Distributions 16 | Distributions with Compact Support

TL;DR

A distribution T vanishes on test functions supported entirely outside supp(T), so only overlap with supp(T) can affect ⟨T,·⟩.

Briefing Cornell Notes

Briefing

Distributions can be applied not only to compactly supported test functions, but also to a larger class of smooth functions—provided the distribution itself has compact support. That extension matters because it creates the right framework for extending convolution beyond the classical setting, which is essential for working with partial differential equations.

Start with a distribution T as a linear functional on test functions. The key structural fact is that T “vanishes outside its support”: if a test function is supported entirely away from the closed set supp(T), then T acting on it gives zero. This observation already hints that test functions don’t need compact support in every situation—only the part overlapping supp(T) can affect the outcome. The limitation is that this only becomes a genuine extension when supp(T) is compact. If supp(T)=R^N, there’s no room to enlarge the domain; if supp(T) is compact, the overlap region stays compact and the functional can be defined more broadly.

To formalize this, the construction introduces a new space of smooth functions, denoted E_T. A function γ is in E_T if it is C^∞ and its effective support inside supp(T) is compact—equivalently, the intersection supp(γ) ∩ supp(T) is compact. When supp(T) itself is compact, E_T becomes strictly larger than the usual test function space: it includes smooth functions without compact support, as long as they only “matter” on the compact region where they meet supp(T). This gives a natural goal: define ⟨T,γ⟩ for every γ in E_T.

The definition proceeds by multiplying γ by a carefully chosen cutoff test function f. The function f is selected so that f(x)=1 on an open neighborhood of supp(γ) ∩ supp(T) (the neighborhood condition prevents boundary pathologies), while f is a standard test function that eventually becomes zero outside a compact set. Then f·γ becomes an ordinary test function, so T can act on it. The extended pairing is defined by ⟨T,γ⟩ := ⟨T, f·γ⟩.

A crucial step is proving this is well-defined: different choices of f must lead to the same value. If f1 and f2 both equal 1 on a neighborhood of the relevant intersection, then f1−f2 vanishes there. Multiplying (f1−f2)·γ produces a test function whose support lies away from supp(T), forcing T to give zero on that difference. Linearity then shows ⟨T,f1·γ⟩ = ⟨T,f2·γ⟩.

Once the extension is set, two properties carry over. Linearity survives because the pairing remains linear in γ. Derivatives also remain compatible: distributional derivatives satisfy the usual integration-by-parts sign rule, with ⟨∂^α T, γ⟩ = (−1)^{|α|}⟨T, ∂^α γ⟩ for every γ in E_T.

With these tools, distributions with compact support get a special role. If supp(T) is compact, the extension reaches the full space E = C^∞(R^N), and such distributions are denoted E′(R^N) (instead of D′(R^N), where the domain is test functions). This is exactly what enables convolution of distributions to be extended when at least one factor has compact support—setting up the next step in the series.

Cornell Notes

A distribution T can be extended from acting on compactly supported test functions to acting on a larger class of smooth functions, as long as T has compact support. The construction defines a space E_T of C^∞ functions γ whose intersection supp(γ) ∩ supp(T) is compact, so only the overlap with supp(T) can influence the value. The pairing ⟨T,γ⟩ is defined by multiplying γ with a cutoff test function f that equals 1 on a neighborhood of supp(γ) ∩ supp(T), ensuring f·γ is a test function. Independence from the choice of f follows because any difference f1−f2 vanishes near the overlap, making T evaluate to zero on the resulting term. Linearity and the standard distributional derivative rule remain valid on E_T, enabling convolution extensions later.

Why does the support of a distribution control which test functions matter?

Because T vanishes on test functions supported away from supp(T). If a test function’s support lies entirely outside the closed set where T is “active,” then ⟨T,φ⟩=0. That means only the portion of a function overlapping supp(T) can affect the value, motivating an enlarged domain when supp(T) is compact.

What exactly is the space E_T, and how does it enlarge the domain of T?

E_T consists of C^∞ functions γ such that supp(γ) ∩ supp(T) is compact. Even if γ itself has non-compact support, the overlap with supp(T) stays in a compact region, so T can be made to depend only on that region. If supp(T)=R^N, then E_T collapses to the usual test functions; if supp(T) is compact, E_T is strictly larger.

How is ⟨T,γ⟩ defined for γ in E_T without violating the rule that T normally acts only on test functions?

Pick a test function f with f(x)=1 on an open neighborhood containing supp(γ) ∩ supp(T). Then f·γ becomes a genuine test function (since f is compactly supported), so ⟨T,γ⟩ is defined as ⟨T,f·γ⟩. The neighborhood condition ensures no ambiguity at the boundary of the relevant supports.

Why doesn’t the value ⟨T,f·γ⟩ depend on which cutoff f is chosen?

If f1 and f2 both equal 1 on a neighborhood of supp(γ) ∩ supp(T), then (f1−f2)=0 near that intersection. Consequently, (f1−f2)·γ is a test function whose support lies away from supp(T), so T gives zero on it. Linearity yields ⟨T,f1·γ⟩=⟨T,f2·γ⟩.

How do linearity and distributional derivatives behave under this extension?

Linearity persists because the extended pairing is defined through multiplication by f and the original linear functional. For derivatives, the usual sign rule remains: for multi-index α, ⟨∂^α T, γ⟩ = (−1)^{|α|}⟨T, ∂^α γ⟩ for every γ in E_T. This comes from applying the product rule inside the definition.

What is the special class E′(R^N), and why does compact support matter for convolution?

When supp(T) is compact, the extension reaches all smooth functions on R^N, so T acts on E=C^∞(R^N). Such distributions are denoted E′(R^N). This maximal extension is what makes it straightforward to extend convolution when one factor has compact support, which is the next topic.

Review Questions

Given γ in E_T, what condition on supp(γ) ∩ supp(T) ensures the extension is possible?
Why is it necessary that the cutoff f equals 1 on an open neighborhood (not just on the closed intersection)?
State the sign rule for distributional derivatives after extending T to act on E_T.

Key Points

1
A distribution T vanishes on test functions supported entirely outside supp(T), so only overlap with supp(T) can affect ⟨T,·⟩.
2
If supp(T) is compact, T can be extended from test functions to smooth functions whose overlap with supp(T) is compact.
3
Define E_T as the set of C^∞ functions γ with supp(γ) ∩ supp(T) compact; then ⟨T,γ⟩ can be defined for all γ in E_T.
4
Choose a cutoff test function f with f=1 on a neighborhood of supp(γ) ∩ supp(T); then f·γ is a test function and ⟨T,γ⟩ := ⟨T,f·γ⟩.
5
The definition is well-defined because any two such cutoffs differ by a function that vanishes near the overlap, forcing T to evaluate the difference to zero.
6
Linearity and the standard distributional derivative identity (−1)^{|α|}⟨T,∂^α γ⟩ remain valid on the extended domain.
7
Distributions with compact support form E′(R^N), and this framework enables extending convolution when one factor has compact support.

Highlights

The extension works because only the part of a smooth function overlapping supp(T) can influence the distribution’s value.

Cutoff functions f are chosen to equal 1 on an open neighborhood of supp(γ) ∩ supp(T), preventing boundary ambiguity.

Well-definedness follows from the fact that (f1−f2)·γ has support away from supp(T), so T kills the difference.

Distributional derivatives keep the usual sign rule after extension: ⟨∂^α T, γ⟩ = (−1)^{|α|}⟨T,∂^α γ⟩.

Distributions with compact support are identified as E′(R^N), setting up convolution extensions in the next step.