Does $L^1$ imply $L^p$ on finite measure spaces?

Question

If $(\Omega,\mu)$ is a finite measure space, i.e., if $\mu(\Omega)<\infty$, then does $f\in L^1(\Omega)$ imply that $f\in L^p(\Omega)$ for every $p$?

This is just a statement that I feel like I've heard before, but I don't have a lot of great intuition for $L^p$ spaces yet, unfortunately.

Answer 1

No, it's the other way around. If $p>1$ and the measure space is finite then $f\in L^p$ implies $f\in L^1$. The proof uses Holder's inequality. Take $q$ such that $\frac{1}{p}+\frac{1}{q}=1$. Then:

$$\int_X |f|\leq \left(\int_X |f|^p\right)^{\frac{1}{p}} \left(\int_X 1^q\right)^{\frac{1}{q}}<\infty$$

Where $\int_X 1<\infty$ exactly follows from the fact that $X$ is a finite measure space.

The converse is false. For example, consider the restriction of the Lebesgue measure to $[0,1]$. Then $\frac{1}{\sqrt{x}}$ is in $L^1$ but not in $L^2$, because $\int_0^1 \frac{1}{\sqrt{x}}\,\mathrm{d}x<\infty$ but $\int_0^1\frac{1}{x}\,\mathrm{d}x=\infty$.

Answer 2

As indicated in the other answer, it is the other way round.

However, your desired implication holds on measure spaces for which there exists a constant $\tau > 0$ such that every null set has measure at least $\tau$. A typical example is $\mathbb N$ equipped with the counting measure. Then, $L^p = \ell^p$ and in these spaces you have $\ell^1 \hookrightarrow \ell^p$.