Show that $L^p(U) \subseteq L^q(U)$ for $p > q \geq1$. [duplicate]

Let $U \in \mathbb{R^n}$ be a closed space and $p > q \geq1$. Show that $L^p(U) \subseteq L^q(U)$.

marked as duplicate by Did, user147263, Tim Raczkowski, graydad, Claude LeiboviciNov 21 '15 at 6:37

• @MikePierce I haven't any idea so I needed some thing to start with. – Melina Nov 20 '15 at 19:30
• Are you certain of the question? For $n=1$, $U=\mathbb{R}$, $p=2$ and $q=1$, I am skeptical. – Clement C. Nov 20 '15 at 19:32
• Then $f$ defined by $f(x)= \mathbb{1}_{[1,\infty)}(x)\frac{1}{x}$ may be an issue: $f\in L^2(\mathbb{R})$, but $f\notin L^1(\mathbb{R})$. – Clement C. Nov 20 '15 at 19:37
• Two words, Holder's inequality. – IAmNoOne Nov 20 '15 at 20:09
• Melina: When people ask for explanations, you ought to be more specific than "Yes, I am sure!" Here the trouble is that $U\subseteq\mathbb R^n$, not $U\in\mathbb R^n$, and that "closed space" probably means "bounded subset" (a quite different concept). Then, and only then, a hint "to start with this" (more than a hint, actually) is the pointwise inequality $$|f|^q\leqslant1+|f|^p.$$ – Did Nov 20 '15 at 20:13

So we know that: $$\int_U|f|^pd\mu (x)<\infty$$ where $\mu$ is our measure.

Then, by Holder's inequality:

$$\int_U|f|^qd\mu (x)\leq \left(\int_U|f|^{q\frac{p}{q}}d\mu (x)\right)^{\dfrac{q}{p}}\left(\int_U 1^{a}d\mu (x)\right)^{\frac{1}{a}}=\left(\int_U |f|^pd\mu (x)\right)^{\frac{q}{p}}(\mu (U))^{\frac{1}{a}}<\infty$$

Where $a$ is such that $\frac{q}{p}+\frac{1}{a}=1.$

And $U$ have to be such that $\mu (U)<\infty.$

• Or our measure $\mu$ should be $\sigma$-finite. – Mesmerized student Nov 20 '15 at 20:09
• It is little confusing to me since in case of $d\mu(x)$ we use dx – Melina Nov 20 '15 at 20:12
• Then just interpret that $\mu(x)=x$ and it all will be fine again. But then you need your $U$ to be bounded. – Mesmerized student Nov 20 '15 at 20:31