# A generalized Hölder inequality

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

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

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

$$$$\| uv \|_{L^{r}} \leq \| u \|_{L^{p}} \| v \|_{L^{q}}.$$$$

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 $$u \in L^{r}(\mathbb{R^{n}})$$.