Let
- $(\Omega,\mathcal A,\mu)$ be a measure space
- $(M,d)$ be a metric space
- $\Lambda\subseteq M$
- $E$ be a $\mathbb R$-Banach space
- $f:\Omega\times\Lambda\to E$ with $$f(\;\cdot\;,x)\in\mathcal L^1(\mu;E)\;\;\;\text{for all }x\in\Lambda\tag1$$ and $$f(\omega,\;\cdot\;)\in C^{0+\alpha}(\Lambda,E)\;\;\;\text{for }\mu\text{-almost all }\omega\in\Omega\tag2$$ for some $\alpha\in(0,1]$ and $$F(x):=\int f(\;\cdot\;,x)\:{\rm d}\mu\;\;\;\text{for }x\in\Lambda$$
Are we able to show that $F\in C^{0+\alpha}(\Lambda,E)$?
My idea is to impose the assumption that $$\int\left\|\left.f(\omega,\;\cdot\;)\right|_K\right\|_{C^{0+\alpha}(K,\:E)}\:{\rm d}\mu(\omega)<\infty\;\;\;\text{for all compact }K\subseteq\Lambda\tag3,$$ where $$\left\|g\right\|_{C^{0+\alpha}(A,\:E)}:=\sup_{x\in A}\left\|g(x)\right\|_E+\sup_{\substack{x,\:y\:\in\:A\\x\:\ne\:y}}\frac{\left\|g(x)-g(y)\right\|_E}{d(x,y)^\alpha}\;\;\;\text{for }g:A\to E$$ for $A\subseteq M$. Can we show the desired claim under this assumption?