Calculus of Variations: The Fundamental Lemma

CV02 ended with the identity

∫ b ( d ) δJ[y;η] = Fy − --Fy ′ η(x )dx a dx

(1)

for fixed-endpoint variations, after integration by parts. If y is stationary, this integral vanishes for every admissible variation η.

The decisive question is then

If an integral against every sufficiently localized test function is zero, what can be concluded about the function multiplying that test function?

The answer is the Fundamental Lemma of the Calculus of Variations. In its classical form it says that a continuous function g satisfying

∫ b g(x)η(x)dx = 0 a

(2)

for every admissible test function η must satisfy

g (x ) = 0

(3)

throughout the interval.

This theorem is the logical hinge of the Euler–Lagrange derivation. It is not valid because one can “cancel” η from an integral. It is valid because the family of test functions is rich enough to probe arbitrarily small regions. If g were positive or negative anywhere, one could localize a test function near that region and force the integral to have the same sign.

1 Learning objectives

After this entry, the reader should be able to

define the support and compact support of a test function;
explain the localization principle behind variational test functions;
state the classical Fundamental Lemma with its hypotheses;
prove the lemma by contradiction using a localized nonnegative bump function;
identify exactly where continuity of g enters the proof;
derive the equivalent fixed-endpoint formulation;
explain why checking only one or finitely many variations is insufficient;
understand why the conclusion becomes “almost everywhere” for merely integrable functions;
apply the lemma to an integral identity arising from a first variation; and
distinguish an integral identity from the pointwise differential equation that follows from the lemma.

2 Test functions and support

The proof depends on the ability to choose variations that are concentrated in small parts of the interval.

2.1 Support

For a function η : (a,b) → ℝ, the support is the closure of the set on which η is nonzero:

---------------------- supp η = {x ∈ (a,b) : η(x) ⁄= 0}.

(4)

A function has compact support in (a,b) if its support is a compact subset of the open interval. Such a function vanishes not only at the endpoints but in entire neighborhoods of the endpoints.

The standard notation

C ∞ (a,b) c

(5)

denotes infinitely differentiable functions with compact support in (a,b). These are often called test functions or bump functions.

2.2 Why compact support is useful

Suppose one wants to determine whether an unknown continuous function g is positive near some point x₀. A test function can be chosen so that

η(x) ≥ 0 everywhere;
η is not identically zero; and
η vanishes outside a small interval surrounding x₀.

Then the integral

∫ b g(x)η(x)dx a

(6)

receives contributions only from that small neighborhood. Test functions therefore provide a mathematical microscope: they allow an integral statement to detect local behavior.

Figure. Localization mechanism behind the Fundamental Lemma. If a continuous g is positive at x₀, then continuity makes it positive on a whole neighborhood of x₀. A nonnegative bump function supported inside that neighborhood forces ∫ gη dx > 0.

3 The classical Fundamental Lemma

Theorem: Fundamental Lemma of the Calculus of Variations. Let

g ∈ C ([a,b]).

(7)

Suppose

∫ b g(x)η(x)dx = 0 a

(8)

for every test function

η ∈ C ∞c (a, b).

(9)

Then

g(x) = 0 for every x ∈ (a,b).

(10)

Because g is continuous on [a,b], the same conclusion extends to the endpoints by continuity, so g ≡ 0 on [a,b].

4 Proof by localization and contradiction

The proof is short, but each step carries important information.

4.1 Step 1: assume the conclusion is false

Suppose, for contradiction, that g is not identically zero. Then there is some point x₀ ∈ (a,b) such that

g(x ) ⁄= 0. 0

(11)

There are two sign cases. It is enough to prove one, because the other is identical after changing signs. Assume

g(x0) > 0.

(12)

4.2 Step 2: use continuity

Let

1 m = -g(x0) > 0. 2

(13)

Because g is continuous at x₀, there exists r > 0 small enough that

(x0 − r,x0 + r) ⊂ (a,b)

(14)

and

g(x) > m

(15)

for every x ∈ (x₀ − r,x₀ + r).

This is the exact point at which continuity enters the classical proof. A positive value at one point is promoted to a positive lower bound on an entire neighborhood.

4.3 Step 3: choose a localized nonnegative test function

Choose

η ∈ C∞c (x0 − r,x0 + r)

(16)

such that

η (x ) ≥ 0

(17)

and η is not identically zero.

For example, after rescaling, one may use a standard smooth bump of the form

( ( ) { --------1-------- η(x) = exp − 1 − ((x − x0)∕r)2 , |x − x0 | < r, ( 0, |x − x | ≥ r. 0

(18)

Its exact formula is not important. What matters is that it is nonnegative, nonzero, smooth, and supported entirely inside the region where g > m.

4.4 Step 4: determine the sign of the integral

Since η vanishes outside (x₀ − r,x₀ + r),

∫ b ∫ x0+r g(x)η(x )dx = g(x)η(x) dx. a x0− r

(19)

On this interval,

g(x) > m

(20)

and

η(x) ≥ 0.

(21)

Therefore

∫ ∫ b x0+r g(x)η(x) dx ≥ m η(x) dx. a x0−r

(22)

Because m > 0 and η is nonnegative and not identically zero,

∫ x0+r η(x) dx > 0. x0− r

(23)

Hence

∫ b g (x )η(x)dx > 0. a

(24)

But the hypothesis says that this integral equals zero for every test function. This is a contradiction.

4.5 Step 5: exclude the negative case

If instead

g(x0) < 0,

(25)

continuity gives a neighborhood on which g remains strictly negative. The same nonnegative bump function then yields

∫ b g (x )η(x)dx < 0, a

(26)

again contradicting the hypothesis.

Therefore no point with g(x₀)≠0 can exist. Thus

g (x ) = 0

(27)

throughout (a,b), and continuity extends the result to [a,b].

Figure. Logical structure of the classical proof. A single nonzero value of a continuous g creates a same-sign neighborhood. A localized nonnegative test function then makes the integral nonzero, contradicting the assumption that it vanishes for every test function.

5 Equivalent fixed-endpoint formulation

Many introductory variational derivations use variations satisfying

η(a) = η(b) = 0

(28)

rather than explicitly introducing compact support.

A useful corollary is therefore the following.

Corollary. Let g ∈ C([a,b]). Suppose

∫ b g(x)η(x)dx = 0 a

(29)

for every continuously differentiable function η satisfying

η(a) = η(b) = 0.

(30)

Then g ≡ 0 on [a,b].

Reason. Every smooth compactly supported test function in (a,b) also satisfies the endpoint conditions. Thus the hypothesis includes the test functions required by the Fundamental Lemma, so the theorem applies immediately.

This illustrates a general logical rule: if an integral identity is known for a large class of variations, it is valid for any smaller test-function class contained within it. One may therefore restrict attention to especially useful localized variations.

6 Why the test function cannot be cancelled

A common informal statement is

Since η is arbitrary, ∫ gη dx = 0 implies g = 0.

The conclusion is correct under the hypotheses of the lemma, but the language can hide the actual logic. There is no algebraic cancellation rule

∫ g (x )η(x)dx = 0 =⇒ g(x) = 0

(31)

for one fixed η.

For example, on [0, 1] let

g(x) = sin(2πx)

(32)

and choose only

η1(x) = sin (πx ).

(33)

Orthogonality gives

∫ 1 sin(2πx )sin(πx )dx = 0, 0

(34)

although g is certainly not zero.

But if one is also allowed to choose

η2(x) = sin(2πx ),

(35)

then

∫ 1 ∫ 1 g (x )η (x)dx = sin2(2πx) dx = 1-> 0. 0 2 0 2

(36)

The theorem works because the identity holds for every admissible test function, including localized functions specifically chosen to expose any nonzero region of g.

Figure. One test direction can miss a nonzero function through orthogonality. The Fundamental Lemma requires the integral identity for an entire separating class of test functions, not for one selected variation.

7 A useful weighted-square test

Sometimes the nonzero character of an integral can be demonstrated without a compactly supported bump.

Suppose on [0, 1]

g(x) = 3x − 1.

(37)

Choose the endpoint-zero test function

η(x) = x(1 − x)g(x).

(38)

Then

η(0) = η(1) = 0

(39)

and

∫ ∫ 1 1 2 g(x)η (x )dx = x(1 − x)g(x) dx. 0 0

(40)

Since x(1 −x) > 0 for 0 < x < 1 and g² ≥ 0, the integrand is nonnegative and not identically zero. Therefore

∫ 1 g (x )η(x)dx > 0. 0

(41)

Thus this g cannot satisfy the hypothesis of the Fundamental Lemma.

This argument is useful pedagogically, but it is not a replacement for the localized proof. Its admissibility depends on the regularity of g, whereas the bump-function proof keeps the test function independently smooth and makes the localization mechanism explicit.

8 Where continuity matters

The classical statement concluded

g(x) = 0 for every x.

(42)

That pointwise conclusion uses continuity. If a function is allowed to differ from zero only on a set of measure zero, ordinary integration cannot detect those exceptional values.

For example, consider the function

{ 1, x = x , g (x ) = 0 0, x ⁄= x0.

(43)

For every continuous test function η,

∫ b g (x )η(x)dx = 0, a

(44)

because changing the value of an integrand at one point does not change its Riemann or Lebesgue integral. Yet g(x₀) = 1.

Thus without continuity one should not expect pointwise equality everywhere. The natural modern conclusion is equality almost everywhere.

9 Modern weak form

The functional-analytic version of the lemma is often stated as follows.

Weak Fundamental Lemma. Let

g ∈ L1 (a,b). loc

(45)

∫ b g(x)η(x)dx = 0 a

(46)

for every

η ∈ C ∞c (a, b),

(47)

then

g (x ) = 0

(48)

for almost every x ∈ (a,b).

In the language of distributions, the hypothesis says that the distribution induced by g is the zero distribution. Two locally integrable functions represent the same distribution exactly when they agree almost everywhere.

The classical theorem is recovered immediately when g is continuous: if a continuous function is zero almost everywhere but nonzero at one point, continuity would make it nonzero on an interval of positive length, a contradiction. Hence continuity upgrades almost-everywhere equality to pointwise equality.

A full proof of the L_loc¹ version belongs to real analysis and distribution theory. The classical localized-bump proof above contains the physical intuition needed for the variational calculations in this course.

10 Vector-valued extension

Physics often produces several coupled dependent variables. Suppose

⌊ ⌋ g1(x) | . | g(x) = ⌈ .. ⌉ gn(x)

(49)

is continuous and

∫ b T g(x) η (x)dx = 0 a

(50)

for every smooth compactly supported vector test function η.

Choose a test function with only its kth component nonzero. The integral reduces to

∫ b gk(x)ηk(x)dx = 0 a

(51)

for every scalar test function η_k. The scalar Fundamental Lemma implies

g (x) = 0. k

(52)

Since this holds for every component,

g(x) = 0.

(53)

This componentwise argument is the basis for coupled Euler–Lagrange equations in systems with many generalized coordinates or fields.

11 Application to the first variation

Return to the first-order functional

∫ b J[y] = F (x,y,y′)dx. a

(54)

CV02 derived

∫ b ′ δJ[y;η] = a (Fy η + Fy′η )dx.

(55)

After integration by parts,

∫ ( ) b b -d- δJ[y;η] = [Fy ′η]a + Fy − dx Fy′ η dx. a

(56)

For fixed endpoints,

η(a) = η(b) = 0,

(57)

∫ b( ) δJ [y;η ] = Fy − d-Fy ′ η dx. a dx

(58)

If y is stationary, then

δJ [y;η] = 0

(59)

for every admissible variation. Define

d g(x ) = Fy − --Fy ′. dx

(60)

Under the regularity assumptions needed to make g continuous, the Fundamental Lemma gives

-d- Fy − dx Fy′ = 0.

(61)

This is the Euler–Lagrange equation.

CV04 will derive this result as a theorem with the regularity assumptions, boundary conditions, and necessary-condition logic presented in one place. Here the important point is narrower: the Fundamental Lemma is exactly the mathematical step that turns the integral statement

∫ b g(x)η(x)dx = 0 for all η a

(62)

into the pointwise statement

g(x) = 0.

(63)

12 Necessary condition, not a minimum proof

The Fundamental Lemma does not say that a stationary curve minimizes a functional. It only allows one to derive a pointwise necessary condition from stationarity.

The logical chain is

local extremum =⇒ δJ [y; η] = 0 for all admissible η =⇒ Euler–Lagrange equation.

(64)

The reverse implications do not hold in general. A solution of the Euler–Lagrange equation can be a maximum, a saddle-type stationary curve, or an extremal that loses minimality after a conjugate point. Those distinctions require the second variation and the Legendre, Jacobi, and Weierstrass theories later in the series.

13 Common misconceptions

13.1 Misconception 1: arbitrary means one can divide by η

No. The theorem is not an algebraic division rule. “Arbitrary” means the integral identity holds for a sufficiently rich family of independent test functions.

13.2 Misconception 2: one nonzero variation is enough to prove g = 0

No. One test function can be orthogonal to a nonzero g. The hypothesis must hold for every test function in the stated class.

13.3 Misconception 3: a test function must be positive everywhere

No. The proof only needs the ability to choose a nonnegative test function localized inside a region where g has a definite sign.

13.4 Misconception 4: continuity is merely cosmetic

No. Continuity is what turns a nonzero point value into a same-sign neighborhood and lets the classical theorem conclude pointwise equality. For locally integrable functions, the correct conclusion is almost-everywhere equality.

13.5 Misconception 5: the Fundamental Lemma proves a minimum

No. It derives a pointwise stationarity equation. Minimum classification is a separate problem.

14 Compact theorem map


Statement	Conclusion

g ∈ C([a,b]), ∫ gη = 0 for all η ∈ C_c^∞	g = 0 everywhere

same identity for all endpoint-zero C¹ variations	g = 0 everywhere

g ∈ L_loc¹, ∫ gη = 0 for all C_c^∞ tests	g = 0 almost everywhere

identity for one or finitely many selected tests	no such conclusion in general

15 What CV04 adds

CV04 will assemble the ingredients developed so far:

a functional J[y] = ∫ _a^bF(x,y,y′) dx;
an admissible perturbation y_𝜖 = y + 𝜖η;
the first variation δJ[y; η];
integration by parts; and
the Fundamental Lemma.

The result will be a full proof of the Euler–Lagrange necessary condition, including a careful distinction between stationarity and minimization.

16 Summary

The Fundamental Lemma is a localization theorem. If a continuous function g satisfies

∫ b g(x)η(x)dx = 0 a

(65)

for every smooth compactly supported test function, then g must vanish identically.

The proof is by contradiction. If g were positive or negative at some point, continuity would preserve that sign on a neighborhood. A nonnegative bump function supported in that neighborhood would then force the integral to be nonzero.

The family of all test functions is essential: a single test direction can be orthogonal to a nonzero function. In the modern weak formulation, locally integrable g is determined only up to sets of measure zero, so the conclusion becomes g = 0 almost everywhere.

For the calculus of variations, the lemma is the bridge from the stationary integral identity

∫ b( ) Fy − -d-Fy′ ηdx = 0 a dx

(66)

for every admissible variation to the pointwise Euler–Lagrange equation

d Fy − ---Fy′ = 0. dx

(67)

17 References and further reading

I. M. Gelfand and S. V. Fomin, Calculus of Variations.
B. van Brunt, The Calculus of Variations.
C. Fox, An Introduction to the Calculus of Variations.
L. C. Evans, Partial Differential Equations, for test functions, weak derivatives, and distributions.
R. Courant and D. Hilbert, Methods of Mathematical Physics, for the broader variational-method viewpoint.

"Calculus of Variations: The Fundamental Lemma" is owned by bloftin.
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