Proving $\Phi_u(p)=|\Omega|^{-1 / p}\|u\|_{L^p(\Omega)}$ is Non-Decreasing for $u \in L^q(\Omega)$

Question

Question: Let $\Omega \subset \mathbb{R}^d$ be measurable with $|\Omega|<\infty$, and let $u: \Omega \rightarrow \mathbb{R}$ be measurable. Define $$ \Phi_u:[1, \infty) \rightarrow[0, \infty], \quad \Phi_u(p):=|\Omega|^{-\frac{1}{p}}\|u\|_{L^p(\Omega)} . $$

Prove the following: $\Phi_u$ is non-decreasing. In particular, for $p<q$, if $u \in L^q(\Omega)$, then $u \in L^p(\Omega)$.

My solution:

1. Definition and Setting:

• The function $\Phi_u$ is defined as $\Phi_u(p)=|\Omega|^{-1 / p}\|u\|_{L^p(\Omega)}$, where $\Omega \subset \mathbb{R}^d$ is measurable with finite measure $|\Omega|$ and $u: \Omega \rightarrow \mathbb{R}$ is measurable.

2. Proving $\Phi_u$ is Non-Decreasing:

• To prove $\Phi_u$ is non-decreasing, we need to show that if $p<q$, then $\Phi_u(p) \leq$ $\Phi_u(q)$.
• Given $p<q$, we use Hölder's inequality. Let $r=\frac{q}{p}>1$ and $\frac{1}{r}+\frac{1}{s}=1$, so $s=$ $\frac{q}{q-p}$.
• We have $\|u\|_{L^p(\Omega)}^p=\int_{\Omega}|u|^p d x$. By Hölder's inequality, for $f=|u|^p$ and $g=1$, we get: $$ \int_{\Omega}|u|^p d x=\int_{\Omega}|u|^p \cdot 1 d x \leq\left(\int_{\Omega}\left(|u|^p\right)^r d x\right)^{1 / r} \cdot\left(\int_{\Omega} 1^s d x\right)^{1 / s} $$
• Simplifying, we have: $$ \|u\|_{L^p(\Omega)}^p \leq\left(\int_{\Omega}|u|^{p r} d x\right)^{1 / r} \cdot|\Omega|^{1 / s}=\left(\int_{\Omega}|u|^q d x\right)^{p / q} \cdot|\Omega|^{(q-p) / q} $$
• Taking the $p$ th root, we get: $$ \|u\|_{L^p(\Omega)} \leq\left(\int_{\Omega}|u|^q d x\right)^{1 / q} \cdot|\Omega|^{(q-p) / p q}=\|u\|_{L^q(\Omega)} \cdot|\Omega|^{1 / q-1 / p} $$
• Now considering $\Phi_u(p)=|\Omega|^{-1 / p}\|u\|_{L^p(\Omega)}$, we have: $$ \Phi_u(p) \leq|\Omega|^{-1 / p}\|u\|_{L^q(\Omega)} \cdot|\Omega|^{1 / q-1 / p}=\Phi_u(q) $$
• Therefore, $\Phi_u$ is non-decreasing as $p<q$ implies $\Phi_u(p) \leq \Phi_u(q)$.

3. Implication $u \in L^q(\Omega) \Rightarrow u \in L^p(\Omega)$ :

• Since $\Phi_u$ is non-decreasing, if $u \in L^q(\Omega)$ (meaning $\Phi_u(q)$ is finite), then for any $p<q, \Phi_u(p)$ is also finite.
• This implies $\|u\|_{L^p(\Omega)}$ is finite for $p<q$, and thus $u \in L^p(\Omega)$.

-- I would be happy if you could proofread and let me know if I have done it correctly! :)