48 2. FOUR IMPORTANT LINEAR PDE Then if |x − x0| δ 2 , we have, according to the lemma, |u(x, t) − g(x0)| = Rn Φ(x − y, t)[g(y) − g(x0)] dy ≤ B(x0,δ) Φ(x − y, t)|g(y) − g(x0)| dy + Rn−B(x0,δ) Φ(x − y, t)|g(y) − g(x0)| dy =: I + J. Now I ≤ ε Rn Φ(x − y, t) dy = ε, owing to (11) and the lemma. Furthermore, if |x − x0| ≤ δ 2 and |y − x0| ≥ δ, then |y − x0| ≤ |y − x| + δ 2 ≤ |y − x| + 1 2 |y − x0|. Thus |y − x| ≥ 1 2 |y − x0|. Consequently J ≤ 2 g L∞ Rn−B(x0,δ) Φ(x − y, t) dy ≤ C tn/2 Rn−B(x0,δ) e− |x−y|2 4t dy ≤ C tn/2 Rn−B(x0,δ) e− |y−x0|2 16t dy = C Rn−B(0,δ/ √ t) e− |z|2 16 dz → 0 as t → 0+. Hence if |x − x0| δ 2 and t 0 is small enough, |u(x, t) − g(x0)| 2ε. Interpretation of fundamental solution. In view of Theorem 1 we sometimes write Φt − ΔΦ = 0 in Rn × (0, ∞) Φ = δ0 on Rn × {t = 0}, δ0 denoting the Dirac measure on Rn giving unit mass to the point 0. Infinite propagation speed. Notice that if g is bounded, continuous, g ≥ 0, g ≡ 0, then u(x, t) = 1 (4πt)n/2 Rn e− |x−y|2 4t g(y) dy

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