I am required to solve this:
Let $\Omega \subset \mathbb{R}^d$ be open and let $p, q \in[1, \infty]$ with $p \leq q$. Let $f \in L^p(\Omega) \cap L^q(\Omega)$. Show that for $r \in(p, q)$ it holds $$ \|f\|_{L^r(\Omega)} \leq\|f\|_{L^p(\Omega)}^\theta\|f\|_{L^q(\Omega)}^{1-\theta}, $$
And this is my attempted solution:
- Given Conditions:
- Let $\Omega \subset \mathbb{R}^d$ be open, and let $p, q \in[1, \infty]$ with $p \leq q$.
- Let $f \in L^p(\Omega) \cap L^q(\Omega)$ and $r \in(p, q)$.
- We are given that $\frac{1}{r}=\frac{\theta}{p}+\frac{1-\theta}{q}$ for some $\theta \in[0,1]$.
- Finding $\theta$ :
- Since $\frac{1}{r}=\frac{\theta}{p}+\frac{1-\theta}{q}$, we solve for $\theta$ : $$ \frac{1}{r}=\frac{\theta}{p}+\frac{1-\theta}{q} \Longrightarrow \theta=\frac{\frac{1}{r}-\frac{1}{q}}{\frac{1}{p}-\frac{1}{q}} $$
- This $\theta$ lies between 0 and 1 because $r$ is between $p$ and $q$.
- Applying Hölder's Inequality:
- Hölder's inequality states that for any non-negative measurable functions $g$ and $h$ on $\Omega$ and for any exponents $a, b \geq 1$ with $\frac{1}{a}+\frac{1}{b}=1$, it holds that $$ \int_{\Omega} g(x) h(x) d x \leq\left(\int_{\Omega} g(x)^a d x\right)^{1 / a}\left(\int_{\Omega} h(x)^b d x\right)^{1 / b} $$
- We choose $g(x)=|f(x)|^{r \theta}$ and $h(x)=1$, with $a=\frac{p}{r \theta}$ and $b=\frac{q}{r(1-\theta)}$. Note that $\frac{1}{a}+\frac{1}{b}=\frac{\theta}{p}+\frac{1-\theta}{q}=\frac{1}{r}$, satisfying the condition for Hölder's inequality.
- Applying to $\|f\|_{L^r(\Omega)}$ :
- We have $\|f\|_{L^r(\Omega)}=\left(\int_{\Omega}|f(x)|^r d x\right)^{1 / r}=\left(\int_{\Omega}|f(x)|^{r \theta} \cdot 1 d x\right)^{1 / r}$.
- Using Hölder's inequality: $$ \left(\int_{\Omega}|f(x)|^{r \theta} d x\right)^{1 / r} \leq\left(\int_{\Omega}|f(x)|^p d x\right)^{\theta / p} \cdot\left(\int_{\Omega} 1^q d x\right)^{(1-\theta) / q} $$
- Simplifying, we get: $$ \|f\|_{L^r(\Omega)} \leq\|f\|_{L^p(\Omega)}^\theta \cdot 1^{(1-\theta)}=\|f\|_{L^p(\Omega)}^\theta $$
- Conclusion:
- Therefore, it holds that $\|f\|_{L^r(\Omega)} \leq\|f\|_{\downarrow^p(\Omega)}^\theta\|f\|_{L^q(\Omega)}^{1-\theta}$, which is the desired inequality.
-- Please proofread my answer and let me know if you spot any errors! :D