HOMEWORK ASSIGNMENT 1 SOLUTIONS

Exercise 1. (S(Rn) is closed under convolution)

Given f, g ∈ S(Rn), show that f ∗ g ∈ S(Rn):

a) Directly from the definition.

b) Using the Fourier transform.

Solution:

a) Given α, β ∈ Nn, we want to show that:

sup∣∣xα∂βx (f ∗ g)

∣∣ <∞.We know that ∂βx (f ∗ g) = (∂βxf) ∗ g and ∂βxf ∈ S(Rn). Hence, we can reduce to the case whenβ = 0. Now, we observe that:

sup∣∣xα(f ∗ g)

∣∣ ≤ sup ∣∣(1 + |x|)|α| · (|f | ∗ |g|)∣∣.

Here, we used the triangle inequality to observe that |f ∗g| ≤ |f |∗|g| and we used the notation |α| :=∑nj=1 αj whenever α = (α1, α2, . . . , αn). By the triangle inequality, we know that |x| ≤ |x−y|+ |y|.

Hence:(1 + |x|) ≤ (1 + |x− y|) · (1 + |y|).

Consequently:

(1 + |x|)|α| · (|f | ∗ |g|)(x) = (1 + |x|)α ·∫Rn|f(x− y)| · |g(y)| dy

≤∫Rn

((1 + |x− y|)α · |f(x− y)|

)·((1 + |y|)|α| · |g(y)|

)dy

= (f1 ∗ g1)(x),

where:f1(y) := (1 + |y|)α · |f(y)|, g1(y) := (1 + |y|)α · |g(y)|.

Since f ∈ S(Rn), it follows that f1 ∈ L1(Rn). Moreover, since g ∈ S(Rn), it follows thatg1 ∈ L∞(Rn). Consequently, by Young’s inequality, we deduce that f1 ∗ g1 ∈ L∞(Rn). Theclaim now follows.

b) We know that (f ∗ g) = f · g. (We recall from class that this identity holds in general for

L1 functions.) Since f, g ∈ S(Rn), it follows that f , g ∈ S(Rn). Hence f · g ∈ S(Rn). Now, weknow that f ∗g is an L2 function (from Young’s inequality) whose Fourier transform is an element of

S(Rn). We know that there exists φ ∈ S(Rn) such that (f ∗g) = φ. By the injectivity of the Fouriertransform on L2 (we saw two different proofs of this fact in class), it follows that f ∗ g = φ ∈ S(Rn).�

Exercise 2. (Approximations of the identity and the Fourier transform)

Let f ∈ L2(Rn) and let φ ∈ L1(Rn) with∫Rn φ(x) dx = 1 be given. We recall that, given ε > 0, we

define φε(x) := 1εnφ(xε ). Using only properties of the Fourier transform, show that:

f ∗ φε → f in L2 as ε→ 0.

1

2 HOMEWORK ASSIGNMENT 1 SOLUTIONS

Solution:

Let us note that, for F ∈ L2(Rn) and for G ∈ L1(Rn), it is the case that:

(1)(F ∗G

)= F · Gas L2 functions. We show (1) by density. Namely, we find sequences (φk) and (ψ`) in S(Rn) suchthat:

φk → F in L2

ψ` → G in L1.

Namely, for fixed k and `, we know that:

(2)(φk ∗ ψ`

)= φk · ψ`If we fix `, we know from Young’s inequality that:

φk ∗ ψ` → F ∗ ψ` in L2 as k →∞.Another application of Young’s inequality shows that:

F ∗ ψ` → F ∗G in L2 as `→∞.Hence, by L2-unitarity of the Fourier transform:

(3) lim`→∞

limk→∞

(φk ∗ ψ`

) → (F ∗G

) in L2.

On the other hand, if we fix `, Holder’s inequality shows that:

φk · ψ` → F · ψ` in L2 as k →∞.Namely:

‖φk · ψ` − F · ψ`‖L2 ≤ ‖φk − F‖L2 · ‖ψ`‖L∞≤ ‖φk − F‖L2 · ‖ψ`‖L1 .

Here we used that the Fourier transform is an isometry on L2 and that it is a contraction from L1

to L∞. The above quantity indeed converges to zero as k →∞.Another application of Holder’s inequality shows that:

F · ψ` → F · G in L2 as `→∞.Namely,

‖F · ψ` − F · G‖L2 ≤ ‖F‖L2 · ‖ψ` − G‖L∞ ≤ ‖F‖L2 · ‖ψ` −G‖L1 .

We used the same boundedness properties of the Fourier transform as in the above estimate. Hence:

(4) lim`→∞

limk→∞

φk · ψ` → F · G in L2.

The claim (1) now follows from (2), (3) and (4).Having proved (1), we can compute:(

f ∗ φε)(ξ) = f(ξ) · φε(ξ) = f(ξ) · φ(ε ξ).

Since φ ∈ L1, we know that φ is continuous, hence:

limε→0

φ(ε ξ) = φ(0) =

∫Rnφ(y) dy = 1.

In other words, it follows that f(ξ) · φ(ε ξ) converges to f(ξ) pointwise almost surely as ε→ 0.

Moreover, we know that φ is bounded. Hence, by the Dominated convergence theorem, it followsthat: (

f ∗ φε) = f(ξ) · φ(ε ξ)→ f in L2 as ε→ 0.

By the L2-unitarity of the Fourier transform, it follows that:

f ∗ φε → f in L2 as ε→ 0. �

HOMEWORK ASSIGNMENT 1 SOLUTIONS 3

Exercise 3. (An explicit formula for the Fourier transform on L2)

Given f ∈ L2(Rn), show that the following limits hold in the L2 sense:

(5) f(ξ) = limM→∞

∫|x|≤M

f(x) · e−2πix·ξ dx

and

(6) f(x) = limM→∞

∫|ξ|≤M

f(ξ) · e2πix·ξ dξ.

We note that (5) gives us an explicit formula for the Fourier transform on L2 (which doesn’t involvefinding an approximating sequence in S(Rn) ). The identity in (6) is the analogue of the Fourierinversion formula for the Fourier transform on L2.

Solution: Let us first prove (5). We know that, for all M ∈ N, it is the case that

fM := f · χ{|x|≤M} ∈ L1(Rn)

by using the Cauchy-Schwarz inequality. Moreover:

fM → f in L2 as M →∞.Consequently, by the L2-unitarity of the Fourier transform:

fM → f in L2 as M →∞.Furthermore, since fM ∈ L1(Rn), we know that:

fM (ξ) =

∫|x|≤M

f(x)e−2πix·ξ.

The claim (5) follows.We now prove (6). Let us find a sequence (φk) in S(Rn) such that φk → f in L2 as k →∞. We

observe that:

f−∫|ξ|≤M

f(ξ)e2πix·ξ dξ = (f−φk)+(φk−

∫|ξ|≤M

φk(ξ)e2πix·ξ dξ)

+(∫|ξ|≤M

φk(ξ)e2πix·ξ dξ−∫|ξ|≤M

f(ξ)e2πix·ξ dξ).

We can use the Fourier inversion formula on S(Rn) to write:

φk =

∫Rnφk(ξ)e2πix·ξ dξ.

Hence, the above expression equals:

(f − φk) +

∫|ξ|>M

φk(ξ) dξ +

∫|ξ|≤M

(φk(ξ)− f(ξ)

)e2πix·ξ dξ =

= (f − φk) +(φk · χ{|ξ|>M}

)(−x) +(

(φk − f ) · χ{|ξ|≤M})(−x).

Hence:

∥∥f − ∫{|ξ|≤M}

f(ξ)e2πix·ξ dξ∥∥L2x≤ ‖f −φk‖L2

x+∥∥(φk ·χ{|ξ|>M})∥∥L2

x+∥∥((φk − f ) ·χ{|ξ|≤M}

)∥∥L2x.

By unitarity, this equals:

‖f − φk‖L2x

+ ‖φk · χ{|ξ|>M}‖L2ξ

+ ‖(φk − f ) · χ{|ξ|≤M}‖L2ξ,

which is:

≤ ‖f − φk‖L2x

+ ‖φk · χ{|ξ|>M}‖L2ξ

+ ‖φk − f‖L2ξ

= 2‖f − φk‖L2x

+ ‖φk · χ{|ξ|>M}‖L2ξ.

4 HOMEWORK ASSIGNMENT 1 SOLUTIONS

Given ε > 0, we can find k ∈ N sufficiently large so that ‖f −φk‖L2x≤ ε

4 . For this k, we find M0 > 0

such that for all M ≥M0, it is the case that ‖φk · χ{|ξ|>m}‖L2ξ≤ ε

2 .

In particular, it follows that, for M ≥M0, it is the case that:

‖f −∫|ξ|≤M

f(ξ)e2πix·ξ dξ‖L2x≤ ε. �

Exercise 4. (The Fourier inversion formula on L1)

Given f ∈ L1(Rn), we will show that:

(7)

∫Rnf(ξ) · e2πix·ξ · e−επ|ξ|

2

dξ → f in L1 as ε→ 0.

a) Explain why, given f ∈ L1(Rn) and ε > 0, the left-hand side of (7) makes sense. What problemswould one face in general for ε = 0?

b) Use the formula∫φ · ψ dx =

∫φ · ψ dx to write the left-hand side of (7) as an integral in-

volving f . Make sure to rigorously justify each step.

c) By writing the result obtained in b) in the appropriate way, deduce the claim in (7).

d) Can one replace e−επ|ξ|2

in (7) by some other function (which is not a rescaled Gaussian) ?

Solution:

a) If f ∈ L1(Rn), then f ∈ L∞(Rn). Since e−επ|ξ|2 ∈ L1

ξ , it follows that∫Rnf(ξ) · e2πix·ξ · e−επ|ξ|

2

dξ

makes sense.

b) We can use the product formula∫φ ·ψ dx =

∫φ · ψ dx when φ ∈ L1 and ψ ∈ S. This was shown

in class by using a density argument (in fact, we showed that it was possible to take φ, ψ ∈ L1).Consequently: ∫

Rnf(ξ) · e2πix·ξ · e−επ|ξ|

2

dξ =

=

∫Rnf(y) · Fξ 7→y

(e2πix·ξ · e−επ|ξ|

2)(y) dy.

Here, by Fξ 7→y we denote the Fourier transform from the ξ variables to the y variables. In other

words, for a Schwartz function G = G(ξ), we let(F−1ξ 7→yG

)(y) :=

∫Rn G(ξ) ·e−2πiy·ξ dξ. We note that

the above application of the product formula was indeed rigorous since f ∈ L1 and e2πix·ξ ·e−επ|ξ|2 ∈S(Rn).We can simplify the above expression as:∫

Rnf(y) · Fξ 7→y

(e−επ|ξ|

2)(y − x) dy

=

∫Rnf(y) · 1

εn2· e−

π|y−x|2ε dy

=

∫Rnf(y) · 1

εn2· e−

π|x−y|2ε dy

HOMEWORK ASSIGNMENT 1 SOLUTIONS 5

=

∫Rnf(y)φ√ε (x− y) dy

where

φ(y) := e−π|y|2

and φλ(y) :=1

λnφ( yλ

).

In the above calculations, we used the modulation and dilation properties of the Fourier transform.

c) We recognize that the integral in part b) is f ∗ φ√ε (x). Since φ ∈ L1(Rn) and∫Rn φ(y) dy = 1,

it follows from properties of approximations of the identity that f ∗ φ√ε → f in L1 as ε→ 0.

d) We can consider more general even Schwartz functions which are equal to 1 at the origin. Forexample, let Φ ∈ S(Rn) be such that Φ(0) = 1. Then φ := F(Φ) ∈ S(Rn) is even and

∫φ(y) dy = 1.

We would like to show that:∫Rnf(ξ) · e2πix·ξ · Φ(

√ε ξ) dξ → f in L1 as ε→ 0.

We define the dilation φλ analogously as in part b). By using the product formula as in part b) andrecalling the Fourier inversion formula, it follows that:∫

Rnf(ξ) · e2πix·ξ · Φ(

√ε ξ) dξ =

∫Rnf(y) · 1

εn2· φ(y − x√

ε

)dy = f ∗ φ√ε(x)

=

∫Rnf(y) · 1

εn2· φ(x− y√

ε

)dy = f ∗ φ√ε(x).

As before, we note that f ∗ φ√ε → f in L1 as ε→ 0. �

Exercise 5. (The heat equation on Rn)

a) Given f ∈ S(Rn), use the Fourier transform (in x) to solve the following initial value prob-lem: {

∂u∂t −∆u = 0 on Rnx × (0,+∞)t

u∣∣t=0

= f on Rnx .

Write the solution as u(x, t) = Kt ∗ f(x), for some function Kt ∈ S(R). Determine the function Kt

explicitly.

b) Show that ‖u(x, t)− f(x)‖L2(Rnx ) → 0 as t→ 0+.

Solution:

a) We take the Fourier transform in the x variable of the equation to deduce that:{∂u∂t + 4π2|ξ|2u = 0 on Rnξ × (0,+∞)t

u∣∣t=0

= f on Rnξ .

More precisely, here u(ξ, t) :=∫Rn u(x, t) ·e−2πix·ξ dx. For fixed ξ ∈ Rn, we can consider the problem

as an ODE in the t variable. It has the solution

u(ξ, t) = A(ξ) · e−4π2|ξ|2t

for some function A = A(ξ). Substituting t = 0, it follows that A(ξ) = f(ξ).Consequently:

u(ξ, t) = e−4π2|ξ|2t · f(ξ).

Hence, by the convolution property of the Fourier transform, we can write u = Kt ∗ f where:

6 HOMEWORK ASSIGNMENT 1 SOLUTIONS

Kt(x) := F−1(e−4π

2|ξ|2t) = F−1(e−π|

√4πtξ|2) =

1

(4πt)n2e−|x|24t .

In other words:

u(x, t) = (Kt ∗ f)(x) =( 1

(4πt)n2e−|·|24t ∗ f

)(x).

Remark: One can check that u ∈ C(Rt,S(Rn)). The easiest way to do this is to use the formula foru(ξ, t). Having done this, the calculations in part a) make sense rigorously.

b) By construction, we know that K := K1 is an L1 function whose integral equals to 1. Hence,by properties of approximations of the identity, it follows that, for all 1 ≤ p ≤ ∞, it is the casethat: u(·, t) = Kt ∗ f → f in Lp as t→ 0+. In particular, this holds for p = 2. Let us remark thatthe convergence result holds in L∞ since we assumed that f ∈ S(Rn). In particular, f is an L∞

function that vanishes at infinity.

Alternatively, we can work on the Fourier side, and observe that, as in Exercise 2, that by the

Dominated convergence theorem u(ξ, t) = e−4π2|ξ|2tf(ξ) converges to f in L2(ξ) as t → 0+. The

claim (when p = 2) then follows from the L2-unitarity of the Fourier transform. �

Exercise 6. (The uncertainty principle revisited)

In this exercise, we give a quantitative version of the uncertainty principle. We recall that this

principle states that it is not possible for a function f and for its Fourier transform f to be simul-taneously localized in space and in frequency, respectively. Let us work in dimension n = 1.

Let f ∈ S(R) be a function such that∫ +∞−∞ |f |

2 dx = 1.

a) Show that: (∫ +∞

−∞x2 · |f(x)|2 dx

)·(∫ +∞

−∞ξ2 · |f(ξ)|2 dξ

)≥ 1

16π2.

b) More generally, show that for all x0, ξ0 ∈ R, it is the case that:(∫ +∞

−∞(x− x0)2 · |f(x)|2 dx

)·(∫ +∞

−∞(ξ − ξ0)2 · |f(ξ)|2 dξ

)≥ 1

16π2.

c) For which functions does equality hold in b)?

d) Give a one-sentence interpretation of b) as a version of the uncertainty principle.

Solution:

a) Let f ∈ S(Rn) with ‖f‖L2 = 1 be given. Then, using integration by parts, it follows that:

1 =

∫ +∞

∞|f(x)|2 dx = −

∫ +∞

−∞−x ·

(|f |2

)′(x) dx =

∫ +∞

−∞

(− x · f · f ′ − x · f ′ · f

)dx.

The boundary terms at infinity vanish since f ∈ S(Rn). We now use the Cauchy-Schwarz inequality

for the inner product on L2(R) given by (φ, ψ) 7→∫ +∞−∞ φ · ψ dx to deduce that this expression is:

(8) ≤ 2 ·(∫ +∞

−∞x2 · |f(x)|2 dx

) 12 ·(∫ +∞

−∞|f ′(x)|2 dx

) 12

.

HOMEWORK ASSIGNMENT 1 SOLUTIONS 7

By the L2-unitarity of the Fourier transform, this is:

= 2 ·(∫ +∞

−∞x2 · |f(x)|2 dx

) 12 ·(∫ +∞

−∞|f ′ (ξ)|2 dξ

) 12

= 2 ·(∫ +∞

−∞x2 · |f(x)|2 dx

) 12 ·(∫ +∞

−∞4π2ξ2 · |f(ξ)|2 dξ

) 12

.

In the last line, we used the fact that f ′ (ξ) = 2πiξ · f(ξ). Squaring both sides of the inequality andcollecting terms, it follows that:(∫ +∞

−∞x2 · |f(x)|2 dx

)·(∫ +∞

−∞ξ2 · |f(ξ)|2 dξ

)≥ 1

16π2.

b) Let us define:

g(x) := f(x+ x0) · e−2πix·ξ0 .Then g ∈ S(Rn) and ‖g‖L2 = 1. Moreover, by the dilation and translation properties of the Fouriertransform, it follows that:

g(ξ) = f(ξ + ξ0) · e2πix0·ξ.

Hence: ∫ +∞

−∞x2 · |g(x)|2 dx =

∫ +∞

−∞x2 · |f(x+ x0)|2 dx =

∫ +∞

−∞(x− x0)2 · |f(x)|2 dx

and ∫ +∞

−∞ξ2 · |g(ξ)|2 dξ =

∫ +∞

−∞ξ2 · |f(ξ + ξ0)|2 dξ =

∫ +∞

−∞(ξ − ξ0)2 · |f(ξ)|2 dξ.

The claim now follows from part a) when applied to the function g.

c) By using the correspondence in part b), we can reduce to the case when x0 = 0 and ξ0 = 0.Namely, if the function g achieves the minimum for x0 = 0 and ξ0 = 0, then the functionf(x) := g(x− x0) · e2πix·ξ0 achieves the minimum for general x0 and ξ0 and vice versa. We now goback to the proof in part a). The only place in which there is a potential inequality is in step (8).We know that equality holds in this step if and only if the functions x · f and f ′ are proportional.In other words, equality holds if and only if there exists an α ∈ R such that:

f ′ = −2παx · f.(The choice for such a normalization of α will simplify the calculations that follow.)

In other words, we want:

f = Ae−απx2

forA ∈ C with |A| = 1

‖e−απx2‖L2

.

Now, ‖e−απx2‖L2 = ‖e−π(x√α)2‖L2 = α−

14 by a change of variables and since ‖e−πx2‖L2 = 1.

Hence, when x0 = 0 and ξ0 = 0, equality holds in b) if and only if:

f = α14 · eiθ · e−απx

2

for some α > 0 and for some θ ∈ R.For general x0, ξ0, equality holds in b) if and only if:

f = α14 · eiθ · e−απ(x−x0)

2

· e2πix·ξ0

for some α > 0 and for some θ ∈ R.

d) The inequality in b) tells us that, for all x0 and ξ0, it is not possible for f to be concentrated

near x0 and for f to be concentrated near ξ0 at the same time. �

Remark: We see from part c) that the functions which are the “most concentrated” are obtained

8 HOMEWORK ASSIGNMENT 1 SOLUTIONS

from the Gaussian by dilation, translation and modulation. For fixed x0, ξ0, the Gaussian is centeredat x0 and oscillates with frequency ξ0.