function with finite measure support in $L^p(\mathbb{R})$

Question

Let $p > q$ be fixed numbers in $[1,\infty]$. Give a proof or a counterexample to each of the statements below.

(a) If $f \in L^p(\mathbb{R})$ has finite measure support, then $f \in L^q(\mathbb{R})$.

(b) If $f \in L^q(\mathbb{R})$ has finite measure support, then $f \in L^p(\mathbb{R})$.

(c) If $f \in L^p(\mathbb{R})$ is bounded, then $f \in L^q(\mathbb{R}$).

(d) If $f \in L^q(\mathbb{R})$ is bounded, then $f \in L^p(\mathbb{R})$.

For (a) I think it is true, since $f \in L^p(\mathbb{R})$ then $|f|^p$ is finite almost everywhere so is f. So let $$K_o=\lbrace x \in \mathbb{R} \; ; \; f(x)=0\rbrace \qquad K=\lbrace x \in \mathbb{R} \; ; \; f(x)\neq 0 \; \& \; \infty \rbrace \qquad K_{\infty}=\lbrace x \in \mathbb{R} \; ; \; f(x)=\infty\rbrace$$ Then $\mu(K_{\infty})=0$ $$\int|f|^q =\int_{K_0}|f|^q+\int_K |f|^q +\int_{K_{\infty}} |f|^q=\int_K |f|^q=M^q \mu(K)<\infty$$

For (b) my guess is false but I could not find any counterexample.

Also no idea about (c) and (d)

Answer 1

For (a), the function $f$ may not me bounded by $M$. Instead, you could split over the set where $\left|f\right|$ is greater/smaller than $1$.

For (b), you can consider $t\mapsto t^a$ for $t\in (0,1]$ and $0$ outside for a well-chosen $a$.

For (c), consider $t\mapsto t^{-p/ q}$ for $t\geqslant 1$ and $0$ otherwise.

For (d), use $|f|^p\leqslant |f|^q M^{p-q} $, where $M$ is a bound for $|f(t)|$ (independent of $t$).