Calculus of Variations Examples: Localization and Test Functions

The Fundamental Lemma works because test functions can be localized. This companion set concentrates on that mechanism: support, smooth bump functions, normalized narrowing probes, finite-dimensional blind spots, and the way local tests reveal behavior that a global average can hide.

1 Exercises

Exercise 1: Support and compact support

On the interval (0, 1), consider

η (x) = x(1 − x ), 1

(1)

( ( ) { 1 exp − 1 −-16-(x −-1∕2)2 , |x − 1∕2| < 1∕4, η2(x ) = ( 0, |x − 1∕2| ≥ 1∕4,

(2)

and

η3(x) = sin (πx ).

(3)

For each function, determine its support relative to [0, 1]. Which of these functions belongs to C_c^∞(0, 1)?

Exercise 2: Build a localized bump

Construct explicitly a nonnegative smooth bump function centered at

x = 0.60 0

(4)

with support contained in

(0.50,0.70).

(5)

State the properties of the function that are important in the proof of the Fundamental Lemma.

Exercise 3: Quantitative contradiction estimate

Suppose g ∈ C([0, 1]) and

g(x0) = 4

(6)

at some x₀ ∈ (0, 1). Continuity provides a radius r > 0 such that

g (x ) > 3

(7)

whenever |x − x₀| < r.

Let η ∈ C_c^∞(x₀ − r,x₀ + r) satisfy

∫ 1 η ≥ 0, η(x) dx = 1. 0

(8)

Prove that

∫ 1 g (x )η(x)dx > 3. 0

(9)

Explain why this contradicts the hypothesis of the Fundamental Lemma.

Exercise 4: A normalized bump sequence acts like a local probe

Let ρ ∈ C_c^∞(−1, 1) satisfy

∫ 1 ρ ≥ 0, −1ρ(s) ds = 1.

(10)

For x₀ ∈ (a,b) and sufficiently small 𝜖 > 0, define

1 ( x − x0 ) η𝜖(x ) = -ρ ------- . 𝜖 𝜖

(11)

Show that

∫ b η (x) dx = 1 a 𝜖

(12)

and prove that for every continuous g,

∫ b li𝜖m→0 g(x)η𝜖(x)dx = g(x0). a

(13)

Exercise 5: A finite collection of tests has a blind spot

On [−1, 1], define two test functions

η1(x) = 1 − x2, η2(x ) = x(1 − x2).

(14)

Find a nonzero quadratic function of the form

2 g(x) = x − c

(15)

such that

∫ 1 g (x )η1(x )dx = 0 −1

(16)

and

∫ 1 g(x)η2(x) dx = 0. −1

(17)

What does this demonstrate about testing against finitely many directions?

Exercise 6: A global average can hide a local sign

Let

g (x ) = x − 1, 0 ≤ x ≤ 1. 2

(18)

Show that

∫ 1 g(x)dx = 0. 0

(19)

Now let η be any nonnegative, nonzero smooth bump supported in (3∕4, 9∕10). Determine the sign of

∫ 1 g (x)η(x)dx. 0

(20)

Explain the localization principle illustrated by this comparison.

Exercise 7: Why one cannot cancel the test function

Suppose

∫ b g (x )η(x)dx = 0. a

(21)

Give two distinct mathematical reasons why the formal step

g(x)η(x) = 0 =⇒ g (x ) = 0

(22)

is not justified from this single integral equation. Use a concrete example to support your explanation.

Exercise 8: Zero weak derivative implies a constant

Suppose g ∈ C¹([a,b]) and

∫ b g (x )η′(x) dx = 0 a

(23)

for every η ∈ C_c^∞(a,b). Prove that g is constant on [a,b].

2 Solutions

Solution 1: Support and compact support

The support of a function is the closure of the set on which it is nonzero.

For

η1(x) = x(1 − x ),

(24)

the function is nonzero for every 0 < x < 1. Therefore

supp η = [0,1]. 1

(25)

It vanishes at the endpoints, but its support reaches the boundary of the open interval (0, 1). Hence it is not compactly supported inside (0, 1).

For η₂, the function is nonzero precisely for

1 3 --< x < --, 4 4

(26)

[ ] 1- 3- supp η2 = 4 ,4 .

(27)

The standard exponential bump joins smoothly to zero with all derivatives vanishing at the endpoints of its support. Thus

|------∞-------| -η2 ∈-C-c-(0,1).

(28)

For

η3(x) = sin (πx ),

(29)

the function is again nonzero throughout (0, 1) and vanishes only at the two endpoints. Therefore

supp η3 = [0,1],

(30)

so η₃ is not compactly supported inside (0, 1) even though it satisfies fixed-endpoint conditions.

This distinction is why the compact-support formulation and the fixed-endpoint formulation are related but not identical descriptions of the test class.

Solution 2: Build a localized bump

Take

r = 0.10

(31)

and define

( [ ] { ----------1----------- η(x) = exp − 1 − ((x − 0.60 )∕0.10 )2 , |x − 0.60 | < 0.10, ( 0, |x − 0.60 | ≥ 0.10.

(32)

Then

supp η = [0.50,0.70].

(33)

The properties important to the Fundamental-Lemma proof are:

η ∈ C_c^∞(0, 1);
η(x) ≥ 0;
η is not identically zero; and
its support lies entirely inside the neighborhood in which g has a known sign.

The exact exponential formula is secondary. The localization and sign properties are what drive the proof.

Solution 3: Quantitative contradiction estimate

Because η is supported inside (x₀ − r,x₀ + r),

∫ 1 ∫ x0+r g(x)η (x )dx = g(x)η(x) dx. 0 x0− r

(34)

On this support,

g (x ) > 3

(35)

and

η(x) ≥ 0.

(36)

Therefore

∫ 1 ∫ x0+r g (x)η(x)dx > 3 η(x)dx. 0 x0−r

(37)

Since η vanishes outside that interval and is normalized,

∫ x0+r η(x) dx = 1. x0− r

(38)

Hence

|--------------------| |∫ 1 | | g (x )η(x)dx > 3.| --0------------------

(39)

This cannot be reconciled with a hypothesis asserting that the same integral is zero for every test function. A nonzero value of continuous g has therefore been detected by a local probe.

Solution 4: A normalized bump sequence acts like a local probe

First compute the integral of η_𝜖. Substitute

x-−-x0- s = 𝜖 , dx = 𝜖 ds.

(40)

Because ρ is supported in (−1, 1),

∫ b ∫ 1 ∫ 1 η𝜖(x )dx = 1-ρ(s)𝜖ds = ρ(s)ds = 1. a − 1𝜖 − 1

(41)

Now consider

∫ b I𝜖 = g (x)η𝜖(x )dx. a

(42)

The same substitution gives

∫ 1 I𝜖 = g (x0 + 𝜖s)ρ(s) ds. −1

(43)

Subtract g(x₀), using the normalization of ρ:

∫ 1 I𝜖 − g(x0) = [g(x0 + 𝜖s) − g(x0)]ρ(s)ds. −1

(44)

Hence

∫ 1 |I𝜖 − g(x0)| ≤ sup |g(x0 + 𝜖s) − g(x0)| ρ(s)ds. |s|≤1 −1

(45)

The last integral equals one, so

|I𝜖 − g(x0 )| ≤ sup |g(x) − g(x0)|. |x− x0|≤𝜖

(46)

Continuity of g at x₀ makes the right-hand side tend to zero as 𝜖 → 0. Therefore

|----------------------------| | ∫ b | |lim g(x)η𝜖(x)dx = g(x0).| -𝜖→0--a----------------------

(47)

Figure. A normalized family of bumps becomes narrower and taller while retaining unit area. Integrating a continuous g against these test functions increasingly samples only the behavior near x₀.

This is a precise mathematical version of the statement that test functions act as local probes.

Solution 5: A finite collection of tests has a blind spot

We seek

g(x) = x2 − c.

(48)

For the first test function,

∫ 1 2 2 0 = (x − c)(1 − x )dx. −1

(49)

Thus

∫ −11 x2(1 − x2)dx c = -∫-1-------2-----. − 1(1 − x )dx

(50)

Compute

∫ 1 2 4- −1(1 − x )dx = 3

(51)

and

∫ 1 ( 1 1) 4 x2(1 − x2)dx = 2 --− -- = --. − 1 3 5 15

(52)

Therefore

|------| | 1-| |c = 5.| --------

(53)

g (x ) = x2 − 1. 5

(54)

For the second test function,

( ) 2 1 2 g(x)η2(x) = x − -- x(1 − x ). 5

(55)

The first and third factors are even and the middle factor x is odd, so the entire product is odd. Its integral over the symmetric interval [−1, 1] therefore vanishes automatically:

∫ 1 −1g(x)η2(x) dx = 0.

(56)

Figure. The nonzero quadratic g = x² − 1∕5 is orthogonal to both selected test functions. Finitely many test directions leave an infinite-dimensional space of possible undetected functions.

The Fundamental Lemma avoids this blind spot by requiring the identity for an entire separating class of localized test functions.

Solution 6: A global average can hide a local sign

Directly,

∫ 1( ) [ 2 ]1 x − 1- dx = x--− x- = 0. 0 2 2 2 0

(57)

Thus the global average does not reveal that g changes sign.

On the interval (3∕4, 9∕10),

1 g(x) = x − --> 0. 2

(58)

If η is nonnegative, nonzero, and supported there, then

g(x )η(x) ≥ 0

(59)

and is positive on a set of positive measure. Hence

|--------------------| |∫ 1 | | g (x )η(x)dx > 0.| --0------------------

(60)

Figure. The unweighted global integral cancels positive and negative contributions. A localized nonnegative bump placed entirely in a positive region cannot experience that cancellation.

This is the essence of localization: the test function can suppress irrelevant parts of the domain and interrogate one chosen neighborhood.

Solution 7: Why one cannot cancel the test function

There are at least two independent problems with the proposed cancellation.

First, an integral equation is not a pointwise equation. From

∫ b gηdx = 0 a

(61)

one cannot conclude

g (x )η(x) = 0

(62)

at every point. Positive and negative contributions may cancel.

Second, test functions typically have zeros. Even if the product were known to vanish pointwise, division by η(x) would not be valid where η(x) = 0.

For a concrete example on [0, 1], take

1 g (x ) = x − 2, η(x) = x(1 − x).

(63)

As shown in CV03E1,

∫ 1 g (x )η(x)dx = 0, 0

(64)

but neither g nor gη is identically zero.

The correct argument is not cancellation. It is the ability to choose many localized test functions, including ones supported entirely inside any neighborhood where a continuous g has a definite sign.

Solution 8: Zero weak derivative implies a constant

Start with

∫ b g(x)η′(x)dx = 0. a

(65)

Integrate by parts:

∫ ∫ b ′ b b ′ g η dx = [gη]a − g ηdx. a a

(66)

Because η has compact support inside (a,b), it vanishes near both endpoints, so

b [gη]a = 0.

(67)

Therefore

∫ b g ′(x)η(x) dx = 0 a

(68)

for every η ∈ C_c^∞(a,b). Since g′∈ C([a,b]), the Fundamental Lemma gives

′ g (x) = 0

(69)

throughout (a,b). Hence

|----------| -g(x)-=-C--|

(70)

for some constant C on [a,b].

This exercise is the simplest example of a weak-derivative statement: a function whose distributional derivative is zero is constant (with suitable connectedness assumptions).

3 Summary

Localization is the operational content of the Fundamental Lemma. Smooth bump functions can be placed inside arbitrarily small neighborhoods, normalized to behave as local averaging probes, and varied independently enough to eliminate the blind spots left by any finite collection of test directions. That is why an integral identity valid for all test functions can force a pointwise or almost-everywhere conclusion.

"Calculus of Variations Examples: Localization and Test Functions" is owned by bloftin.
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