Distributions 6 | The Delta Distribution Is Not Regular [dark version]

Q: How does the choice of a bump test function Φ concentrated near 0 drive the contradiction?

For δ, the value is determined only by Φ(0): ⟨δ,Φ⟩ = Φ(0), which is positive for the constructed bump. The bump is arranged to be zero outside the ε-ball and to have a fixed positive profile inside (an exponential shape scaled to the ε-ball). Under the assumed representation δ = T_f, the integral ∫ f(x)Φ(x) dx is forced to depend only on f inside that same ε-ball, where the earlier step guarantees ∫_{|x|≤ε} |f(x)| dx is tiny.

Q: Where exactly does the factor “smaller than 1/2” enter the inequality?

After taking absolute values, the integral is bounded by ||Φ||_∞ times ∫_{|x|≤ε} |f(x)| dx. The construction ensures ∫_{|x|≤ε} |f(x)| dx = B can be chosen 0.

TL;DR

No locally integrable function f can satisfy ⟨δ,Φ⟩ = ∫ f(x)Φ(x) dx for all test functions Φ.

Briefing Cornell Notes

Briefing

The delta distribution cannot be represented as a “regular” distribution coming from an ordinary locally integrable function. In practical terms, there is no locally integrable function f on ℝ^N such that, for every test function Φ, the identity ⟨δ,Φ⟩ = ⟨T_f,Φ⟩ holds—where ⟨T_f,Φ⟩ is the usual integral of f(x)Φ(x). This matters because it marks a hard boundary between distributions that behave like functions and the genuinely singular objects needed to model point masses.

The proof starts by assuming the opposite: suppose such a locally integrable f exists. Since f is locally integrable, its integral over the unit ball is finite; call this value a. The argument then zooms in toward the origin by decomposing the punctured unit ball into a countable union of “annular shells” (described as rings/wings) with radii 1/k. Because these shells are disjoint, the integral over the whole region becomes an infinite sum of the integrals over each shell. Since the total sum is finite, the shell integrals must form a sequence that tends to 0. By choosing the starting index far enough, the proof ensures that the integral over a sufficiently small punctured neighborhood of 0 can be made arbitrarily small—specifically, smaller than 1/2.

Next comes the key move: pick a test function Φ that is concentrated near 0. A bump function is constructed to be zero outside a small ε-ball around the origin and to match a fixed exponential profile inside that ε-ball (with scaling by 1/ε to preserve the intended shape under zooming). For the delta distribution, ⟨δ,Φ⟩ equals Φ(0), which is positive. Under the assumed regular representation ⟨T_f,Φ⟩ = ∫ f(x)Φ(x) dx, the proof takes absolute values and uses the triangle inequality to bound the integral by the supremum norm of Φ times the integral of |f| over the ε-ball.

Because Φ is supported inside the ε-ball, the bound reduces to a product: ||Φ||_∞ multiplied by B, where B is the integral of |f| over that ε-ball. The earlier step guarantees B can be chosen smaller than 1/2. But ||Φ||_∞ is tied to Φ(0), which is positive, so the inequality forces a positive quantity to be less than half of itself—an impossibility. This contradiction shows no such locally integrable f can exist, so δ is not a regular distribution.

The takeaway is structural: delta’s action cannot be reproduced by integrating a genuine function against test functions. That singularity is exactly why distributions extend beyond classical functions and why later methods are needed to compute with them.

Cornell Notes

The delta distribution δ cannot be written in the form ⟨δ,Φ⟩ = ∫ f(x)Φ(x) dx for any locally integrable function f. Assuming such an f exists leads to a contradiction. Local integrability implies the integral of |f| over sufficiently small neighborhoods of 0 can be made arbitrarily small; in particular, it can be forced below 1/2. A test function Φ is then chosen to be a positive bump supported inside a tiny ε-ball around 0, so ⟨δ,Φ⟩ = Φ(0) > 0. Bounding ∫ f(x)Φ(x) using the triangle inequality and the supremum of Φ yields an inequality that would make a positive number smaller than half of itself, which cannot happen. Therefore δ is not a regular distribution.

Why does local integrability of f imply the integral of |f| over small balls can be made arbitrarily small?

Local integrability means ∫_{|x|≤1} |f(x)| dx is finite. The proof then partitions the region near 0 into disjoint shells with radii 1/k (a countable union). Since the total integral is finite, the sum of the shell integrals is finite, so the shell integrals must go to 0. That lets one choose a small ε-ball (excluding the single point 0, which doesn’t affect the integral) so that ∫_{|x|≤ε} |f(x)| dx = B is as small as desired, in particular B < 1/2.

How does the choice of a bump test function Φ concentrated near 0 drive the contradiction?

For δ, the value is determined only by Φ(0): ⟨δ,Φ⟩ = Φ(0), which is positive for the constructed bump. The bump is arranged to be zero outside the ε-ball and to have a fixed positive profile inside (an exponential shape scaled to the ε-ball). Under the assumed representation δ = T_f, the integral ∫ f(x)Φ(x) dx is forced to depend only on f inside that same ε-ball, where the earlier step guarantees ∫_{|x|≤ε} |f(x)| dx is tiny.

Where exactly does the factor “smaller than 1/2” enter the inequality?

After taking absolute values, the integral is bounded by ||Φ||_∞ times ∫_{|x|≤ε} |f(x)| dx. The construction ensures ∫_{|x|≤ε} |f(x)| dx = B can be chosen < 1/2. Since ||Φ||_∞ is essentially Φ(0) for the bump, the inequality becomes something like Φ(0) ≤ B·Φ(0) with B < 1/2, forcing Φ(0) to be less than itself by a factor less than 1—contradicting Φ(0) > 0.

Why is it legitimate that the ε-ball argument “doesn’t include 0 as a point,” yet the calculation still works?

The proof notes that integrals ignore single points in ℝ^N for locally integrable functions: changing the integrand on a measure-zero set (like removing the point {0}) does not change the value of the integral. So whether the ε-ball is punctured or not, the bound on ∫ |f| over that region remains valid.

What role does the triangle inequality play in the contradiction?

The triangle inequality converts ∫ f(x)Φ(x) dx into an upper bound involving absolute values: |∫ fΦ| ≤ ∫ |f||Φ|. Then, because Φ is bounded, ||Φ||_∞ can be pulled out, leaving only ∫_{|x|≤ε} |f(x)| dx. That reduction is what makes the earlier “small neighborhood” estimate directly applicable.

Review Questions

What specific property of locally integrable functions is used to guarantee that ∫_{|x|≤ε} |f(x)| dx can be made smaller than 1/2?
How does the identity ⟨δ,Φ⟩ = Φ(0) interact with a test function supported in an ε-ball to force a contradiction?
Which inequality step turns the assumed representation ∫ f(x)Φ(x) dx into a bound involving ||Φ||_∞ and ∫ |f| over the ε-ball?

Key Points

1
No locally integrable function f can satisfy ⟨δ,Φ⟩ = ∫ f(x)Φ(x) dx for all test functions Φ.
2
Assuming such an f exists, partitioning the region near 0 into disjoint shells forces the integrals over those shells to approach 0.
3
This yields a small ε-ball where B = ∫_{|x|≤ε} |f(x)| dx can be chosen less than 1/2.
4
A bump test function Φ is constructed to be zero outside the ε-ball and positive at the origin, so ⟨δ,Φ⟩ = Φ(0) > 0.
5
Bounding ∫ f(x)Φ(x) using absolute values and the triangle inequality reduces the integral to ||Φ||_∞·B.
6
With B < 1/2, the resulting inequality would require a positive number to be smaller than half of itself, which is impossible.
7
Therefore δ is not a regular distribution and must be treated as a genuinely singular distribution.

Highlights

Delta’s action depends only on Φ(0), so any attempt to realize δ via an ordinary integrable function must reproduce a point evaluation using only neighborhood integrals.

Finite integrals near 0 force the “shell integrals” to shrink to zero, enabling a bound B < 1/2 on sufficiently small balls.

A carefully scaled bump test function supported in an ε-ball turns the abstract contradiction into a concrete inequality involving Φ(0) and B.

The contradiction comes from combining support near 0 with the triangle inequality and the ability to make ∫|f| arbitrarily small.

Topics

Delta Distribution
Regular Distributions
Locally Integrable Functions
Test Functions
Singularities