Prove that $\|f\|_{L^r}\leq \|f\|_{L^p}^\alpha \|f\|_{L^q}^{1-\alpha }$ if $\frac{1}{r}=\frac{\alpha }{p}+\frac{1-\alpha }{q}$

Question

How can I prove that $$\|f\|_{L^r}\leq \|f\|_{L^p}^\alpha \|f\|_{L^q}^{1-\alpha }$$ if $\alpha \in[0,1]$, $\frac{1}{r}=\frac{\alpha }{p}+\frac{1-\alpha }{q}$ and $f\in L^p\cap L^q$. It looks to be something with Holder, but I can' conclude.

Answer 1

One can show using Hölder's inequality that if $r,p,q\in [1,\infty]$ with $$ \frac{1}{r} = \frac{1}{p} + \frac{1}{q}, $$ then $$ \|fg\|_{L^r}\le \|f\|_{L^p}\|g\|_{L^q} $$ for every $f\in L^p$, $g\in L^q$. Now, since we are given $$ \frac{1}{r} = \frac{\alpha}{p} + \frac{1-\alpha}{q} = \frac{1}{p/\alpha} + \frac{1}{q/(1-\alpha)}, $$ applying the result I just mentioned yields \begin{align*} \|f\|_{L^r} = \|f^\alpha f^{1-\alpha}\|_{L^r} & \le \|f^\alpha\|_{L^{p/\alpha}}\|f^{1-\alpha}\|_{L^{q/(1-\alpha)}} \\ & = \left(\int |f^\alpha|^{p/\alpha}\right)^{\alpha/p}\left(\int |f^{1-\alpha}|^{q/(1-\alpha)}\right)^{(1-\alpha)/q} \\ & = \|f\|_{L^p}^\alpha\|f\|_{L^q}^{1-\alpha} \end{align*} for any $f\in L^p\cap L^q$.