I am currently studying $L^p$ spaces and am trying to prove the following inequality, which I just can't seem to work out:
Let $1 \leq p < r < \infty$. Then if $$f \in L^p(\mathbb{R}) \cap L^\infty(\mathbb{R}), \,\,\|f\|_r \leq (\|f\|_p)^\frac{p}{r}(\|f\|_\infty)^{1 - \frac{p}{r}}.$$
I have tried proving this via the generalized Hölder inequality to no avail, and I also tried just straight-out expanding everything out. Since $p < r < \infty$ and $f \in L^p(\mathbb{R}) \cap L^\infty(\mathbb{R})$, we have that $f \in L^r(\mathbb{R})$ as well. $$\|f\|_r = \left(\int_\mathbb{R}|f(x)|^r \, dx \right)^\frac{1}{r}, \,\, (\|f\|_p)^\frac{p}{r} = \left(\int_\mathbb{R} |f(x)|^p \, dx\right)^{\frac{1}{p} \cdot \frac{p}{r}} = \left( \int_\mathbb{R} |f(x)|^r \, dx\right)^\frac{1}{p}$$ Now this implies that $\|f\|_r \leq (\|f\|_p)^\frac{p}{r}$, so all I have to do is show that $(\|f\|_\infty)^{1 - \frac{p}{r}} \geq 1$, then this would be proven. But I am not sure where to go here. I tried using the infimum definition of $L^\infty$ without success, and I also tried using the fact that in this case we can write $\|f\|_\infty = \lim_{n \to \infty} \|f\|_n$, again with no success.
Am I approaching this the wrong way, or have I made a mistake? Any hints, corrections, or tips would be appreciated! Thanks!
This is much simpler. $\int |f|^{r} =\int |f|^{p} |f|^{r-p} \leq \|f\|_{\infty}^{r-p} \int |f|^{p}$ Just rise both side to power $\frac 1 r$ to finish.