# Tricky $L^p$ Spaces Inclusions

analysisfunctional-analysisnormed-spaces

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

1. neither $$L^1(\mathbb{R}) \subset L^{\infty}(\mathbb{R})$$ nor $$L^{\infty}(\mathbb{R}) \subset L^1(\mathbb{R})$$
2. 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 ðŸ™‚

$$\mathbf{1}_{\Bbb{R}}$$ 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.