I want to show that for a unit norm weight $h > 0$ in $L^1$ with $\lVert h \rVert_1 = 1$ and a function $f \in L^{\infty}$ both defined on $\mathbb{R}^n$ that:
$$\Bigl(\int_{\mathbb{R}^n} |f|^p h\Bigl)^{1/p} \to \lVert f \rVert_{\infty} \ \text{as } p \to \infty$$
I thought of using Hölder's inequality with $q=1$ and $p=\infty$, but I have never done so with a weighted-norm.
Expanding on Jochen's comment, the key insight is that if $\lambda$ is the Lebesgue measure and $h > 0$ is in $L^1(\lambda)$, then the measure $\nu$, defined by $$ \nu(E) := \int_E h \ d\lambda$$ satisfies the following the properties:
Since $\lambda$ and $\nu$ have the same null sets, we have $$\|f \|_{L^\infty(\lambda)} = \| f \|_{L^\infty(\nu)}.$$ In particular, $f \in L^{\infty}(\lambda) \iff f \in L^{\infty}(\nu)$.
Since $\nu$ is a finite measure, you can use this argument to show that $$ \lim_{p \to \infty} \| f \|_{L^p(\nu)} = \| f \|_{L^\infty(\nu)}.$$ for all $f \in L^{\infty}(\nu)$.
Now, $$\|f \|_{L^p(\nu)} = \left(\int_{\mathbb R} |f|^p h \ d\lambda\right)^{\frac 1 p},$$ by the definition of $\nu$.
Furthermore, we've already explained that any $f $ in $L^{\infty}(\lambda)$ is also in $L^{\infty}(\nu)$, and $\|f \|_{L^\infty(\nu)} = \| f \|_{L^\infty(\lambda)}$.
Thus $$ \lim_{p \to \infty} \left(\int_{\mathbb R} |f|^p h \ d\lambda\right)^{\frac 1 p} = \| f \|_{L^\infty(\lambda)}$$ for all $f \in L^\infty(\lambda)$.
