# $\Vert f\Vert_p\leq \Vert f\Vert_r^\lambda \Vert f \Vert_s^{1-\lambda}$

Let $(X;A;\mu)$ be a measure space, choose real numbers $1 \leq r < p < s < \infty,$ and let $0 < \lambda < 1$ such that $$\frac \lambda r+ \frac {1-\lambda} {s}=\frac1p$$ Prove that every measurable function $f : X \rightarrow \mathbb{R}$ satisfies the inequality: $$\Vert f\Vert_p\leq \Vert f\Vert_r^\lambda \Vert f \Vert_s^{1-\lambda}.$$

I assume I have to use Hölder's inequality, but I don't see exactly how. Any help would be great.

Rewrite the equality as $$\frac{1}{\left(\frac{r}{\lambda p}\right)}+\frac{1}{\left(\frac{s}{p(1-\lambda)}\right)}=1$$ Now apply Holder with $f^p=f^{p\lambda}f^{p(1-\lambda)}$ and the numbers above: $$\Vert f^p\Vert_1\leq \Vert f^{p\lambda}\Vert_{r/\lambda p}\cdot\Vert f^{p(1-\lambda)}\Vert_{s/p(1-\lambda)}$$ Now note that $\Vert f^p\Vert_1=\Vert f\Vert_p^p$, $\Vert f^{p\lambda}\Vert_{r/\lambda p}=\Vert f\Vert_r^{\lambda p}$ and $\Vert f^{p(1-\lambda)}\Vert_{s/p(1-\lambda)}=\Vert f\Vert_s^{(1-\lambda)p}$. Substitute in the inequality above and take $p$-th root to obtain what we want.