I have the following execize and I would like to know if it is right:

Let $u\in L^{p}(\mathbb{R^{n}}) \cap L^{q}(\mathbb{R^{n}})$ with $1\leq p \leq q\leq +\infty$. Prove that $u \in L^{r}(\mathbb{R^{n}})$ for any $r \in [p,q]$ and

\begin{equation} \| u \|_{L^{r}} \leq \| u \|^{\alpha}_{L^{p}} \| u \|^{1-\alpha}_{L^{q}} \quad \text{where} \quad \frac{1}{r}=\frac{\alpha}{p}+\frac{1-\alpha}{q} \end{equation}

My solution is the following. Consider $u\in L^{p}(\mathbb{R^{n}})$,$v\in L^{q}(\mathbb{R^{n}})$ and let $\frac{1}{r}=\frac{1}{p}+\frac{1}{q}$, hence the generalized Hölder inequality tells us that if $uv\in L^{r}(\mathbb{R^{n}})$ we have that

\begin{equation} \| uv \|_{L^{r}} \leq \| u \|_{L^{p}} \| v \|_{L^{q}}. \end{equation}

Now if we consider $u\in L^{p}(\mathbb{R^{n}}) \cap L^{q}(\mathbb{R^{n}})$ and let $\frac{1}{r}=\frac{1}{p'}+\frac{1}{q'}$ where $p'=\frac{p}{\alpha}$ and $q'=\frac{q}{1-\alpha}$ we can apply the genaralized Hölder inequality to the function $u$ as follows

\begin{split} \| u \|_{L^{r}} &=\left( \int_{\mathbb{R^{n}}}u^{r}dx\right)^{\frac{1}{r}} \\ & = \left( \int_{\mathbb{R^{n}}}\left(u^{\alpha}\right)^{r}\left(u^{1-\alpha}\right)^{r}dx\right)^{\frac{1}{r}} \\ & \leq \left( \int_{\mathbb{R^{n}}}\left(u^{\alpha}\right)^{p'}dx\right)^{\frac{1}{p'}} \left( \int_{\mathbb{R^{n}}}\left(u^{1-\alpha}\right)^{q'}dx\right)^{\frac{1}{q'}} \\ & = \left( \int_{\mathbb{R^{n}}}\left(u^{\alpha}\right)^{\frac{p}{\alpha}}dx\right)^{\frac{\alpha}{p}} \left( \int_{\mathbb{R^{n}}}\left(u^{1-\alpha}\right)^{\frac{q}{1-\alpha}}dx\right)^{\frac{1-\alpha}{q}} \\ &= \left(\left( \int_{\mathbb{R^{n}}}u^{p}dx\right)^{\frac{1}{p}}\right)^{\alpha} \left( \left( \int_{\mathbb{R^{n}}}u^{q}dx\right)^{\frac{1}{q}}\right)^{1-\alpha} \\ & = \| u \|^{\alpha}_{L^{p}} \| u \|^{1-\alpha}_{L^{q}} \end{split}

hence we have $\| u \|_{L^{r}} \leq \| u \|^{\alpha}_{L^{p}} \| u \|^{1-\alpha}_{L^{q}}$.

Now given the proven inequality and the fact that $u\in L^{p}(\mathbb{R^{n}}) \cap L^{q}(\mathbb{R^{n}})$ we have also that $\| u \|_{L^{r}} < +\infty$, that means $u \in L^{r}(\mathbb{R^{n}})$.


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