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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedaudited 2026-09-14
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Polynomial growth functions define tempered distributions

Statement

Let f:Rn→C be locally integrable. If, for some integer N≥0,

∫Rn∣f(x)∣(1+∣x∣)−N dx<∞,

then the regular functional uf(φ)=∫fφ is a tempered distribution. Consequently this holds if f has pointwise polynomial growth outside a compact set, and every class in complex Lp(Rn), 1≤p≤∞, has a representative defining a tempered distribution. These are sufficient conditions; pointwise polynomial growth is not asserted to characterize all regular tempered distributions.

Facts & Assumptions

Given: A locally integrable complex function f on Rn.

[F1]
[F2]

One finite rectangular Schwartz-seminorm estimate characterizes tempered functionals (Finite seminorm bound characterizes tempered distributions).

[F3]

Hölder's inequality, including the L1 and L∞ endpoints, applies to the real nonnegative functions ∣f∣ and a weight (Holder's inequality for integrals, including the endpoint cases).

[F4]

Complex Lp classes, their moduli, and their locally integrable representatives use the conventions of Complex Lp classes and Euclidean test-function conventions.

[F5]

A real p-series converges when its exponent exceeds one (The p-series for a real exponent p converges exactly when p is greater than one).

Proof

technique · weighted integral estimate
1.1algebra

For each integer N≥0, expansion of (1+∣x1∣+⋯+∣xn∣)N gives a finite constant An,N for which the following estimate holds.

(1+∣x∣)N∣φ(x)∣≤An,Nmax⁡∣α∣≤Npα,0(φ).

Thus the weighted hypothesis implies ∣uf(φ)∣≤An,N(∫∣f∣(1+∣x∣)−N)max⁡∣α∣≤Npα,0(φ). [F1, algebra]

2.1F1F2step 1.1

The estimate in step 1.1 makes the integral absolutely convergent for every Schwartz test and proves that uf is tempered. It also shows directly that changing f on a null set changes no pairing.

3.1F5step 2.1

The integer shells m≤∣x∣<m+1 have measure at most (2m+2)n. Hence ∫(1+∣x∣)−sdx<∞ whenever s>n+1, by comparison with ∑m≥1mn−s. If ∣f(x)∣≤C(1+∣x∣)d off a compact set, choose an integer N>d+n+1. Local integrability handles the compact part, and the shell estimate handles its complement, so step 2.1 applies.

4.1F3F4step 3.1

If f∈L1, take N=0. If 1<p<∞ and q is conjugate to p, choose N with Nq>n+1; Hölder gives ∫∣f∣(1+∣x∣)−N≤∥f∥p∥(1+∣⋅∣)−N∥q<∞. If p=∞, choose N>n+1 and use any finite essential bound for ∣f∣. The same estimates on bounded balls give local integrability of the chosen representatives.

5.1F1step 2.1step 3.1∎

For n≥1, as a boundary calculation, define fr(x)=∣x∣r for x≠0 and assign any finite value, say fr(0)=0, when r<0; set f0≡1, and for r>0 use the usual value fr(0)=0. The value at this measure-zero point has no effect on local integrability or the induced distribution. The function fr is locally integrable at the origin exactly when r>−n: for r<0, the dyadic annuli 2−j−1≤∣x∣<2−j give a series comparable to ∑j2−j(n+r); for r≥0 there is no singularity, while the reverse bound on a fixed cone gives divergence when r≤−n. At infinity it has polynomial growth. Therefore fr is among the tempered regular examples precisely for the locally meaningful range r>−n.

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