In sampling theory, $L^p$ spaces are very important mathematical constructs. To this, we are currently studying some of their properties in class. In this regard, one exercise that I just cannot wrap my head around is to find examples for complex functions on $\mathbb{R}$ which show that
- neither $L^1(\mathbb{R}) \subset L^{\infty}(\mathbb{R})$ nor $L^{\infty}(\mathbb{R}) \subset L^1(\mathbb{R})$
- neither $L^1(\mathbb{R}) \subset L^{2}(\mathbb{R})$ nor $L^2(\mathbb{R}) \subset L^1(\mathbb{R})$
I am currently stuck in finding examples for such functions $f$. As a refresher, the $p$ norm is defined as
$$
\begin{align*}
\left\|f\right\|_p =
\begin{cases}
\left ( \int_{\mathbb{R}} |f(t)|^p dt \right )^{\frac{1}{p}} & \text{, $1 \leq p < \infty$} \\
\text{sup}_{t \in \mathbb{R}} |f(t)| & \text{, $p = \infty$}
\end{cases}
\end{align*}
$$
and we want to have that $\left\|f\right\|_p < \infty$ (bounded).
I would be very happy if you could help me! Maybe I am just a bit overworked right now 🙂
Best Answer
$\mathbf{1}_{\Bbb{R}}$ is $L^{\infty}$ but not $L^{1}$.
$\frac{1}{\sqrt{x}}\mathbf{1}_{(0,1)}$ is $L^{1}$ but not $L^{\infty}$.
$\frac{1}{\sqrt{x}}\mathbf{1}_{(0,1)}$ is $L^{1}$ but not $L^{2}$.
$\frac{1}{x}\mathbf{1}_{[1,\infty)}$ is $L^{2}$ but not $L^{1}$.
However for finite measures $\mu$ you have that $L^{q}(\mu)\subset L^{p}(\mu)$ if $1\leq p\leq q\leq \infty$ . Eg if you replace $\Bbb{R}$ with say a finite interval like $(0,1)$. Then $L^{\infty}((0,1))\subset L^{2}((0,1))\subset L^{1}((0,1))$ with the Lebesgue measure.