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Put ${\epsilon }_n={\sup }_{[-1,1]}|F_n-F|$. Uniform convergence means ${\epsilon }_n\to 0$, and continuity ensures that the functions have Riemann integrals. The integral triangle inequality givesThis proves that uniform convergence preserves integrals on a compact interval.

For the derivative assertion, take $F_n(x)=\sin (n^{2}x)/n$ and $F(x)=0$. These functions are differentiable, with $\sup |F_n|\le 1/n$, so their convergence is uniform. However, $F_n^{\prime }(0)=n$ whereas $F^{\prime }(0)=0$. The derivatives therefore need not converge even at one interior point.

The Laplace-transform formula for a semigroup resolvent is locally uniformly convergent in the half-plane $\mathrm{Re}z>\omega$, so it may be differentiated under the Bochner integral:The resolvent identity gives $R^{\prime }(z)=-R(z{)}^2$ and, by mathematical induction,ThereforeFor real $\lambda >\omega$, the integral triangle inequality and the Gamma integral giveConsequentlyThe exponent $-n+1$ printed in the question is a typographical error: already at $n=0$ it contradicts the first-resolvent bound.

With the notation from part (i), the total-variation process of $[M,N]$ isFor the centered bivariate normal distribution $(U,V)$ used there, $c=\mathbb{E}[UV]$. The integral triangle inequality givesIntegrating this pointwise inequality against $dC$ proves $V_t\le {\tilde{V}}_t$ for every $t\ge 0$.