$L^p(\Omega) \subsetneq L^q(\Omega)$ for all $1\leq q< p$ and $\Omega$?

Question

Let $\Omega \subset \mathbb{R^n}$ be a bounded open set.

It is known that $L^p(\Omega) \subset L^q(\Omega)$ for $1 \leq q \leq p$.

Does it hold that $L^p(\Omega) \subsetneq L^q(\Omega)$ for all $1\leq q< p$ and $\Omega$?

Namely, is there a $f\in L^q(\Omega) \setminus L^p(\Omega)$ for all $1\leq q< p$ and $\Omega$?

Answer 1

This question shows that $x \mapsto \frac{1}{\|x\|^\alpha}$ is integrable on a neighbourhood of $0$ in $\mathbb{R}^n$ if and only if $\alpha < n$.

Assume $1 \le q < p$.

Let $x_0 \in \Omega$ and define $$f(x) = \frac{1}{\|x - x_0\|^{\frac{n}p}}, \text{ for } x \in \Omega$$

We have:

$$\int_\Omega f^q \,d\lambda = \int_\Omega \left(\frac{1}{|x - x_0\|^{\frac{n}p}}\right)^q \,d\lambda(x) = \int_\Omega \frac{1}{\|x - x_0\|^\frac{qn}{p}} \,d\lambda(x)= \int_{\Omega-x_0} \frac{1}{\|x\|^\frac{qn}{p}} \,d\lambda(x) < +\infty$$

because $\frac{q}{p} < 1$. Therefore, $f \in L^q(\Omega)$.

$$\int_\Omega f^p \,d\lambda = \int_\Omega \left(\frac{n}{|x - x_0\|^{\frac{n}p}}\right)^p \,d\lambda(x) = \int_\Omega \frac{1}{\|x - x_0\|^n} \,d\lambda(x)= \int_{\Omega-x_0} \frac{1}{\|x\|^n} \,d\lambda(x) = +\infty$$

so $f \notin L^p(\Omega)$.