I am struggling to find a proof of the following inequality which is a modification of the generalized Holder inequality
Let $(\Omega, \mathcal F, \mu)$ a mesurable space. Let $ \alpha \in [0,1]$ and $p,q,r\in ]0,\infty]$ such that $\frac{\alpha}{p} + \frac{1- \alpha}{q} = \frac{1}{r}$, $f \in L^r(\Omega)$. Then $$||f||_{L^r} \leq ||f||_{L^p}^{\alpha}||f||_{L^q}^{1-\alpha}.$$ I followed the method shown in the post Question.
No results anticipated.
Thank you for your help!
Let's assume $q < \infty$; we can write $$ \lvert f \rvert^r =\lvert f \rvert^{p \frac{\alpha r}{p}}\cdot \lvert f \rvert^{q\frac{(1-\alpha)r}{q}} $$ since $\frac{p}{\alpha r} > 1$, we apply Holder's inequality $$ \lVert \lvert f \rvert^r \rVert_1 \leqslant \lVert \lvert f \rvert^{\alpha r} \rVert_{\frac{p}{\alpha r}} \cdot \lVert \lvert f \rvert^{(1-\alpha) r} \rVert_{\frac{q}{(1-\alpha) r}} $$ Hence $$ \lVert \lvert f \rvert \rVert_r^r \leqslant \lVert \lvert f \rvert \rVert_{p}^{\alpha r} \cdot \lVert \lvert f \rvert \rVert_q^{(1-\alpha)r} \quad \text{and} \quad \lVert \lvert f \rvert \rVert_r \leqslant \lVert \lvert f \rvert \rVert_{p}^{\alpha } \cdot \lVert \lvert f \rvert \rVert_q^{(1-\alpha)}. $$ If $q = \infty$ then $\lvert f \rvert^r =\lvert f \rvert^p \cdot \lvert f \rvert^{r-p} \leqslant \lvert f \rvert^p \cdot \lvert f \rvert_{\infty}^{r-p}$ almost every where since $\alpha=\frac{p}{r}$, then $$ \lVert \lvert f \rvert \rVert_r \leqslant \lVert \lvert f \rvert \rVert_{p}^{\alpha } \cdot \lVert \lvert f \rvert \rVert_{\infty}^{(1-\alpha)}. $$