mathbb{R}^n$ by Fourier transform and recover the heat kernel.

An image measured in frequency

In the world In use MRI scanners sample the Fourier transform

A magnetic resonance imaging scanner doesn't photograph the body. In a strong magnetic field, hydrogen nuclei precess at a frequency proportional to the field strength. Adding a magnetic field gradient makes that frequency depend on position, so the signal received from a slice, in the simplest model (ignoring relaxation and noise), is

S(k) = \int\rho(x)\,e^{-2\pi ik\cdot x}\,dx = \hat\rho(k),

the Fourier transform of the density $\rho$ of hydrogen nuclei, at a spatial frequency $k$ that the scanner steers by switching the gradients. The scanner thus samples $\hat\rho$ on a grid in " $k$ -space", and the image is computed by an inverse discrete Fourier transform. Paul Lauterbur and Peter Mansfield shared the 2003 Nobel Prize in Physiology or Medicine for the development of MRI.

The mathematics of this chapter predicts what the images look like. Sampling $k$ -space on a grid of spacing $\Delta k$ makes the reconstructed image periodic with period $1/\Delta k$ , so if the field of view is too small, parts of the body outside it wrap around into the image (aliasing, the Poisson summation formula below). Sampling only low frequencies gives a blurred image; sampling only high frequencies keeps the edges and loses the overall brightness (Figure 5.1). And since the scan time is proportional to the number of samples, a large research effort goes into reconstructing images from fewer samples than the grid requires.

Figure 5.1. A synthetic phantom and reconstructions from parts of its Fourier transform (computed by a discrete Fourier transform on a

48 \times 48

grid). Keeping only the low frequencies (centre of

k

-space) gives the overall shapes, blurred; keeping only the high frequencies gives the edges. Smoothness lives in low frequencies, edges in high ones.

The Fourier transform on $L^1$

For $f \in L^1(\mathbb{R}^n)$ the integral defining $\hat f$ converges absolutely, so $|\hat f(\xi)| \leq \|f\|_1$ , and by dominated convergence $\hat f$ is continuous. More is true: $\hat f(\xi) \to 0$ as $|\xi| \to \infty$ (the Riemann–Lebesgue lemma). For the indicator of a box it is a direct computation; for general $f$ , approximate in $L^1$ by finite combinations of box indicators (3A.7 Lᵖ Spaces and Jensen’s Inequality) and use $|\hat f - \hat g| \leq \|f - g\|_1$ .

The transform turns structure on one side into different structure on the other. Each line is a change of variables or an integration by parts (Exercise 5.7):

$f(x)$	$\hat f(\xi)$
$f(x - a)$ (translate)	$e^{-2\pi ia\cdot\xi}\hat f(\xi)$ (modulate)
$f(\lambda x)$ , $\lambda > 0$ (dilate)	$\lambda^{-n}\hat f(\xi/\lambda)$
$\partial_jf(x)$ (differentiate)	$2\pi i\xi_j\,\hat f(\xi)$
$-2\pi ix_jf(x)$	$\partial_j\hat f(\xi)$
$(f * g)(x)$ (convolve)	$\hat f(\xi)\,\hat g(\xi)$ (multiply)

The third line is the reason the transform exists. A constant-coefficient differential operator $P(\partial)$ becomes multiplication by the polynomial $P(2\pi i\xi)$ , so the equation $P(\partial)u = f$ becomes $\hat u = \hat f/P(2\pi i\xi)$ , wherever that makes sense. The Laplacian becomes multiplication by $-4\pi^2|\xi|^2$ .

The second line is the scaling of thread S: concentrating $f$ (large $\lambda$ ) spreads $\hat f$ out by the same factor. A function and its transform can't both be concentrated, which is the uncertainty principle below.

The Gaussian

Proposition 5.1 The Gaussian is its own Fourier transform

For $g(x) = e^{-\pi|x|^2}$ on $\mathbb{R}^n$ , $\hat g = g$ . More generally, for $a > 0$ , the transform of $e^{-\pi a|x|^2}$ is $a^{-n/2}e^{-\pi|\xi|^2/a}$ .

Proof. By Tonelli and Fubini (3A.5 Product Measures and Change of Variables), $e^{-\pi|x|^2} = \prod_je^{-\pi x_j^2}$ and the transform factorises, so take $n = 1$ . Then $g' = -2\pi xg$ . Transform both sides using the dictionary: $2\pi i\xi\,\hat g = -i\,\hat g'$ , that is, $\hat g' = -2\pi\xi\,\hat g$ . So $\hat g$ satisfies the same differential equation as $g$ , and $\hat g(0) = \int e^{-\pi x^2}dx = 1 = g(0)$ (3A.5 Product Measures and Change of Variables). By uniqueness for $y' = -2\pi\xi y$ (2B.6 Power Series, Exponentials and Bump Functions), $\hat g = g$ . (Differentiating $\hat g$ under the integral sign is justified by domination by $|x|e^{-\pi x^2}$ , 3A.3 The Lebesgue Integral.) The general case follows by dilation with $\lambda = \sqrt a$ .

Figure 5.2. Gaussians

e^{-\pi a x^2}

for

a = 4, 1, \tfrac14

(left) and their Fourier transforms

a^{-1/2}e^{-\pi\xi^2/a}

(right). The narrower the function, the wider its transform. The case

a = 1

is its own transform.

Inversion and Plancherel

The natural setting is the Schwartz class $\mathcal{S}(\mathbb{R}^n)$ : smooth functions that, together with all their derivatives, decay faster than any power of $|x|$ . Gaussians and smooth compactly supported functions belong to it. By the dictionary, the Fourier transform maps $\mathcal{S}$ to itself: decay of $f$ gives smoothness of $\hat f$ , and smoothness of $f$ gives decay of $\hat f$ .

The key lemma moves the transform from one factor to another: for $f, g \in L^1$ ,

\int\hat f(\xi)\,g(\xi)\,d\xi = \int f(x)\,\hat g(x)\,dx \tag{$*$}

(both equal $\iint f(x)g(\xi)e^{-2\pi ix\cdot\xi}\,dx\,d\xi$ , by Fubini).

Theorem 5.2 Fourier inversion

For $f \in \mathcal{S}(\mathbb{R}^n)$ , $\;f(x) = \displaystyle\int\hat f(\xi)\,e^{2\pi ix\cdot\xi}\,d\xi$ .

Proof. Insert a Gaussian damping factor and remove it in the limit. For $\varepsilon > 0$ let $g_\varepsilon(\xi) = e^{2\pi ix\cdot\xi}e^{-\pi\varepsilon^2|\xi|^2}$ . By the Gaussian computation and the dictionary, $\hat g_\varepsilon(y) = \varepsilon^{-n}e^{-\pi|y - x|^2/\varepsilon^2} = K_\varepsilon(y - x)$ , a Gaussian approximate identity centred at $x$ (3A.8 Convolution and Mollifiers). By $(*)$ ,

\int\hat f(\xi)\,e^{2\pi ix\cdot\xi}e^{-\pi\varepsilon^2|\xi|^2}\,d\xi = \int f(y)\,K_\varepsilon(y - x)\,dy.

As $\varepsilon \to 0$ the left side tends to $\int\hat fe^{2\pi ix\cdot\xi}$ by dominated convergence ( $\hat f \in L^1$ ), and the right side to $f(x)$ , since $f$ is continuous and bounded.

So the inverse transform is the same operation with the sign of the exponent changed. Combined with $(*)$ and the conjugation rule $\widehat{\bar f} = \overline{\hat f(-\xi)}$ , inversion gives the most important property of the transform.

Theorem 5.3 Plancherel

For $f, g \in \mathcal{S}(\mathbb{R}^n)$ , $\;\int f\bar g\,dx = \int\hat f\,\overline{\hat g}\,d\xi$ , and in particular $\|\hat f\|_2 = \|f\|_2$ . The Fourier transform therefore extends uniquely to a bijective isometry of $L^2(\mathbb{R}^n)$ .

Proof. Apply $(*)$ with $g$ replaced by $\overline{\hat g}$ , whose transform is $\bar g$ by inversion. Then the transform is an isometry on the dense subspace $\mathcal{S}$ of $L^2$ (3A.8 Convolution and Mollifiers), so it extends by continuity to $L^2$ (Exercise 5.9); the inverse transform extends the same way, so the extension is onto.

For $f \in L^2$ that isn't integrable, $\hat f$ is defined as an $L^2$ limit, for instance $\hat f = \lim_{R\to\infty}\widehat{f1_{B(0,R)}}$ in $L^2$ , not by the integral formula. Plancherel is the continuous analogue of Parseval's identity for Fourier series (2B.7 Fourier Series and the First Heat Equation), and like it says that energy can be computed in either domain.

Poisson summation

The Fourier series of 2B.7 Fourier Series and the First Heat Equation and the transform of this chapter are linked by a formula that was promised there.

Theorem 5.4 Poisson summation formula

For $f \in \mathcal{S}(\mathbb{R})$ , $\;\displaystyle\sum_{n\in\mathbb{Z}}f(x + n) = \sum_{k\in\mathbb{Z}}\hat f(k)\,e^{2\pi ikx}$ , and in particular $\sum_nf(n) = \sum_k\hat f(k)$ .

Proof. The left side $F(x) = \sum_nf(x + n)$ converges uniformly with all derivatives (rapid decay), and is $1$ -periodic. Its Fourier coefficients are

\hat F(k) = \int_0^1\sum_nf(x + n)e^{-2\pi ikx}\,dx = \int_{\mathbb{R}}f(y)e^{-2\pi iky}\,dy = \hat f(k),

splitting $\mathbb{R}$ into the intervals $[n, n+1]$ . A smooth periodic function equals its Fourier series (2B.7 Fourier Series and the First Heat Equation).

Applied to the heat kernel $f(x) = (4\pi t)^{-1/2}e^{-x^2/4t}$ , whose transform is $e^{-4\pi^2\xi^2t}$ (Gaussian with $a = 1/(4\pi t)$ ), it gives

\sum_{n\in\mathbb{Z}}\frac{1}{\sqrt{4\pi t}}e^{-(x+n)^2/4t} = \sum_{k\in\mathbb{Z}}e^{-4\pi^2k^2t}e^{2\pi ikx}:

the periodic heat kernel of 2B.7 Fourier Series and the First Heat Equation is the Gaussian heat kernel of the line wrapped around the circle, and in particular it is positive. The same formula, in $k$ -space, explains MRI aliasing: sampling $\hat\rho$ at spacing $\Delta k$ reconstructs $\sum_n\rho(x + n/\Delta k)$ , the image plus its translates.

The uncertainty principle

Theorem 5.5 Heisenberg's uncertainty principle

For $f \in \mathcal{S}(\mathbb{R})$ ,

\|f\|_2^2 \leq 4\pi\,\|xf\|_2\,\|\xi\hat f\|_2,

with equality exactly for Gaussians $f(x) = Ce^{-ax^2}$ , $a > 0$ .

Proof. Integrate by parts: $\int|f|^2dx = \int x\cdot\frac{d}{dx}\big(-|f|^2\big)\,dx$ , so $\|f\|_2^2 = -2\operatorname{Re}\int x\bar ff'\,dx \leq 2\|xf\|_2\|f'\|_2$ by Cauchy–Schwarz. By Plancherel and the dictionary, $\|f'\|_2 = 2\pi\|\xi\hat f\|_2$ . Equality in Cauchy–Schwarz requires $f' = -cxf$ for a real $c > 0$ , which gives Gaussians.

If $|f|^2$ and $|\hat f|^2$ are thought of as probability densities (after normalising), $\|xf\|_2$ and $\|\xi\hat f\|_2$ measure their spreads about $0$ , and the inequality says the spreads can't both be small. In quantum mechanics, with $\hbar$ restoring the units, it is the uncertainty relation between position and momentum.

In the world Model A click has no pitch

A pure tone lasting a long time is a long wave train, and its transform is concentrated near one frequency: we hear a definite pitch. A short click, lasting a millisecond, has a transform spread over a band of width about a kilohertz, so it has no definite pitch: we hear a percussive sound. Musicians meet the same limit as a trade-off between rhythm and pitch: a note too short can't be heard as clearly in tune. Spectrogram software, which shows how the frequency content of a sound changes in time, has to choose a window length, and the uncertainty principle says it can't have fine resolution in time and in frequency at once. Dennis Gabor analysed this trade-off for communication signals in 1946 and showed that Gaussian-windowed wave packets achieve the minimum.

Smoothness is decay

The dictionary says $\widehat{\partial^\alpha f} = (2\pi i\xi)^\alpha\hat f$ , so by Plancherel, $f$ has $k$ derivatives in $L^2$ exactly when $(1 + |\xi|^2)^{k/2}\hat f \in L^2$ . That suggests measuring smoothness, even fractional smoothness, by decay:

\|f\|_{H^s}^2 = \int_{\mathbb{R}^n}(1 + |\xi|^2)^s\,|\hat f(\xi)|^2\,d\xi.

Proposition 5.6 Sobolev embedding, the Fourier way

If $s > n/2$ and $f \in H^s(\mathbb{R}^n)$ , then $\hat f \in L^1$ , and $f$ is (equal almost everywhere to) a bounded continuous function, with $\sup|f| \leq C_{n,s}\|f\|_{H^s}$ .

Proof. By Cauchy–Schwarz, $\int|\hat f| = \int(1 + |\xi|^2)^{-s/2}\cdot(1 + |\xi|^2)^{s/2}|\hat f| \leq \Big(\int(1 + |\xi|^2)^{-s}d\xi\Big)^{1/2}\|f\|_{H^s}$ , and the first integral converges exactly when $2s > n$ (polar coordinates, 3A.5 Product Measures and Change of Variables). Then $f$ is the inverse transform of an integrable function, which is bounded by $\|\hat f\|_1$ and continuous by dominated convergence.

The threshold $s > n/2$ is dimensional: in more dimensions, more derivatives are needed to guarantee continuity. That is the first instance of the scaling exponents of 4A.10 Sobolev Embeddings and Critical Exponents, where the same question is answered for $L^p$ -based spaces without Fourier methods.

The heat equation by Fourier transform

Solve $\partial_tu = \Delta u$ on $\mathbb{R}^n$ with $u(\cdot, 0) = f \in \mathcal{S}$ . Transforming in $x$ , the equation becomes, for each frequency $\xi$ separately, the ODE $\partial_t\hat u(\xi, t) = -4\pi^2|\xi|^2\hat u(\xi, t)$ , so

\hat u(\xi, t) = e^{-4\pi^2|\xi|^2t}\,\hat f(\xi).

By the Gaussian proposition (with $a = 1/(4\pi t)$ ), $e^{-4\pi^2|\xi|^2t}$ is the transform of the heat kernel $H(x, t) = (4\pi t)^{-n/2}e^{-|x|^2/4t}$ , and a product of transforms is the transform of a convolution. So $u(\cdot, t) = f * H(\cdot, t)$ , the solution found directly in 3A.8 Convolution and Mollifiers. Every property of heat flow is visible in the factor $e^{-4\pi^2|\xi|^2t}$ : high frequencies decay fastest, so the solution is smooth for $t > 0$ (Exercise 5.13); the $L^2$ norm decreases (Plancherel); and running the equation backwards would multiply high frequencies by huge factors, so it is ill-posed (2B.7 Fourier Series and the First Heat Equation, 3A.8 Convolution and Mollifiers).

In the world In use Spectrum analysers and the fast Fourier transform

Audio equalisers, spectrum analysers, radio receivers and image compressors compute discrete Fourier transforms of sampled signals millions of times a second. That is possible because of the fast Fourier transform, published by James Cooley and John Tukey in 1965, which computes the transform of $N$ samples in about $N\log N$ operations instead of $N^2$ . (Gauss had found the same idea around 1805, in unpublished work on asteroid orbits.) An equaliser is exactly the dictionary's convolution rule: boosting or cutting frequency bands is multiplication of $\hat f$ by a chosen function, which is convolution of the signal with that function's inverse transform.

Where this goes Where the Fourier transform is used

The Sobolev spaces $H^s$ (4A.9 Sobolev Spaces, 4A.10 Sobolev Embeddings and Critical Exponents); the heat equation and its kernel on $\mathbb{R}^n$ (6A.3 The Heat Equation on ℝⁿ), and the principal symbol of a differential operator, the polynomial $P(2\pi i\xi)$ of its highest-order terms, which decides whether it is elliptic or parabolic (6A.1 What a PDE Is). In 11A.3 Short-Time Existence and Uniqueness, computing the principal symbol of the Ricci operator shows exactly which directions fail to be parabolic (the directions generated by diffeomorphisms), which is what DeTurck's trick repairs. On a compact manifold there is no Fourier transform, and its role is played by the eigenfunction expansion of the Laplacian (4A.7 Compact Operators and Spectra, 9B.7 The Heat Equation on a Manifold).

History

Fourier introduced integral transforms for heat flow on infinite domains in his 1807 memoir and his 1822 book. Michel Plancherel proved his theorem in 1910. Poisson's summation formula appears in his work of the 1820s. The uncertainty principle was formulated by Werner Heisenberg in 1927 and proved as an inequality by Earle Kennard and Hermann Weyl in 1927–28. Laurent Schwartz introduced the class of rapidly decreasing functions in his theory of distributions around 1950 (4A.8 Distributions and Weak Derivatives). Cooley and Tukey published the fast Fourier transform in 1965. Lauterbur (1973) and Mansfield (in the 1970s) developed spatial encoding by magnetic field gradients, the basis of MRI.

Recall Where we stand

The Fourier transform $\hat f(\xi) = \int fe^{-2\pi ix\cdot\xi}$ turns derivatives into multiplications, convolutions into products, and dilations into inverse dilations. The Gaussian $e^{-\pi|x|^2}$ is its own transform. On the Schwartz class the transform is inverted by flipping the sign, and it extends to an isometry of $L^2$ (Plancherel). Poisson summation links it to Fourier series. A function and its transform can't both be concentrated (uncertainty). Smoothness is decay of $\hat f$ , and $H^s \subseteq C$ for $s > n/2$ . The heat equation multiplies $\hat f$ by $e^{-4\pi^2|\xi|^2t}$ , recovering the Gaussian heat kernel. 4A.6 Weak Convergence and the Direct Method returns to the problem of compactness, and recovers it in a weaker sense.

Exercises

Exercise 5.7 The dictionary

Prove the translation, dilation and differentiation rules of the table for $f \in \mathcal{S}(\mathbb{R}^n)$ , and the convolution rule for $f, g \in L^1$ .

Exercise 5.8 The box and the sinc

Show that the transform of $1_{[-1/2, 1/2]}$ is $\operatorname{sinc}\xi = \frac{\sin\pi\xi}{\pi\xi}$ , which is not in $L^1$ . Use Plancherel to show $\int_{\mathbb{R}}\big(\frac{\sin\pi\xi}{\pi\xi}\big)^2d\xi = 1$ . Explain why a sharp cutoff in frequency (a "brick-wall" filter) produces ringing in time.

Exercise 5.9 Extending an isometry

Let $D$ be a dense subspace of a Banach space $X$ , $Y$ a Banach space, and $T : D \to Y$ linear with $\|Tx\| = \|x\|$ . Show $T$ extends uniquely to an isometry $X \to Y$ .

Exercise 5.10 A theta identity

Apply Poisson summation to $f(x) = e^{-\pi tx^2}$ to show $\sum_ne^{-\pi tn^2} = t^{-1/2}\sum_ke^{-\pi k^2/t}$ for $t > 0$ . (This is the functional equation of Jacobi's theta function, which Riemann used to prove the functional equation of the zeta function.) Check it numerically at $t = 1$ and $t = 4$ .

Solution

$\hat f(\xi) = t^{-1/2}e^{-\pi\xi^2/t}$ by the Gaussian proposition. At $t = 1$ the identity is trivially true. At $t = 4$ : left $1 + 2e^{-4\pi} + \cdots \approx 1.0000069$ ; right $\tfrac12(1 + 2e^{-\pi/4} + 2e^{-\pi} + 2e^{-9\pi/4} + \cdots) \approx \tfrac12(1 + 0.91178 + 0.08643 + 0.00170 + 0.00001) \approx 1.0000$ .

Exercise 5.11 The threshold is sharp

Show that $f(x) = \log\log(1 + 1/|x|)\,\phi(x)$ (with $\phi$ a smooth cutoff equal to $1$ near $0$ ) is in $H^1(\mathbb{R}^2)$ but is unbounded. So $s > n/2$ can't be weakened to $s = n/2$ . (Compute $\int|\nabla f|^2$ in polar coordinates.)

Exercise 5.12 Gaussians are optimal

Check directly that $f(x) = e^{-\pi x^2}$ gives equality in the uncertainty principle: compute $\|f\|_2^2 = 2^{-1/2}$ and $\|xf\|_2 = \|\xi\hat f\|_2 = 2^{-5/4}\pi^{-1/2}$ .

Exercise 5.13 Rehearsal: how fast the heat equation smooths

Let $u(\cdot, t) = f * H(\cdot, t)$ with $f \in L^2(\mathbb{R}^n)$ . Using Plancherel and $\widehat{\partial^\alpha u} = (2\pi i\xi)^\alpha\hat u$ , show that for every $k \geq 0$

\|\nabla^ku(\cdot, t)\|_2 \leq C_k\,t^{-k/2}\,\|f\|_2,

by maximising $|\xi|^ke^{-4\pi^2|\xi|^2t}$ over $\xi$ . The factor $t^{-k/2}$ is forced by parabolic scaling (thread S): $x$ scales like $\sqrt t$ , so each derivative costs a factor $t^{-1/2}$ . Shi's derivative estimates for Ricci flow (11A.3 Short-Time Existence and Uniqueness) have exactly this form, $|\nabla^k\mathrm{Rm}| \leq C_kKt^{-k/2}$ , for exactly this reason.

Solution

$\|\nabla^ku\|_2^2 \leq \int(2\pi|\xi|)^{2k}e^{-8\pi^2|\xi|^2t}|\hat f|^2 \leq \sup_\xi\big((2\pi|\xi|)^{2k}e^{-8\pi^2|\xi|^2t}\big)\|f\|_2^2$ . With $s = 4\pi^2|\xi|^2t$ , the supremum is $\sup_s(s/t)^ke^{-2s} = t^{-k}(k/2)^ke^{-k}$ .

© 2026 NeckPinch (www.neckpinch.com). All content in the guidebook (text, mathematics, figures and exercises) is protected by copyright. All rights reserved. No part may be copied, republished or redistributed without written permission.

5.1An image measured in frequency

5.2The Fourier transform on L1L^1L1

5.3The Gaussian

5.4Inversion and Plancherel

5.4.1Poisson summation

5.5The uncertainty principle

5.6Smoothness is decay

5.7The heat equation by Fourier transform

5.8History

5.9Exercises