Week 9 Tutorial Solutions

1. We calculate $$\begin{array}{rl}\underset{-T}{\overset{T}{\int }}{\displaystyle \frac{N}{\pi }}{\displaystyle \frac{1}{1+(Nx{)}^2}}\mathsf{d}x& ={\left[{\displaystyle \frac{\arctan (Nx)}{\pi }}\right]}_{-T}^T\\ & ={\displaystyle \frac{\arctan (NT)}{\pi }}-{\displaystyle \frac{\arctan (-NT)}{\pi }}\end{array}$$ which converges to $1$ as $T\to \mathrm{\infty }$.
2. For $x\ne 0$ we have $$0\le f_N(x)={\displaystyle \frac{1}{N}}{\displaystyle \frac{1}{\pi }}{\displaystyle \frac{1}{\frac{1}{N^2}+x^2}}\le {\displaystyle \frac{1}{N}}\to 0$$ as $N\to \mathrm{\infty }$.
3. Fix a bound $B$ on $g$. We calculate that $$\begin{array}{rl}|\underset{\delta }{\overset{T}{\int }}{f}_N(x)g(x)\mathsf{d}x|& \le B\underset{\delta }{\overset{T}{\int }}{f}_N(x)\mathsf{d}x\\ & =B({\displaystyle \frac{\arctan (NT)}{\pi }}-{\displaystyle \frac{\arctan (N\delta )}{\pi }})\end{array}$$ so that $$0\le |\underset{\delta }{\overset{\mathrm{\infty }}{\int }}{f}_N(x)g(x)\mathsf{d}x|\le {\displaystyle \frac{B}{2}}-{\displaystyle \frac{B\arctan (N\delta )}{\pi }}$$ and the right-hand side converges to zero as $N\to \mathrm{\infty }$.
4. Fix $ϵ>0$. As $g$ is continuous there is $\delta >0$ such that $$|x|<\delta \Rightarrow |g(x)-g(0)|<ϵ/4$$ and we have $$\begin{array}{rl}& \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}_N(x)g(x)\mathsf{d}x-g(0)\\ =& \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}(g(x)-g(0))f_N(x)\mathsf{d}x\\ =& \underset{-\delta }{\overset{\delta }{\int }}(g(x)-g(0))f_N(x)\mathsf{d}x+\underset{\delta }{\overset{\mathrm{\infty }}{\int }}(g(x)-g(0))f_N(x)\mathsf{d}x\\ & +\underset{-\delta }{\overset{-\mathrm{\infty }}{\int }}(g(x)-g(0))f_N(x)\mathsf{d}x\end{array}$$ which, combined with $$|\underset{-\delta }{\overset{\delta }{\int }}(g(x)-g(0))f_N(x)\mathsf{d}x|\le {\displaystyle \frac{ϵ}{4}}\underset{-\delta }{\overset{\delta }{\int }}{f}_N(x)\mathsf{d}x\to {\displaystyle \frac{ϵ}{4}}$$ as $N\to \mathrm{\infty }$ and the conclusion of 3., implies that if $N$ is large enough then $$|\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}{f}_N(x)g(x)\mathsf{d}x-g(0)|<ϵ$$ as desired.
5. We can follow the same outline provided the sequence $F_N$ has the necessary properties. First, since $$\underset{0}{\overset{1}{\int }}{e}^{2\pi ikx}\mathsf{d}x=\{\begin{array}{ll}1& k=0\\ 0& k\ne 0\end{array}$$ we see that $$\underset{0}{\overset{1}{\int }}{F}_N(x)\mathsf{d}x=1$$ for all $N\in \mathbb{N}$. Secondly, using lots of trigonometric identities, we can write $$F_N(x)={\displaystyle \frac{1}{N}}{\displaystyle \frac{1-\cos (2\pi Nx)}{1-\cos (2\pi x)}}$$ from which we see that $F_N(x)\ge 0$ for all $x\in [0,1)$ and that $F_N(x)\to 0$ as $N\to \mathrm{\infty }$ for all $x\ne 0$. Thus, if $g$ is continuous on $[0,1)$ then $$\underset{\delta }{\overset{1-\delta }{\int }}{F}_N(x)g(x)\mathsf{d}x\to 0$$ as $N\to \mathrm{\infty }$ and we can repeat the argument in 4.