Manuscript XVI-B@1 · XVI-B.tex · lemma · 6a0abffd

Lemma: An integrated kernel can have an affine past tail

Update from the supplied later sheets › Auxiliary evaluation and corrected support: XVI-F-v2 · lines 1442–1481 · proof included or linked

XVI-B@1#lem:affine-tail, XVI-B@1#eq:affine-tail

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Let $τ \in C_{c}^{\infty} (R)$ be supported in $[0, t_{*}]$ and define

K_{I} (u) = \int_{u}^{\infty} (t - u) τ (t) d t .

Then $K_{I}$ is smooth, $K_{I}^{''} = τ$ , and $K_{I} = K_{I}^{'} = 0$ for $u > t_{*}$ . For $u < 0$ ,

K_{I} (u) = M_{1} - u M_{0}, M_{j} = \int_{0}^{\infty} t^{j} τ (t) d t (j = 0, 1) .

Thus compact support of $τ$ does not imply compact support of $K_{I}$ .

Nevertheless $σ K_{I}$ and $σ^{'} K_{I}$ belong to $C_{c}^{\infty} (R)$ . The same statements hold componentwise for tensor coefficients.

Proof. Differentiate under the integral:

K_{I}^{'} (u) = - \int_{u}^{\infty} τ (t) d t, K_{I}^{''} (u) = τ (u) .

The future vanishing follows from the source support. For $u < 0$ , expand $(t - u)$ over the fixed source support to obtain (eq:affine-tail).

The product $σ K_{I}$ is smooth, vanishes for $u \leq 0$ , and vanishes for $u > t_{*}$ . The product $σ^{'} K_{I}$ is smooth and supported in the fixed ramp.

Displayed formulas in this passage

XVI-B@1#eq:affine-tail · line 1452
$K_{I} (u) = M_{1} - u M_{0}, M_{j} = \int_{0}^{\infty} t^{j} τ (t) d t (j = 0, 1) .$
unlabeled · line 1446
$K_{I} (u) = \int_{u}^{\infty} (t - u) τ (t) d t .$
unlabeled · line 1469
$K_{I}^{'} (u) = - \int_{u}^{\infty} τ (t) d t, K_{I}^{''} (u) = τ (u) .$

Lemma: An integrated kernel can have an affine past tail

Displayed formulas in this passage

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