Decide if functions are in $L^1(\mathbb R),L^2(\mathbb R),L^\infty(\mathbb R)$

Question

Let $c>0$ fixed.

Let $f(x)=\begin{cases}1,&& x\in[-c,c]\\0&& \mathrm {otherwise}\end{cases}$

$g(x)=\begin{cases}2-|x|,&& x\in[-2c,2c]\\0&& \mathrm {otherwise}\end{cases}$

$h(x)=\int_{-\infty}^{\infty}f(t)e^{-itx}dt$

$i(x)=\int_{-\infty}^{\infty}g(t)e^{-itx}dt$

I need to decide if $f,g,h,i$ are in $L^1(\mathbb R),L^2(\mathbb R),L^\infty(\mathbb R)$.

$f$ clearly is in $L^1(\mathbb R),L^2(\mathbb R),L^\infty(\mathbb R)$.

For $g$ I would say that $|g|$ and $|g^2|$ are continuous functions with compact support so $g$ is in $L^1(\mathbb R),L^2(\mathbb R)$. And $g(x)\le 2+|x|\le2+2c$ so it is in $L^\infty(\mathbb R)$.

Is that correct so far?

What's the best way to the same for $h$ and $i?$

Answer 1

If you integrate you have that $h(x)=\frac{\sin{cx}}{x}$

So $h$ is not in $L^1$

But $\int_|h|^2 \leq \int_{\{|x|>1\}}|h|^2+\int_{\{|x| \leq 1\}}|h|^2<+\infty$

because $h(x)$ is bounded in $[-1,1]$(Note the limit of $h$ exist at zero)

Clearly $h \in L^{\infty}$

$i(x)=C_1\frac{\sin{(2cx)}}{x}+2\frac{1-\cos{(2cx)}}{x^2}=C_1\frac{\sin{(2cx)}}{x}+\frac{2\sin^2{(cx)}}{x^2}$ for some constant $C_1 \in \Bbb{R}$

Similarly examine $i(x)$