这是一份GLA格拉斯哥大学MATHS3021_1/MATHS2035_1作业代写的成功案例

which is bounded above by $4 H K$ where
$$
H=\limsup {k, \delta} \frac{\delta}{c(\delta)} \int{0}^{\infty} \psi_{\delta}(v) \sigma_{k \delta-v}^{2} \mathrm{~d} v
$$
and
$$
K=\frac{\delta}{c(\delta)} \sum_{k=1}^{n}\left(\int_{0}^{\infty} \chi_{\delta}(v) a_{k \delta-v} \mathrm{~d} v\right)^{2}
$$
Here
$$
\int_{0}^{\infty} \psi_{\delta}(v) \sigma_{k \delta-v}^{2} \mathrm{~d} v \leq C c(\delta)
$$
where the constant $C$ depends on $t$ and $\sigma$. Hence $H \rightarrow 0$. Furthermore,
$$
\begin{aligned}
\sum_{k=1}^{n}\left(\int_{0}^{\infty} \chi_{\delta}(v) a_{k \delta-v} \mathrm{~d} v\right)^{2} & \leq C \sum_{k=1}^{n}\left(\int_{0}^{\infty}\left|\chi_{\delta}(v)\right| \mathrm{d} v\right)^{2} \
&=C \delta^{-1}\left(\int_{0}^{\infty}\left|\chi_{\delta}(v)\right| \mathrm{d} v\right)^{2}
\end{aligned}
$$
where $C$, again, depends on $t$ and $a$. Hence Condition A implies $K \rightarrow 0$.

MATHS3021_1/MATHS2035_1 COURSE NOTES ：

Here we find
$$
c_{1}(\delta)=\frac{1}{2 \lambda}\left(1-e^{-2 \lambda \delta}\right) \sim \delta
$$
while for $k=2, \ldots, n$
$$
c_{k}(\delta)=\frac{1}{2 \lambda}\left(e^{\lambda \delta}-1\right)^{3} e^{-2 k \lambda} \sim \frac{\lambda^{2}}{2} e^{-2 k \lambda} \delta^{3} .
$$
Moreover we have
$$
c_{n+1}(\delta)=\frac{1}{2 \lambda}\left(1-e^{-2 \lambda \delta}\right) \sim e^{-2 \lambda l} \delta,
$$
whereas $c_{k}(\delta)=0$ for $k>n+1$. Finally, $c(\delta) \sim \delta\left(1+e^{-2 \lambda l}\right)$ and
$$
\bar{c}{n+1}(\delta) c(\delta)^{-1} \int{l}^{l+\delta} g^{2}(v-\delta) \mathrm{d} v \rightarrow\left(1+e^{2 \lambda l}\right)^{-1} .
$$
So, Conditions A, C and D are met. But Condition B is not and we have that $\pi_{\delta} \rightarrow \pi$, where
$$
\pi=\frac{1}{1+e^{-2 \lambda l}} \delta_{0}+\frac{1}{1+e^{2 \lambda l}} \delta_{1},
$$