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Week 9 Tutorial Solutions
- We calculate
\[
\begin{align*}
\int\limits_{-T}^T \dfrac{N}{\pi} \dfrac{1}{1 + (Nx)^2} \intd x &{} = \left[ \dfrac{\arctan(Nx)}{\pi} \right]_{-T}^T \\
&{} = \dfrac{\arctan(NT)}{\pi} - \dfrac{\arctan(-NT)}{\pi} \\
\end{align*}
\]
which converges to $1$ as $T \to \infty$.
- For $x \ne 0$ we have
\[
0 \le f_N(x) = \dfrac{1}{N} \dfrac{1}{\pi} \dfrac{1}{\tfrac{1}{N^2} + x^2} \le \dfrac{1}{N} \to 0
\]
as $N \to \infty$.
- Fix a bound $B$ on $g$. We calculate that
\[
\begin{align*}
\left| \int\limits_\delta^T f_N(x) g(x) \intd x \right|
&{} \le
B \int\limits_\delta^T f_N(x) \intd x
\\
&{} = B \left(\dfrac{\arctan(NT)}{\pi} - \dfrac{\arctan(N\delta)}{\pi} \right)
\end{align*}
\]
so that
\[
0 \le \left| \int\limits_\delta^\infty f_N(x) g(x) \intd x \right| \le \dfrac{B}{2} - \dfrac{B \arctan(N\delta)}{\pi}
\]
and the right-hand side converges to zero as $N \to \infty$.
- Fix $\epsilon > 0$. As $g$ is continuous there is $\delta \g 0$ such that
\[
|x| \l \delta \Rightarrow |g(x) - g(0)| \l \epsilon/4
\]
and we have
\[
\begin{align*}
&{} \int\limits_{-\infty}^\infty f_N(x) g(x) \intd x - g(0) \\
= {}& \int\limits_{-\infty}^\infty (g(x) - g(0)) f_N(x) \intd x \\
= {}& \int\limits_{-\delta}^\delta (g(x) - g(0)) f_N(x) \intd x + \int\limits_\delta^\infty (g(x) - g(0)) f_N(x) \intd x \\
& \qquad + \int\limits_{-\delta}^{-\infty} (g(x) - g(0)) f_N(x) \intd x
\end{align*}
\]
which, combined with
\[
\left| \int\limits_{-\delta}^\delta (g(x) - g(0)) f_N(x) \intd x \right| \le \dfrac{\epsilon}{4} \int\limits_{-\delta}^\delta f_N(x) \intd x \to \dfrac{\epsilon}{4}
\]
as $N \to \infty$ and the conclusion of 3., implies that if $N$ is large enough then
\[
\left| \int\limits_{-\infty}^\infty f_N(x) g(x) \intd x - g(0) \right| \l \epsilon
\]
as desired.
- We can follow the same outline provided the sequence $F_N$ has the necessary properties. First, since
\[
\int\limits_0^1 e^{2 \pi i k x} \intd x = \begin{cases} 1 & k = 0 \\ 0 & k \ne 0 \end{cases}
\]
we see that
\[
\int\limits_0^1 F_N(x) \intd x = 1
\]
for all $N \in \N$. Secondly, using lots of trigonometric identities, we can write
\[
F_N(x) = \dfrac{1}{N} \dfrac{1 - \cos(2 \pi Nx)}{1 - \cos(2 \pi x)}
\]
from which we see that $F_N(x) \ge 0$ for all $x \in [0,1)$ and that $F_N(x) \to 0$ as $N \to \infty$ for all $x \ne 0$. Thus, if $g$ is continuous on $[0,1)$ then
\[
\int\limits_\delta^{1-\delta} F_N(x) g(x) \intd x \to 0
\]
as $N \to \infty$ and we can repeat the argument in 4.