$\lim_{n \to \infty}\|\hat f_n \cdot g-g\|_1 \to 0$ for $f_n(x)=\sqrt{2\pi} \cdot \max(0,n-n^2 |x|)$

Question

$f_n(x)=\sqrt{2\pi} \cdot \max(0,n-n^2 |x|), \ n \in \mathbb N$.

Show that for $\hat f_n(x):=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f_n(t)e^{-itx}dt$ and $g \in L^1(\mathbb R):$

$$\lim_{n \to \infty}\|\hat f_n \cdot g-g\|_1 \to 0$$

$n-n^2 |x|\ge0 \Leftrightarrow |x|\le \frac{1}{n}$.

Therefore,

\begin{aligned} \hat f_n(x)=\int_{-1/n}^{1/n}(n-n^2|t|)e^{-itx}dt& =n\int_{-1/n}^{1/n} e^{-itx}dt -n^2\int_{-1/n}^{1/n}|t|e^{-itx}dt\\ &=n(\frac{e^{ix/n}}{ix}-\frac{e^{-ix/n}}{ix}) + 2n^2\int^{1/n}_0t\cos(tx)\,dt \end{aligned}

Answer 1

A simple computations gives (integration by parts and what not) $$ \hat{f}_n(t)=\int^{1/n}_{-1/n}(n-n^2|x|)e^{-ixt}\,dx = 2\frac{1-\cos(t/n)}{(t/n)^2} $$ This is an integrable bounded function over the real line, and as $n\rightarrow\infty$, converges to 1. The conclusion is an application of Lebesgue dominated convergence.

Answer 2

Note that $1_{[\frac{-1}{n},\frac{1}{n}]}=1_{[\frac{-1}{n},0]}+1_{(0\frac{1}{n}]}$

From this you can prove that $\int_{\Bbb{R}}f_n(x)dx=\sqrt{2\pi} ,\forall n \in \Bbb{N}$

Now since $g \in L^1$,by Fourier inversion $g=\widehat{g^{\vee}}$ where $g^{\vee}(x)=\frac{1}{\sqrt{2\pi}}\int_{\Bbb{R}}g(\xi)e^{i \xi x}d\xi$

Thus $\hat{f_n}g=\widehat{f_n \ast g^{\vee}}$

Thus $$||\hat{h_n}g-g||_1=\int |\widehat{h_n \ast g^{\vee}-g^{\vee}}|\leq \int |h_n \ast g^{\vee}-g^{\vee}|=||h_n \ast g^{\vee}-g^{\vee}||_1$$

where e $h_n=\frac{f_n}{\sqrt{2 \pi}}$.

Also $h_n$ is a sequence of good kernels, and since $g^{\vee} \in L^1$ then $||f_n \ast g^{\vee}-g^{\vee}||_1 \to 0$