If $f$ is compactly supported and belongs to $L^{p}$ does it also belong to $L^{q}$?

Question

I got to this result myself, just working with my understanding of $L^{p}$ spaces. I wanted to check with you if this is in fact true, or if I am making some mistake. Suppose that, for some $p \in [1,\infty]$, $f \in L^{p}(\mathbb{R}^{d})$ is compactly supported (let us assume that its support lies within a compact set $K \subset \mathbb{R}^{d})$. Is it true that $f \in L^{q}(\mathbb{R}^{d})$ for every $q< p$? It seems true by a simple Hölder's inequality argument: $$\|f\|_{q} = \bigg{(}\int dx |f(x)|^{q}\bigg{)}^{\frac{1}{q}} = \bigg{(}\int_{K}dx |f(x)|^{q}\bigg{)}^{\frac{1}{q}} = \bigg{(}\int_{K} dx 1^{\frac{p}{p-q}}\bigg{)}^{\frac{p}{p-q}}\bigg{(}\int_{K}dx|f(x)|^{q\frac{p}{q}}\bigg{)}^{\frac{q}{p}}$$ and the latter is finite because: $$\int_{K}dx 1^{\frac{p}{p-q}} = m(K) < \infty$$ where $m$ is the Lebesgue measure on $\mathbb{R}^{d}$, which is finite because $K$ is compact and the other integral is finite because $f \in L^{p}(\mathbb{R}^{d})$. Is this correct?

Answer 1

I'm posting as an answer just so you can close the question, but I think you might enjoy some observations made in Exercises 9 through 12 of Terry Tao's Lp space notes. We have certain guarantees as long as the support of $f$ is not “too big” or “too small.” In particular:

a. If the measure of (any subset of) $f$'s support is bounded away from infinity (i.e. $\mu(K)$ is finite), then $$f \in L^p \implies f \in L^q$$ for all $q \le p$.

b. If the measures of (non-null) subsets of $f$'s support are bounded away from 0 (i.e. they all have measure at least $m > 0$), then $$f \in L^p \implies f \in L^q$$ for all $q \ge p$.

The latter is important for $\ell^p$ spaces, where you're using the counting measure (with minimal positive measure at least 1).

(As an aside, I think Tao's notes are really nice in general for this kind of stuff.)