Complete the following alternative argument for Example 7.16. Assume that (left(u_{n} ight)_{n geqslant 1} subset mathcal{C}_{infty}^{2}(mathbb{R})) such that (left(frac{1}{2} u_{n}^{prime prime} ight)_{n geqslant 1}) is

Complete the following alternative argument for Example 7.16. Assume that \(\left(u_{n}\right)_{n \geqslant 1} \subset \mathcal{C}_{\infty}^{2}(\mathbb{R})\) such that \(\left(\frac{1}{2} u_{n}^{\prime \prime}\right)_{n \geqslant 1}\) is a Cauchy sequence in \(\mathcal{C}_{\infty}\) and that \(u_{n}\) converges uniformly to \(u\).

a) Show that \(u_{n}(x)-u_{n}(0)-x u_{n}^{\prime}(0)=\int_{0}^{x} \int_{0}^{y} u_{n}^{\prime \prime}(z) d z d y \rightarrow \int_{0}^{x} \int_{0}^{y} 2 g(z) d z d y\) uniformly. Conclude that \(c^{\prime}:=\lim _{n \rightarrow \infty} u_{n}^{\prime}(0)\) exists.

b) Show that \(u_{n}^{\prime}(x)-u_{n}^{\prime}(0) \rightarrow \int_{0}^{x} 2 g(z) d z\) uniformly. Deduce from this and part a) that \(u_{n}^{\prime}(x)\) converges uniformly to \(\int_{0}^{x} 2 g(z) d z+c^{\prime}\) where \(c^{\prime}=\int_{-\infty}^{0} 2 g(z) d z\).

c) Use a) and b) to show that \(u_{n}(x)-u_{n}(0) \rightarrow \int_{0}^{x} \int_{-\infty}^{y} 2 g(z) d z\) uniformly and argue that \(u_{n}(x)\) has the uniform limit \(\int_{-\infty}^{x} \int_{-\infty}^{y} 2 g(z) d z d y\).

Data From 7.16 Example 

7.16 Example. Let (A, D(A)) be the generator of the transition semigroup (Pt)to of a BMd. We know from Example 7.9 that e2, (Rd) c D(A). If d = 1, then (A) = 2, (R) = {ue Coo(R) : u, u', u" e Coo(R)}. Proof. We know from Proposition 7.13.f) that Ua is the inverse of a id-A. In particular, Ex. 7.12 Ua(Coo(R)) = D(A). This means that any u = D(A) can be written as u = Uaf for some f Coo(R). Using the formula (7.15) for Ua from Example 7.14, we see for all Coo(R) Uaf(x) = ! = e-vzaly-xlf(y) dy. -00 By the differentlability lemma for parameter dependent Integrals, cf. [232, Theorem 12.5], we find Since d dx Uaf(x) = e-zay-x\ sgn(y - x) f(y) dy -00 x - za (y-x)f(y) dy - e-(x-1)f(y) dy dx Uaf(x) = za je -00 Uaf(x) 22aUalf(x) and 2aly-x -00 f(y) dy - 2f(x). Uaf(x) = 2aUaf(x)-2f(x), Proposition 7.13.d) shows that Uaf Coo(R), thus C2, (R) > D(A). If d>1, the arguments of Example 7.16 are no longer valid. !