# Interpolating between the $L^{p}$ norm and the Hölder semi-norm

Set-up.

For a bounded continuous function $u \colon \mathbb{R}^n \to \mathbb{R}$, the $\gamma$-Hölder semi-norm of $u$ is $$\begin{eqnarray} [u]_{C^\gamma} &=& \sup \left\{\frac{|u(x) - u(y)|}{|x-y|^\gamma} : x,y \in U, x \neq y \right\} \\ &=& \inf \left\{ C \geq 0 : |u(x) - u(y)| \leq C |x-y|^{\gamma} \text{ for all } x,y \in U \right\}. \end{eqnarray}$$

The Problem

Fix $1 \leq p < \infty$, $0 < \gamma \leq 1$, and $0 < \lambda < 1$ such that $$0 = \frac{\lambda}{p} - (1-\lambda)\frac{\gamma}{n}.$$ I am trying to prove that there is a constant $C$ such that $$\|u\|_{L^\infty} \leq C \|u\|_{L^p}^{\lambda} [u]_{C^\gamma}^{1-\lambda}$$ for every compactly supported $C^{1}(\mathbb{R}^n)$ function $u$.

My Strategy

My plan is to use the interpolation result for Lebesgue spaces: For every $q,r$ satisfying $p < q < r \leq \infty$ and $$\frac{1}{q} = \frac{\lambda}{p} + \frac{1-\lambda}{r},$$ we have $$\|u\|_{L^q} \leq \|u\|_{L^{p}}^{\lambda} \|u\|_{L^r}^{1-\lambda}$$ for every $u \in L^p \cap L^r$.

Solving for $1/r$ in terms of $q$ and using the relationship between $p$, $\gamma$, and $n$, we find $$\frac{1}{r} = \frac{1/q - \lambda/p}{1-\lambda} = \frac{1/q}{1-\lambda} - \frac{\gamma}{n}$$ So, by letting $q \to \infty$, we have $r \to -n / \gamma$, and $$\frac{1}{q} = \frac{\lambda}{p} + \frac{1-\lambda}{r},$$ goes to $$0 = \frac{\lambda}{p} - (1-\lambda)\frac{\gamma}{n}.$$ Meanwhile, for $u \in L^{\infty}$ (which certainly holds when $u$ is compactly supported and $C^1$), we have that $\lim_{q \to \infty} \|u\|_{L^q} =\|u\|_{L^\infty}$.

So if I could prove that $\|u\|_r \to [u]_{C^{\gamma}}$ as $q \to \infty$ (i.e., as $r \to -n / \gamma$), I'd be done. The problem is that I don't know how to prove this. Moreover, I'm not sure if this is even the right approach to prove the desired inequality.