246B, Notes 2: Some connections with the Fourier transform

What's new 2021-01-24

In Exercise 5 (and Lemma 1) of 246A Notes 4 we already observed some links between complex analysis on the disk (or annulus) and Fourier series on the unit circle:

(i) Functions ${f}$ that are holomorphic on an annulus ${\{ r_- < |z| < r_+ \}}$ are expressed by a convergent Fourier series (and also Laurent series) ${f(re^{i\theta}) = \sum_{n=-\infty}^\infty r^n a_n e^{in\theta}}$ , where
$\displaystyle \limsup_{n \rightarrow +\infty} |a_n|^{1/n} \leq \frac{1}{r_+}; \limsup_{n \rightarrow -\infty} |a_n|^{1/|n|} \leq \frac{1}{r_-}; \ \ \ \ \ (1)$
conversely, every doubly infinite sequence ${(a_n)_{n=-\infty}^\infty}$ of coefficients obeying (1) arises from such a function ${f}$ .
(ii) Functions ${f}$ that are holomorphic on a disk ${\{ |z| < R \}}$ are expressed by a convergent Fourier series (and also Taylor series) ${f(re^{i\theta}) = \sum_{n=0}^\infty r^n a_n e^{in\theta}}$ (so in particular ${a_n = \frac{1}{n!} f^{(n)}(0)}$ ), where
$\displaystyle \limsup_{n \rightarrow +\infty} |a_n|^{1/n} \leq \frac{1}{R}; \ \ \ \ \ (2)$
conversely, every infinite sequence ${(a_n)_{n=0}^\infty}$ of coefficients obeying (2) arises from such a function ${f}$ .
(iii) In the situation of (i), there is a unique decomposition ${f = f_1 + f_2}$ where ${f_1}$ extends holomorphically to ${\{ z: |z| < r_+\}}$ , and ${f_2}$ extends holomorphically to ${\{ z: |z| > r_-\}}$ and goes to zero at infinity, and are given by the formulae
$\displaystyle f_1(z) = \sum_{n=0}^\infty a_n z^n = \frac{2\pi i} \int_\gamma \frac{f(w)}{w-z}\ dw$
where ${\gamma}$ is any anticlockwise contour in ${\{ z: |z| < r_+\}}$ enclosing ${z}$ , and and
$\displaystyle f_2(z) = \sum_{n=-\infty}^{-1} a_n z^n = - = \frac{2\pi i} \int_\gamma \frac{f(w)}{w-z}\ dw$
where ${\gamma}$ is any anticlockwise contour in ${\{ z: |z| > r_-\}}$ enclosing ${0}$ but not ${z}$ .

This connection lets us interpret various facts about Fourier series through the lens of complex analysis, at least for some special classes of Fourier series. For instance, the Fourier inversion formula ${a_n = \frac{1}{2\pi} \int_0^{2\pi} f(e^{i\theta}) e^{-in\theta}\ d\theta}$ becomes the Cauchy-type formula for the Laurent or Taylor coefficients of ${f}$ , in the event that the coefficients are doubly infinite and obey (1) for some ${r_- < 1 < r_+}$ , or singly infinite and obey (2) for some ${R > 1}$ .

It turns out that there are similar links between complex analysis on a half-plane (or strip) and Fourier integrals on the real line, which we will explore in these notes.

We first fix a normalisation for the Fourier transform. If ${f \in L^1({\bf R})}$ is an absolutely integrable function on the real line, we define its Fourier transform ${\hat f: {\bf R} \rightarrow {\bf C}}$ by the formula

$\displaystyle \hat f(\xi) := \int_{\bf R} f(x) e^{-2\pi i x \xi}\ dx. \ \ \ \ \ (3)$

From the dominated convergence theorem ${\hat f}$ will be a bounded continuous function; from the Riemann-Lebesgue lemma it also decays to zero as ${\xi \rightarrow \pm \infty}$ . My choice to place the ${2\pi}$ in the exponent is a personal preference (it is slightly more convenient for some harmonic analysis formulae such as the identities (4), (5), (6) below), though in the complex analysis and PDE literature there are also some slight advantages in omitting this factor. In any event it is not difficult to adapt the discussion in this notes for other choices of normalisation. It is of interest to extend the Fourier transform beyond the ${L^1({\bf R})}$ class into other function spaces, such as ${L^2({\bf R})}$ or the space of tempered distributions, but we will not pursue this direction here; see for instance these lecture notes of mine for a treatment.

Exercise 1 (Fourier transform of Gaussian) If ${a>0}$ and ${f}$ is the Gaussian function ${f(x) := e^{-\pi a x^2}}$ , show that the Fourier transform ${\hat f}$ is given by the gaussian ${\hat f(\xi) = a^{-1/2} e^{-\pi \xi^2/a}}$ .

The Fourier transform has many remarkable properties. On the one hand, as long as the function ${f}$ is sufficiently “reasonable”, the Fourier transform enjoys a number of very useful identities, such as the Fourier inversion formula

$\displaystyle f(x) = \int_{\bf R} \hat f(\xi) e^{2\pi i x \xi} d\xi, \ \ \ \ \ (4)$

the Plancherel identity

$\displaystyle \int_{\bf R} |f(x)|^2\ dx = \int_{\bf R} |\hat f(\xi)|^2\ d\xi, \ \ \ \ \ (5)$

and the Poisson summation formula

$\displaystyle \sum_{n \in {\bf Z}} f(n) = \sum_{k \in {\bf Z}} \hat f(k). \ \ \ \ \ (6)$

On the other hand, the Fourier transform also intertwines various qualitative properties of a function ${f}$ with “dual” qualitative properties of its Fourier transform ${\hat f}$ ; in particular, “decay” properties of ${f}$ tend to be associated with “regularity” properties of ${\hat f}$ , and vice versa. For instance, the Fourier transform of rapidly decreasing functions tend to be smooth. There are complex analysis counterparts of this Fourier dictionary, in which “decay” properties are described in terms of exponentially decaying pointwise bounds, and “regularity” properties are expressed using holomorphicity on various strips, half-planes, or the entire complex plane. The following exercise gives some examples of this:

Exercise 2 (Decay of ${f}$ implies regularity of ${\hat f}$ ) Let ${f \in L^1({\bf R})}$ be an absolutely integrable function.

(i) If ${f}$ has super-exponential decay in the sense that ${f(x) \lesssim_{f,M} e^{-M|x|}}$ for all ${x \in {\bf R}}$ and ${M>0}$ (that is to say one has ${|f(x)| \leq C_{f,M} e^{-M|x|}}$ for some finite quantity ${C_{f,M}}$ depending only on ${f,M}$ ), then ${\hat f}$ extends uniquely to an entire function ${\hat f : {\bf C} \rightarrow {\bf C}}$ . Furthermore, this function continues to be defined by (3).

(ii) If ${f}$ is supported on a compact interval ${[a,b]}$ then the entire function ${\hat f}$ from (i) obeys the bounds ${\hat f(\xi) \lesssim_f \max( e^{a \mathrm{Im} \xi}, e^{b \mathrm{Im} \xi} )}$ for ${\xi \in {\bf C}}$ . In particular, if ${f}$ is supported in ${[-M,M]}$ then ${\hat f(\xi) \lesssim_f e^{-M |\Im(\xi)|}}$ .

(iii) If ${f}$ obeys the bound ${f(x) \lesssim_{f,a} e^{-a|x|}}$ for all ${x \in {\bf R}}$ and some ${a>0}$ , then ${\hat f}$ extends uniquely to a holomorphic function ${\hat f}$ on the horizontal strip ${\{ \xi: |\mathrm{Im} \xi| < a \}}$ , and obeys the bound ${\hat f(\xi) \lesssim_{f,a} \frac{1}{a - |\mathrm{Im}(\xi)|}}$ in this strip. Furthermore, this function continues to be defined by (3).

(iv) If ${f}$ is supported on ${[0,+\infty)}$ (resp. ${(-\infty,0]}$ ), then there is a unique continuous extension of ${f}$ to the upper half-plane ${\{ \xi: \mathrm{Im} \xi \geq 0\}}$ (resp. the lower half-plane ${\{ \xi: \mathrm{Im} \xi \leq 0 \}}$ which is holomorphic in the interior of this half-plane, and such that ${\hat f(\xi) \rightarrow 0}$ uniformly as ${\mathrm{Im} \xi \rightarrow +\infty}$ (resp. ${\mathrm{Im} \xi \rightarrow -\infty}$ ). Furthermore, this function continues to be defined by (3).

Hint: to establish holomorphicity in each of these cases, use Morera’s theorem and the Fubini-Tonelli theorem. For uniqueness, use analytic continuation, or (for part (iv)) the Cauchy integral formula.

Proof:

Later in these notes we will give a partial converse to part (ii) of this exercise, known as the Paley-Wiener theorem; there are also partial converses to the other parts of this exercise.

From (3) we observe the following intertwining property between multiplication by an exponential and complex translation: if ${\xi_0}$ is a complex number and ${f: {\bf R} \rightarrow {\bf C}}$ is an absolutely integrable function such that the modulated function ${f_{\xi_0}(x) := e^{2\pi i \xi_0 x} f(x)}$ is also absolutely integrable, then we have the identity

$\displaystyle \widehat{f_{\xi_0}}(\xi) = \hat f(\xi - \xi_0) \ \ \ \ \ (7)$

whenever ${\xi}$ is a complex number such that at least one of the two sides of the equation in (7) is well defined. Thus, multiplication of a function by an exponential weight corresponds (formally, at least) to translation of its Fourier transform. By using contour shifting, we will also obtain a dual relationship: under suitable holomorphicity and decay conditions on ${f}$ , translation by a complex shift will correspond to multiplication of the Fourier transform by an exponential weight. It turns out to be possible to exploit this property to derive many Fourier-analytic identities, such as the inversion formula (4) and the Poisson summation formula (6), which we do later in these notes. (The Plancherel theorem can also be established by complex analytic methods, but this requires a little more effort; see Exercise 7.

…

The material in these notes is loosely adapted from Chapter 4 of Stein-Shakarchi’s “Complex Analysis”.

— 1. The inversion and Poisson summation formulae —

We now explore how the Fourier transform ${\hat f}$ of a function ${f}$ behaves when ${f}$ extends holomorphically to a strip. For technical reasons we will also impose a fairly mild decay condition on ${f}$ at infinity to ensure integrability. As we shall shortly see, the method of contour shifting then allows us to insert various exponentially decaying factors into Fourier integrals that make the justification of identities such as the Fourier inversion formula straightforward.

Proposition 3 (Fourier transform of functions holomorphic in a strip) Let ${a > 0}$ , and suppose that ${f}$ is a holomorphic function on the strip ${\{ z: |\mathrm{Im} z| < a \}}$ which obeys a decay bound of the form
$\displaystyle |f(x+iy)| \leq \frac{C_b}{1+|x|^2} \ \ \ \ \ (8)$
for all ${x \in {\bf R}}$ , ${0 < b < a}$ , ${y \in [-b,b]}$ , and some ${C>0}$ (or in asymptotic notation, one has ${f(x+iy) \lesssim_{b,f} \frac{1}{1+|x|^2}}$ whenever ${x \in {\bf R}}$ and ${|y| \leq b < a}$ ).

(i) (Translation intertwines with modulation) For any ${w}$ in the strip ${\{ z: |\mathrm{Im}(z)| < a \}}$ , the Fourier transform of the function ${x \mapsto f(x+w)}$ is ${\xi \mapsto e^{2\pi i w \xi} \hat f(\xi)}$ .

(ii) (Exponential decay of Fourier transform) For any ${0 < b < a}$ , there is a quantity ${C}$ such that ${|\hat f(\xi)| \leq C_b e^{-2\pi b|\xi|}}$ for all ${\xi \in {\bf R}}$ (or in asymptotic notation, one has ${\hat f(\xi) \lesssim_{a,b,f} e^{-2\pi b|\xi|}}$ for ${0 < b < a}$ and ${\xi \in {\bf R}}$ ).

(iii) (Partial Fourier inversion) For any ${0 < \varepsilon < a}$ and ${x \in {\bf R}}$ , one has
$\displaystyle \int_0^\infty \hat f(\xi) e^{2\pi i x \xi}\ d\xi = \frac{1}{2\pi i} \int_{-\infty}^\infty \frac{f(y-i\varepsilon)}{y-i\varepsilon - x}\ dy$
and
$\displaystyle \int_{-\infty}^0 \hat f(\xi) e^{2\pi i x \xi}\ d\xi = -\frac{1}{2\pi i} \int_{-\infty}^\infty \frac{f(y+i\varepsilon)}{y+i\varepsilon - x}\ dy.$

(iv) (Full Fourier inversion) For any ${x \in {\bf R}}$ , the identity (4) holds for this function ${f}$ .

(v) (Poisson summation formula) The identity (6) holds for this function ${f}$ .

Proof: We begin with (i), which is a standard application of contour shifting. Applying the definition (3) of the Fourier transform, our task is to show that

$\displaystyle \int_{\bf R} f(x+w) e^{-2\pi i x \xi}\ dx = e^{2\pi i w \xi} \int_{\bf R} f(x) e^{-2\pi i x \xi}\ dx$

whenever ${|\mathrm{Im} w| < a}$ and ${\xi \in {\bf R}}$ . Clearly

$\displaystyle \int_{\bf R} f(x) e^{-2\pi i x \xi}\ d\xi = \lim_{R \rightarrow \infty} \int_{\gamma_{-R \rightarrow R}} f(z) e^{-2\pi i z \xi}\ dz$

where ${\gamma_{z_1 \rightarrow z_2}}$ is the line segment contour from ${z_1}$ to ${z_2}$ , and similarly after a change of variables

$\displaystyle e^{-2\pi i w \xi} \int_{\bf R} f(x+w) e^{-2\pi i x \xi}\ dx = \lim_{R \rightarrow \infty} \int_{\gamma_{-R+w \rightarrow R+w}} f(z) e^{-2\pi i z \xi}\ dz.$

On the other hand, from Cauchy’s theorem we have

$\displaystyle \int_{\gamma_{-R+w \rightarrow R+w}} = \int_{\gamma_{-R \rightarrow R}} + \int_{\gamma_{R \rightarrow R+w}} - \int_{\gamma_{-R \rightarrow -R+w}}$

when applied to the holomorphic integrand ${f(z) e^{-2\pi i z \xi}}$ . So it suffices to show that

$\displaystyle \int_{\gamma_{\pm R \rightarrow \pm R+w}} f(z) e^{-2\pi i z \xi}\ dz \rightarrow 0$

as ${R \rightarrow \infty}$ . But the left hand side can be rewritten as

$\displaystyle w e^{\mp 2\pi i R \xi} \int_0^1 f(\pm R + tw) e^{-2\pi i tw \xi}\ dt,$

and the claim follows from (8) and dominated convergence.

For (ii), we apply (i) with ${w = \pm i b}$ to observe that the Fourier transform of ${x \mapsto f(x \pm ib)}$ is ${\xi \mapsto e^{\mp 2\pi b \xi} \hat f(\xi)}$ . Applying (8) and the triangle inequality we conclude that

$\displaystyle e^{\mp 2\pi b \xi} \hat f(\xi) \lesssim_{f,b} 1$

for both choices of sign ${\mp}$ and all ${\xi \in {\bf R}}$ , giving the claim.

For the first part of (iii), we write ${f_{-\varepsilon}(y) := f(y-i\varepsilon)}$ . By part (i), we have ${\hat f_{-\varepsilon}(\xi) = e^{2\pi \varepsilon \xi} \hat f(\xi)}$ , so we can rewrite the desired identity as

$\displaystyle \int_0^\infty \hat f_{-\varepsilon}(\xi) e^{-2\pi \varepsilon \xi} e^{2\pi i x \xi}\ d\xi = \frac{1}{2\pi i} \int_{-\infty}^\infty \frac{f_{-\varepsilon}(y)}{y-i\varepsilon-x}\ dy.$

By (3) and Fubini’s theorem (taking advantage of (8) and the exponential decay of ${e^{-2\pi \varepsilon \xi}}$ as ${\varepsilon \rightarrow +\infty}$ ) the left-hand side can be written as

$\displaystyle \int_{\bf R} f_{-\varepsilon}(y) \int_0^\infty e^{-2\pi i y \xi} e^{-2\pi \varepsilon \xi} e^{2\pi i x \xi}\ dy d\xi.$

But a routine calculation shows that

$\displaystyle \int_0^\infty e^{-2\pi i y \xi} e^{-2\pi \varepsilon \xi} e^{2\pi i x \xi}\ dy = \frac{1}{2\pi i} \frac{1}{y-i\varepsilon-x}$

giving the claim. The second part of (iii) is proven similarly.

To prove (iv), it suffices in light of (iii) to show that

$\displaystyle \int_{-\infty}^\infty \frac{f(y-i\varepsilon)}{y-i\varepsilon - x}\ dy - \int_{-\infty}^\infty \frac{f(y+i\varepsilon)}{y+i\varepsilon - x}\ dy = 2\pi i f(x)$

for any ${x \in {\bf R}}$ . The left-hand side can be written after a change of variables as

$\displaystyle \lim_{R \rightarrow \infty} \int_{\gamma_{-R-i\varepsilon \rightarrow R-i\varepsilon}} \frac{f(z)}{z-x}\ dz + \int_{\gamma_{R+i\varepsilon \rightarrow R+i\varepsilon}} \frac{f(z)}{z-x}\ dz.$

On the other hand, from dominated convergence as in the proof of (i) we have

$\displaystyle \lim_{R \rightarrow \infty} \int_{\gamma_{R-i\varepsilon \rightarrow R+i\varepsilon}} \frac{f(z)}{z-x}\ dz + \int_{\gamma_{-R+i\varepsilon \rightarrow -R-i\varepsilon}} \frac{f(z)}{z-x}\ dz = 0$

while from the Cauchy integral formula one has

$\displaystyle \lim_{R \rightarrow \infty} \int_{\gamma_{-R-i\varepsilon \rightarrow R-i\varepsilon \rightarrow R+i\varepsilon \rightarrow -R+i\varepsilon \rightarrow -R-i\varepsilon}} \frac{f(z)}{z-x}\ dz = 2\pi i f(x)$

giving the claim.

Now we prove (v). Let ${0 < \varepsilon < a}$ . From (i) we have

$\displaystyle \hat f(k) = \int_{\bf R} f(x-i\varepsilon) e^{-2\pi \varepsilon k} e^{-2\pi i k x}\ dx$

and

$\displaystyle \hat f(k) = \int_{\bf R} f(x+i\varepsilon) e^{2\pi \varepsilon k} e^{-2\pi i k x}\ dx$

for any ${k \in {\bf Z}}$ . If we sum the first identity for ${k=0,1,2,\dots}$ we see from the geometric series formula and Fubini’s theorem that

$\displaystyle \sum_{k=0}^\infty \hat f(k) = \int_{\bf R} \frac{f(x-i\varepsilon)}{1 - e^{-2\pi i (x-i\varepsilon)}}\ dx$

and similarly if we sum the second identity for ${k=-1,-2,\dots}$ we have

$\displaystyle \sum_{k=-\infty}^{-1} \hat f(k) = \int_{\bf R} \frac{f(x+i\varepsilon) e^{2\pi i (x+i\varepsilon)}}{1 - e^{2\pi i (x+i\varepsilon)}}\ dx$

$\displaystyle = - \int_{\bf R} \frac{f(x+i\varepsilon)}{1 - e^{-2\pi i (x+i\varepsilon)}}\ dx.$

Adding these two identities and changing variables, we conclude that

$\displaystyle \sum_{k=-\infty}^\infty \hat f(k) = \lim_{R \rightarrow \infty} \int_{\gamma_{-R-i\varepsilon \rightarrow R-i\varepsilon}} \frac{f(z)}{1 - e^{-2\pi i z}}\ dz + \int_{\gamma_{R+i\varepsilon \rightarrow -R+i\varepsilon}} \frac{f(z)}{1 - e^{-2\pi i z}}\ dz.$

We would like to use the residue theorem to evaluate the right-hand side, but we need to take a little care to avoid the poles of the integrand ${\frac{f(z)}{1 - e^{-2\pi i z}}}$ , which are at the integers. Hence we shall restrict ${R}$ to be a half-integer ${R = N + \frac{1}{2}}$ . In this case, a routine application of the residue theorem shows that

$\displaystyle \int_{\gamma_{-R-i\varepsilon \rightarrow R-i\varepsilon \rightarrow R+i\varepsilon \rightarrow -R+i\varepsilon \rightarrow -R-i\varepsilon}} \frac{f(z)}{1 - e^{-2\pi i z}}\ dz = \sum_{n=-N}^N f(n).$

Noting that ${\frac{1}{1-e^{-2\pi i z}}}$ stays bounded for ${z}$ in ${\gamma_{R-i\varepsilon \rightarrow R+i\varepsilon}}$ or ${\gamma_{-R+i\varepsilon \rightarrow -R-i\varepsilon}}$ when ${R}$ is a half-integer, we also see from dominated convergence as before that

$\displaystyle \lim_{R \rightarrow \infty} \int_{\gamma_{R-i\varepsilon \rightarrow R+i\varepsilon}} \frac{f(z)}{1 - e^{-2\pi i z}}\ dz + \int_{\gamma_{-R+i\varepsilon \rightarrow -R-i\varepsilon}} \frac{f(z)}{1 - e^{-2\pi i z}} = 0.$

The claim follows. $\Box$

Exercise 4 (Hilbert transform and Plemelj formula) Let ${a, f}$ be as in Proposition 3. Define the Cauchy-Stieltjes transform ${{\mathcal C} f: {\bf C} \backslash {\bf R} \rightarrow {\bf C}}$ by the formula
$\displaystyle {\mathcal C} f(z) := \int_{\bf R} \frac{f(x)}{z-x}\ dx.$

(i) Show that ${{\mathcal C} f}$ is holomorphic on ${{\bf C} \backslash {\bf R}}$ and has the Fourier representation
$\displaystyle {\mathcal C} f(z) = -2\pi i \int_0^\infty \hat f(\xi) e^{2\pi i z \xi}\ d\xi$
in the upper half-plane ${\mathrm{Im} z > 0}$ and
$\displaystyle {\mathcal C} f(z) = 2\pi i \int_{-\infty}^0 \hat f(\xi) e^{2\pi i z \xi}\ d\xi$
in the lower half-plane ${\mathrm{Im} z < 0}$ .

(ii) Establish the Plemelj formulae
$\displaystyle \lim_{\varepsilon \rightarrow 0^+} {\mathcal C} f(x+i\varepsilon) = \pi H f(x) - i \pi f(x)$
and
$\displaystyle \lim_{\varepsilon \rightarrow 0^+} {\mathcal C} f(x-i\varepsilon) = \pi H f(x) + i \pi f(x)$
for any ${x \in {\bf R}}$ , where the Hilbert transform ${Hf}$ of ${f}$ is defined by the principal value integral
$\displaystyle Hf(x) := \lim_{\varepsilon \rightarrow 0^+, R \rightarrow \infty} \frac{1}{\pi} \int_{\varepsilon \leq |y-x| \leq R} \frac{f(y)}{x-y}\ dy.$

(iii) Show that ${{\mathcal C} f}$ is the unique holomorphic function on ${{\bf C} \backslash {\bf R}}$ that obeys the decay bound
$\displaystyle \sup_{x+iy \in {\bf C} \backslash {\bf R}} |y| |{\mathcal C} f(x+iy)| < \infty$
and solves the (very simple special case of the) Riemann-Hilbert problem
$\displaystyle \lim_{\varepsilon \rightarrow 0^+} {\mathcal C} f(x+i\varepsilon) - {\mathcal C} f(x-i\varepsilon) = - 2\pi i f(x)$
uniformly in ${x \in {\bf R}}$ .

(iv) Establish the identity
$\displaystyle Hf(x) = -\int_{\bf R} i \mathrm{sgn}(\xi) \hat f(\xi) e^{2\pi i x \xi}\ d\xi,$
where the signum function ${\mathrm{sgn}(\xi)}$ is defined to equal ${+1}$ for ${\xi>0}$ , ${-1}$ for ${\xi < 0}$ , and ${0}$ for ${\xi=0}$ .

(v) Show that ${Hf}$ extends holomorphically to the strip ${\{ z: |\mathrm{Im} z| < a \}}$ and obeys the bound (8) (but possibly with a different constant ${C_b}$ ), with the identity
$\displaystyle {\mathcal C} f(z) = \pi H f(z) - \pi i \mathrm{sgn}(\mathrm{Im}(z)) f(z) \ \ \ \ \ (9)$
holding for ${0 < |\mathrm{Im} z| < a}$ .

(vi) Establish the identities
$\displaystyle H(Hf) = -f$
and
$\displaystyle H( fHf) = \frac{(Hf)^2 - f^2}{2}.$
(Hint: for the latter inequality, square both sides of (9).)

Exercise 5

(i) By applying the Poisson summation formula to the function ${x \mapsto \frac{1}{x^2+a^2}}$ , establish the identity
$\displaystyle \sum_{n \in {\bf Z}} \frac{1}{n^2 + a^2} = \frac{\pi}{a} \frac{e^{2\pi a}+1}{e^{2\pi a}-1}$
for any positive real number ${a}$ . Explain why this is consistent with Exercise 24 from Notes 1.

(ii) By carefully taking limits of (i) as ${a \rightarrow 0}$ , establish yet another alternate proof of Euler’s identity
$\displaystyle \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}.$

Exercise 6 For ${\tau}$ in the upper half-plane ${\{ \mathrm{Im} \tau > 0\}}$ , define the theta function ${\theta(\tau) := \sum_{n \in {\bf Z}} e^{-\pi i n^2 \tau}}$ . Use Exercise 1 and the Poisson summation formula to establish the modular identity
$\displaystyle \theta(\tau) = (-i\tau)^{-1/2} \theta(-1/\tau)$
for such ${\tau}$ , where one takes the standard branch of the square root.

Exercise 7 (Fourier proof of Plancherel identity) Let ${f: {\bf R} \rightarrow {\bf C}}$ be smooth and compactly supported. For any ${\xi \in {\bf C}}$ with ${\mathrm{Im} \xi \geq 0}$ , define the quantity
$\displaystyle A(\xi) := \int_{\bf R} \int_{\bf R} e^{2\pi i \xi |x-y|} \overline{f}(x) f(y)\ dx dy.$

(i) When ${\xi}$ is real, show that ${\mathrm{Re} A(\xi) = |\hat f(\xi)|^2}$ . (Hint: find a way to rearrange the expression ${A(\xi) + \overline{A}(\xi)}$ .)

(ii) For ${\xi}$ non-zero, show that ${A(\xi) = \frac{1}{\pi i \xi} \int_{\bf R} |f(x)|^2\ dx + O(1/|\xi|^2)}$ , where the implied constant in the ${\xi}$ notation can depend on ${f}$ . (Hint: integrate by parts several times.)

(iii) Establish the Plancherel identity (5).

Remarkably, this proof of the Plancherel identity generalises to a nonlinear version involving a trace formula for the scattering transform for either Schrodinger or Dirac operators. For Schrodinger operators this was first obtained (implicitly) by Buslaev and Faddeev, and later more explicitly by by Deift and Killip. The version for Dirac operators more closely resembles the linear Plancherel identity; see for instance the appendix to this paper of Muscalu, Thiele, and myself. The quantity ${A(\xi)}$ is a component of a nonlinear quantity known as the transmission coefficient ${a(\xi)}$ of a Dirac operator with potential ${f}$ and spectral parameter ${\xi}$ (or ${2\pi \xi}$ , depending on normalisations).

The Fourier inversion formula was only established in Proposition 3 for functions that had a suitable holomorphic extension to a strip, but one can relax the hypotheses by a limiting argument. Here is one such example of this:

Exercise 8 (More general Fourier inversion formula) Let ${f: {\bf R} \rightarrow {\bf C}}$ be continuous and obey the bound ${|f(x)| \leq \frac{C}{1+|x|^2}}$ for all ${x \in {\bf R}}$ and some ${C>0}$ . Suppose that the Fourier transform ${\hat f: {\bf R} \rightarrow {\bf C}}$ is absolutely integrable.

(i) Show that for any ${\varepsilon>0}$ , the regularised function
$\displaystyle f_\varepsilon(x) := \int_{\bf R} f(x+\varepsilon y) e^{-\pi y^2}\ dy$
extends to an entire function obeying the hypotheses of Proposition 3 for any ${0 < a < \infty}$ , with ${\hat f_\varepsilon(\xi) = e^{-\pi \varepsilon^2 \xi^2} \hat f(\xi)}$ for any ${\xi \in {\bf R}}$ . (Hint: use Exercise 1.)

(ii) Show that the Fourier inversion formula (4) holds for this function ${f}$ .

Exercise 9 (Laplace inversion formula) Let ${f: [0,+\infty) \rightarrow {\bf C}}$ be a continuously twice differentiable function, obeying the bounds ${|f(x)|, |f'(x)|, |f''(x)| \leq \frac{C}{1+|x|^2}}$ for all ${x \geq 0}$ and some ${C>0}$ .

(i) Show that the Fourier transform ${\hat f}$ obeys the asymptotic
$\displaystyle \hat f(\xi) = \frac{f(0)}{2\pi i \xi} + O( \frac{C}{|\xi|^2} )$
for any non-zero ${\xi \in {\bf R}}$ .

(ii) Establish the principal value inversion formula
$\displaystyle f(x) = \lim_{T \rightarrow +\infty} \int_{-T}^T \hat f(\xi) e^{2\pi i x \xi}\ d\xi$
for any positive real ${x}$ . (Hint: modify the proof of Exercise 8(ii).) What happens when ${x}$ is negative? zero?

(iii) Define the Laplace transform ${{\mathcal L} f(s)}$ of ${f}$ for ${\mathrm{Re}(s) \geq 0}$ by the formula
$\displaystyle {\mathcal L} f(s) := \int_0^\infty f(t) e^{-st}\ dt.$
Show that ${{\mathcal L} f}$ is continuous on the half-plane ${\{ s: \mathrm{Re}(s) \geq 0\}}$ , holomorphic on the interior of this half-plane, and obeys the Laplace-Mellin inversion formula
$\displaystyle f(t) = \frac{1}{2\pi i} \lim_{T \rightarrow +\infty} \int_{\gamma_{\sigma-iT \rightarrow \sigma+iT}} e^{st} {\mathcal L} f(s) \ ds \ \ \ \ \ (10)$
for any ${t>0}$ and ${\sigma \geq 0}$ , where ${\gamma_{\sigma-iT \rightarrow \sigma+iT}}$ is the line segment contour from ${\sigma-iT}$ to ${\sigma+iT}$ . Conclude in particular that the Laplace transform ${{\mathcal L}}$ is injective on this class of functions ${f}$ .

The Laplace-Mellin inversion formula in fact holds under more relaxed decay and regularity hypotheses than the ones given in this exercise, but we will not pursue these generalisations here. The limiting integral in (10) is also known as the Bromwich integral, and often written (with a slight abuse of notation) as ${\frac{1}{2\pi i} \int_{\sigma-i\infty}^{\sigma+i\infty} e^{st} {\mathcal L} f(s)\ ds}$ . The Laplace transform is a close cousin of the Fourier transform that has many uses; for instance, it is a popular tool for analysing ordinary differential equations on half-lines such as ${[0,+\infty)}$ .

Exercise 10 (Mellin inversion formula) Let ${f: (0,+\infty) \rightarrow {\bf C}}$ be a continuous function that is compactly supported in ${(0,+\infty)}$ . Define the Mellin transform ${{\mathcal M} f: {\bf C} \rightarrow {\bf C}}$ by the formula
$\displaystyle {\mathcal M} f(s) := \int_0^\infty x^s f(x) \frac{dx}{x}.$
Show that ${{\mathcal M} f}$ is entire and one has the Mellin inversion formula
$\displaystyle f(x) = \frac{1}{2\pi i} \lim_{T \rightarrow +\infty} \int_{\gamma_{\sigma-iT \rightarrow \sigma+iT}} x^{-s} {\mathcal M} f(s)\ ds$
for any ${x \in {\bf R}}$ and ${\sigma \in {\bf R}}$ . The regularity and support hypotheses on ${f}$ can be relaxed significantly, but we will not pursue this direction here.

Exercise 11 (Perron’s formula) Let ${f: {\bf N} \rightarrow {\bf C}}$ be a function which is of subpolynomial growth in the sense that ${|f(n)| \leq C_\varepsilon n^\varepsilon}$ for all ${n \in {\bf N}}$ and ${\varepsilon>0}$ , where ${C_\varepsilon}$ depends on ${\varepsilon}$ (and ${f}$ ). For ${s}$ in the half-plane ${\{ \mathrm{Re} s > 1 \}}$ , form the Dirichlet series
$\displaystyle F(s) := \sum_{n=1}^\infty \frac{f(n)}{n^s}.$
For any non-integer ${x>0}$ and any ${\sigma>1}$ , establish Perron’s formula
$\displaystyle \sum_{n \leq x} f(n) = \frac{1}{2\pi i} \lim_{T \rightarrow \infty} \int_{\gamma_{\sigma-iT \rightarrow \sigma+iT}} F(s) \frac{x^s}{s}\ ds. \ \ \ \ \ (11)$
What happens when ${x}$ is an integer?

Exercise 12 (Solution to Schrödinger equation) Let ${f, a}$ be as in Proposition 3. Define the function ${u: {\bf R} \times {\bf R} \rightarrow {\bf C}}$ by the formula \{ u(t,x) := \int_R \hat f(\xi) e^{2\pi i x \xi – 4 \pi^2 i \xi^2 t} d\xi.\}

(i) Show that ${f}$ is a smooth function of ${t,x}$ that obeys the Schrödinger equation ${i \partial_t u + \partial_{xx} u = 0}$ with initial condition ${u(0,x) = f(x)}$ for ${x \in {\bf R}}$ .

(ii) Establish the formula
$\displaystyle u(t,x) = \frac{1}{(4\pi i t)^{1/2}} \int_{\bf R} e^{-\frac{|x-y|^2}{4it}} f(y)\ dy$
for ${x \in {\bf R}}$ and ${t \neq 0}$ , where we use the standard branch of the square root.

— 2. Phragmen-Lindelof and Paley-Wiener —

The maximum modulus principle (Exercise 26 from 246A Notes 1) for holomorphic functions asserts that if a function continuous on a compact subset ${K}$ of the plane and holomorphic on the interior of that set is bounded in magnitude by a bound ${M}$ on the boundary ${\partial K}$ , then it is also bounded by ${M}$ on the interior. This principle does not directly apply for noncompact domains ${K}$ : for instance, on the entire complex plane ${{\bf C}}$ , there is no boundary whatsoever and the bound is clearly vacuous. On the half-plane ${\{ \mathrm{Im} z \geq 0 \}}$ , the holomorphic function ${\cos z}$ (for instance) is bounded in magnitude by ${1}$ on the boundary of the half-plane, but grows exponentially in the interior. Similarly, in the strip ${\{ z: -\pi/2 \leq \mathrm{Re} z \leq \pi/2\}}$ , the holomorphic function ${\exp(\exp(iz))}$ is bounded in magnitude by ${1}$ on the boundary of the strip, but is grows double-exponentially in the interior of the strip. However, if one does not have such absurdly high growth, one can recover a form of the maximum principle, known as the Phragmén-Lindelöf principle. Here is one formulation of this principle:

Theorem 13 (Lindelöf’s theorem) Let ${f}$ be a continuous function on a strip ${S := \{ \sigma+it: a \leq \sigma \leq b; t \in {\bf R} \}}$ for some ${b>a}$ , which is holomorphic in the interior of the strip and obeys the bound
$\displaystyle |f(\sigma+it)| \leq A \exp( B \exp( (1-\delta) \frac{\pi}{b-a} |t|) ) \ \ \ \ \ (12)$
for all ${\sigma+it \in S}$ and some constants ${A, \delta > 0}$ . Suppose also that ${|f(a+it)| \leq M}$ and ${|f(b+it)| \leq M}$ for all ${t \in {\bf R}}$ and some ${M>0}$ . Then we have ${|f(\sigma+it)| \leq M}$ for all ${a \leq \sigma \leq b}$ and ${t \in {\bf R}}$ .

Remark 14 The hypothesis (12) is a qualitative hypothesis rather than a quantitative one, since the exact values of ${A, B, \delta}$ do not show up in the conclusion. It is quite a mild condition; any function of exponential growth in ${t}$ , or even with such super-exponential growth as ${O( |t|^{|t|})}$ or ${O(e^{|t|^{O(1)}})}$ , will obey (12). The principle however fails without this hypothesis, as discussed previously.

Proof: By shifting and dilating (adjusting ${A,B,\delta}$ as necessary) we can reduce to the case ${a = -\pi/2}$ , ${b = \pi/2}$ , and by multiplying ${f}$ by a constant we can also normalise ${M=1}$ .

Suppose we temporarily assume that ${f(\sigma+it) \rightarrow 0}$ as ${|\sigma+it| \rightarrow \infty}$ . Then on a sufficiently large rectangle ${\{ \sigma+it: -\pi/2 \leq \sigma \leq \pi/2; -T \leq t \leq T \}}$ , we have ${|f| \leq 1}$ on the boundary of the rectangle, hence on the interior by the maximum modulus principle. Sending ${T \rightarrow \infty}$ , we obtain the claim.

To remove the assumption that ${f}$ goes to zero at infinity, we use the trick of giving ourselves an epsilon of room. Namely, we multiply ${f(z)}$ by the holomorphic function ${g_\varepsilon(z) := \exp( -\varepsilon \exp(i(1-\delta/2) z) )}$ for some ${\varepsilon > 0}$ . A little complex arithmetic shows that the function ${f(z) g_\varepsilon(z)}$ goes to zero at infinity in ${S}$ . Applying the previous case to this function, then taking limits as ${\varepsilon \rightarrow 0}$ , we obtain the claim. $\Box$

Corollary 15 (Phragmén-Lindelöf principle in a sector) Let ${f}$ be a continuous function on a sector ${S := \{ re^{i\theta}: r \geq 0, \alpha \leq \theta \leq \beta \}}$ for some ${\alpha < \beta < \alpha + 2\pi}$ , which is holomorphic on the interior of the sector and obeys the bound
$\displaystyle |f(z)| \leq A \exp( B |z|^a )$
for some ${A,B > 0}$ and ${0 < a < \frac{\pi}{\beta-\alpha}}$ . Suppose also that ${|f(z)| \leq M}$ on the boundary of the sector ${S}$ for some ${M >0}$ . Then one also has ${|f(z)| \leq M}$ in the interior.

Proof: Apply Theorem 13 to the function ${f(\exp(z))}$ on the strip ${\{ \sigma+it: \alpha \leq \sigma \leq \beta\}}$ . $\Box$

Exercise 16 With the notation and hypotheses of Theorem 13, show that the function ${\sigma \mapsto \sup_{t \in {\bf R}} |f(\sigma+it)|}$ is log-convex on ${[a,b]}$ .

Exercise 17 (Hadamard three-circles theorem) Let ${f}$ be a holomorphic function on an annulus ${\{ z \in {\bf C}: R_1 \leq |z| \leq R_2 \}}$ . Show that the function ${r \mapsto \sup_{\theta \in [0,2\pi]} |f(re^{i\theta})|}$ is log-convex on ${[R_1,R_2]}$ .

Exercise 18 (Phragmén-Lindelöf principle) Let ${f}$ be as in Theorem 13 with ${a=0, b=1}$ , but suppose that we have the bounds ${f(0+it) \leq C(1+|t|)^{a_0}}$ and ${f(1+it) \leq C(1+|t|)^{a_1}}$ for all ${t \in {\bf R}}$ and some exponents ${a_0,a_1 \in {\bf R}}$ and a constant ${C>0}$ . Show that one has ${f(\sigma+it) \leq C' (1+|t|)^{(1-\sigma) a_0 + \sigma a_1}}$ for all ${\sigma+it \in S}$ and some constant ${C'}$ (which is allowed to depend on the constants ${A, \delta}$ in (12)). (Hint: it is convenient to work first in a half-strip such as ${\{ \sigma+it \in S: t \geq T \}}$ for some large ${T}$ . Then multiply ${f}$ by something like ${\exp( - ((1-z)a_0+z a_1) \log(-iz) )}$ for some suitable branch of the logarithm and apply a variant of Theorem 13 for the half-strip. A more refined estimate in this regard is due to Rademacher.) This particular version of the principle gives the convexity bound for Dirichlet series such as the Riemann zeta function. Bounds which exploit the deeper properties of these functions to improve upon the convexity bound are known as subconvexity bounds and are of major importance in analytic number theory, which is of course well outside the scope of this course.

Now we can establish a remarkable converse of sorts to Exercise 2(ii) known as the Paley-Wiener theorem, that links the exponential growth of (the analytic continuation) of a function with the support of its Fourier transform:

Theorem 19 (Paley-Wiener theorem) Let ${f: {\bf R} \rightarrow {\bf C}}$ be a continuous function obeying the decay condition
$\displaystyle |f(x)| \leq C/(1+|x|^2) \ \ \ \ \ (13)$
for all ${x \in {\bf R}}$ and some ${C>0}$ . Let ${M > 0}$ . Then the following are equivalent:

(i) The Fourier transform ${\hat f}$ is supported on ${[-M,M]}$ .

(ii) ${f}$ extends analytically to an entire function that obeys the bound ${|f(z)| \leq A e^{2\pi M |z|}}$ for some ${A>0}$ .

(iii) ${f}$ extends analytically to an entire function that obeys the bound ${|f(z)| \leq A e^{2\pi M |\mathrm{Im} z|}}$ for some ${A>0}$ .

The continuity and decay hypotheses on ${f}$ can be relaxed, but we will not explore such generalisations here.

Proof: If (i) holds, then by Exercise 8, we have the inversion formula (4), and the claim (iii) then holds by a slight modification of Exercise 2(ii). Also, the claim (iii) clearly implies (ii).

Now we see why (iii) implies (i). We first assume that we have the stronger bound

$\displaystyle |f(z)| \leq A e^{2\pi M |\mathrm{Im} z|} / (1 + |z|^2) \ \ \ \ \ (14)$

for ${z \in {\bf C}}$ . Then we can apply Proposition 3 for any ${a>0}$ , and conclude in particular that

$\displaystyle \hat f(\xi) = e^{-2\pi b \xi} \int_{\bf R} f(x-ib) e^{-2\pi i x \xi}\ dx$

for any ${\xi \in {\bf R}}$ and ${b \in {\bf R}}$ . Applying (14) and the triangle inequality, we see that

$\displaystyle \hat f(\xi) \lesssim_A e^{-2\pi b \xi} e^{2\pi M |b|}.$

If ${\xi > M}$ , we can then send ${b \rightarrow +\infty}$ and conclude that ${\hat f(\xi)=0}$ ; similarly for ${\xi < -M}$ we can send ${b \rightarrow -\infty}$ and again conclude ${\hat f(\xi) = 0}$ . This establishes (i) in this case.

Now suppose we only have the weaker bound on ${f}$ assumed in (iii). We again use the epsilon of room trick. For any ${\varepsilon>0}$ , we consider the modified function ${f_\varepsilon(z) := f(z) / (1 - i \varepsilon z)^2}$ . This is still holomorphic on the lower half-plane ${\{ z: \mathrm{Im} z \leq 0 \}}$ and obeys a bound of the form (14) on this half-plane. An inspection of the previous arguments shows that we can still show that ${\hat f_\varepsilon(\xi) = 0}$ for ${\xi > M}$ despite no longer having holomorphicity on the entire upper half-plane; sending ${\varepsilon \rightarrow 0}$ using dominated convergence we conclude that ${\hat f(\xi) = 0}$ for ${\xi > M}$ . A similar argument (now using ${1+i\varepsilon z}$ in place of ${1-i\varepsilon z}$ shows that ${\hat f(\xi) = 0}$ for ${\xi < -M}$ . This proves (i).

Finally, we show that (ii) implies (iii). The function ${f(z) e^{2\pi i Mz}}$ is entire, bounded on the real axis by (13), bounded on the upper imaginary axis by (iii), and has exponential growth. By Corollary 15, it is also bounded on the upper half-plane, which gives (iii) in the upper half-plane as well. A similar argument (using ${e^{-2\pi i Mz}}$ in place of ${e^{2\pi i Mz}}$ ) also yields (iii) in the lower half-plane. $\Box$

— 3. The Hardy uncertainty principle —

Informally speaking, the uncertainty principle for the Fourier transform asserts that a function ${f}$ and its Fourier transform cannot simultaneously be strongly localised, except in the degenerate case when ${f}$ is identically zero. There are many rigorous formulations of this principle. Perhaps the best known is the Heisenberg uncertainty principle

$\displaystyle (\int_{\bf R} (\xi-\xi_0)^2 |\hat f(\xi)|^2\ d\xi)^{1/2} (\int_{\bf R} (x-x_0)^2 |f(x)|^2\ dx)^{1/2} \geq \frac{1}{4\pi} \int_{\bf R} |f(x)|^2\ dx,$

valid for all ${f: {\bf R} \rightarrow {\bf C}}$ and all ${x_0,\xi_0 \in {\bf R}}$ , which we will not prove here (see for instance Exercise 47 of this previous set of lecture notes).

Another manifestation of the uncertainty principle is the following simple fact:

Lemma 20

(i) If ${f: {\bf R} \rightarrow {\bf C}}$ is an integrable function that has exponential decay in the sense that one has ${|f(x)| \leq C e^{-a|x|}}$ for all ${x \in {\bf R}}$ and some ${C,a>0}$ , then the Fourier transform ${\hat f: {\bf R} \rightarrow {\bf C}}$ is either identically zero, or only has isolated zeroes (that is to say, the set ${\{ \xi \in {\bf R}: \hat f(\xi) = 0 \}}$ is discrete.

(ii) If ${f: {\bf R} \rightarrow {\bf C}}$ is a compactly supported continuous function such that ${\hat f}$ is also compactly supported, then ${f}$ is identically zero.

Proof: For (i), we observe from Exercise 2(iii) that ${\hat f}$ extends holomorphically to a strip around the real axis, and the claim follows since non-zero holomorphic functions have isolated zeroes. For (ii), we observe from (i) that ${\hat f}$ must be identically zero, and the claim now follows from the Fourier inversion formula (Exercise 8). $\Box$

Lemma 20(ii) rules out the existence of a bump function whose Fourier transform is also a bump function, which would have been a rather useful tool to have in harmonic analysis over the reals. (Such functions do exist however in some non-archimedean domains, such as the ${p}$ -adics.) On the other hand, from Exercise 1 we see that we do at least have gaussian functions whose Fourier transform also decays as a gaussian. Unfortunately this is basically the best one can do:

Theorem 21 (Hardy uncertainty principle) Let ${f}$ be a continuous function which obeys the bound ${|f(x)| \leq C e^{-\pi ax^2}}$ for all ${x \in {\bf R}}$ and some ${C,a>0}$ . Suppose also that ${|\hat f(\xi)| \leq C' e^{-\pi \xi^2/a}}$ for all ${\xi \in {\bf R}}$ and some ${C'>0}$ . Then ${f(x)}$ is a scalar multiple of the gaussian ${e^{-\pi ax^2}}$ , that is to say one has ${f(x) = c e^{-\pi ax^2}}$ for some ${c \in {\bf C}}$ .

Proof: By replacing ${f}$ with the rescaled version ${x \mapsto f(x/a^{1/2})}$ , which replaces ${\hat f}$ with the rescaled version ${\xi \mapsto a^{1/2} \hat f(a^{1/2} \xi)}$ , we may normalise ${a=1}$ . By multiplying ${f}$ by a small constant we may also normalise ${C=C'=1}$ .

From Exercise 2(i), ${\hat f}$ extends to an entire function. By the triangle inequality, we can bound

$\displaystyle |\hat f(\xi+i\eta)| \leq \int_{\bf R} e^{-\pi x^2} e^{2\pi x \eta}\ dx$

for any ${\xi,\eta \in {\bf R}}$ . Completing the square ${e^{-\pi x^2} e^{2\pi x \eta} = e^{-\pi (x-\eta)^2} e^{\pi \eta^2}}$ and using ${\int_{\bf R} e^{-\pi (x-\eta)^2}\ dx = \int_{\bf R} e^{-\pi x^2}\ dx = 1}$ , we conclude the bound

$\displaystyle |\hat f(\xi+i\eta)| \leq e^{\pi \eta^2}.$

In particular, if we introduce the normalised function

$\displaystyle F(z) := e^{\pi z^2} \hat f(z)$

then

$\displaystyle |F(\xi+i\eta)| \leq e^{\pi \xi^2}. \ \ \ \ \ (15)$

In particular, ${|F|}$ is bounded by ${1}$ on the imaginary axis. On the other hand, from hypothesis ${F}$ is also bounded by ${1}$ on the real axis. We can now almost invoke the Phragmén-Lindeöf principle (Corollary 15) to conclude that ${F}$ is bounded on all four quadrants, but the growth bound we have (15) is just barely too weak. To get around this we use the epsilon of room trick. For any ${\varepsilon>0}$ , the function ${F_\varepsilon(z) := e^{\pi i\varepsilon z^2} F(z)}$ is still entire, and is still bounded by ${1}$ in magnitude on the real line. From (15) we have

$\displaystyle |F_\varepsilon(\xi+i\eta)| \leq e^{\pi \xi^2 - 2\varepsilon \pi \xi \eta}$

so in particular it is bounded by ${1}$ on the slightly tilted imaginary axis ${(2\varepsilon + i) {\bf R}}$ . We can now apply Corollary 15 in the two acute-angle sectors between ${(2\varepsilon+i) {\bf R}}$ and ${{\bf R}}$ to conclude that ${|F_\varepsilon(z)| \leq 1}$ in those two sectors; letting ${\varepsilon \rightarrow 0}$ , we conclude that ${|F(z)| \leq 1}$ in the first and third quadrants. A similar argument (using negative values of ${\varepsilon}$ ) shows that ${|F(z)| \leq 1}$ in the second and fourth quadrants. By Liouville’s theorem, we conclude that ${F}$ is constant, thus we have ${\hat f(z) = c e^{-\pi z^2}}$ for some complex number ${c}$ . The claim now follows from the Fourier inversion formula (Proposition 3(iv)) and Exercise 1. $\Box$

One corollary of this theorem is that if ${f}$ is continuous and decays like ${e^{-\pi ax^2}}$ or better, then ${\hat f}$ cannot decay any faster than ${e^{-\pi \xi^2/a}}$ without ${f}$ vanishing identically. This is a stronger version of Lemma 20(ii). There is a more general tradeoff known as the Gel’fand-Shilov uncertainty principle, which roughly speaking asserts that if ${f}$ decays like ${e^{-\pi a x^p}}$ then ${\hat f}$ cannot decay faster than ${e^{-\pi b x^q}}$ without ${f}$ vanishing identically, whenever ${1 < p,q < \infty}$ are dual exponents in the sense that ${\frac{1}{p}+\frac{1}{q}=1}$ , and ${a^{1/p} b^{1/q}}$ is large enough (the precise threshold was established in work of Morgan). See for instance this article of Nazarov for further discussion of these variants.

Exercise 22 If ${f}$ is continuous and obeys the bound ${|f(x)| \leq C (1+|x|)^M e^{-\pi ax^2}}$ for some ${M \geq 0}$ and ${C,a>0}$ and all ${x \in {\bf R}}$ , and ${\hat f}$ obeys the bound ${|\hat f(\xi)| \leq C' (1+|\xi|)^M e^{-\pi \xi^2/a}}$ for some ${C'>0}$ and all ${\xi \in {\bf R}}$ , show that ${f}$ is of the form ${f(x) = P(x) e^{-\pi ax^2}}$ for some polynomial ${P}$ of degree at most ${M}$ .

Remark 23 There are many further variants of the Hardy uncertainty principle. For instance we have the following uncertainty principle of Beurling, which we state in a strengthened form due to Bonami, Demange, and Jaming: if ${f}$ is a square-integrable function such that ${\int_{\bf R} \int_{\bf R} \frac{|f(x)| |\hat f(\xi)|}{(1+|x|+|\xi|)^N} e^{2\pi |x| |\xi|}\ dx d\xi < \infty}$ , then ${f}$ is equal (almost everywhere) to a polynomial times a gaussian; it is not difficult to show that this implies Theorem 21 and Exercise 22, as well as the Gel’fand-Shilov uncertainty principle. In recent years, PDE-based proofs of the Hardy uncertainty principle have been established, which have then been generalised to establish uncertainty principles for various Schrödinger type equations; see for instance this review article of Kenig. I also have some older notes on the Hardy uncertainty principle in this blog post. Finally, we mention the Beurling-Malliavin theorem, which provides a precise description of the possible decay rates of a function whose Fourier transform is compactly supported; see for instance this paper of Mashregi, Nazarov, and Khavin for a modern treatment.