If $f$ is Hölder continuous, are we able to conclude that $(t,x)\mapsto\int_0^tf(x+sh)\:{\rm d}s$ is Hölder continuous?

Question

Let

• $X$ be a normed $\mathbb R$-vector space
• $d$ denote the metric induced by $\left\|\;\cdot\;\right\|_X$
• $h\in X$ with $\left\|x\right\|_X=1$
• $\Lambda\subseteq X$ be open
• $E$ be a $\mathbb R$-Banach space
• $\alpha\in(0,1]$

Now, let $$\left\|g\right\|_{C^{0+\alpha}(A,\:E)}:=\sup_{x\in\Lambda}\left\|g(x)\right\|_E+\sup_{\substack{x,\:y\:\in\:A\\x\:\ne\:y}}\frac{\left\|g(x)-g(y)\right\|_E}{\left\|x-y\right\|_X^\alpha}\;\;\;\text{for }g:\Lambda\to E$$ for $A\subseteq\Lambda$ and $$C^{0+\alpha}(\Lambda,E):=\left\{g:\Lambda\to E\mid\left\|g\right\|_{C^{0+\alpha}(K,\:E)}<\infty\text{ for all compact }K\subseteq\Lambda\right\}.$$ Moreover, let $\varepsilon>0$ and $$\Lambda_\varepsilon:=\left\{x\in M:d(x,\Lambda^c)>\varepsilon\right\}.$$

Let $f\in C^{0+\alpha}(\Lambda,E)$ and $$F(t,x):=\int_0^tf(x+sh)\:{\rm d}s\;\;\;\text{for }(t,x)\in(-\varepsilon,\varepsilon)\times\Lambda_\varepsilon.$$ Are we able to conclude $F\in C^{0+\alpha}((-\varepsilon,\varepsilon)\times\Lambda_\varepsilon)$?

Answer 1

Let us restrict to $X = \mathbb{R}^d$. In another question If Λ is a open subset of a metric space and $K⊆Λ_ε:=\{x:d(x,Λ^c)>ε\}$ is compact, is there a compact $L⊆Λ_ε$ s.t. $B_δ(x)⊆L$ for all $x∈K$? and the comments you have shown that it suffices to do the following: Given a compact $K \subset (-\epsilon,\epsilon) \times \Lambda_\epsilon$, we can clearly find $\delta \in (0,\epsilon)$ and a compact $\tilde{K} \subset \Lambda_\epsilon$ such that $K \subset [-\delta,\delta] \times \tilde{K}$. Next we look for a compact $C \subset \Lambda$ (it seems you do not need the sharper $C \subset \Lambda_\epsilon$ because $f$ is defined on $\Lambda$) such that $C$ contains all points $y$ in a distance of at most $\delta$ from $\tilde{K}$.

Now define $C$ as the set of all $y \in \mathbb{R}^d$ having a distance of at most $\delta$ from $\tilde{K}$. Clearly $C$ is closed and bounded, therefore compact. Since $\delta < \epsilon$, we have $C \subset \Lambda$.

Then you get

$$\lVert F(x,t) \rVert_E \le \int_0^t \lVert f(x+sh) \rVert_E ds \le \delta sup_{y \in C} \lVert f(y) \rVert $$

for $(t,x) \in [-\delta,\delta] \times \tilde{K}$.

Answer 2

Paul Frost's answer is not complete: We still need to show that $$\sup_{\stackrel{(x,\:s),\:(y,\:t)\:\in\:K}{(x,\:s)\:\ne\:(y,\:t)}}\frac{\left\|F(s,x)-F(t,y)\right\|_E}{\left|\begin{pmatrix}s\\x\end{pmatrix}-\begin{pmatrix}t\\y\end{pmatrix}\right|^\alpha}<\infty.\tag1$$ So, let $(x,y),(y,t)\in K$ with $(x,s)\ne(y,t)$. If $x=y$, then (assuming $s<t$) \begin{equation}\begin{split}\frac{\left\|F(s,x)-F(t,y)\right\|_E}{\left|\begin{pmatrix}s\\x\end{pmatrix}-\begin{pmatrix}t\\y\end{pmatrix}\right|^\alpha}&\le\frac1{|s-t|^\alpha}\int_s^t\left\|f(x+rh)-f(y+rh)\right\|_E\:{\rm d}r\\&\le2|s-t|^{1-\alpha}\sup_{z\:\in\:C}\left\|f(z)\right\|_E\end{split}\tag2\end{equation} and clearly $|s-t|\le2\delta$. On the other hand, if $x\ne y$, then (assuming $s\le t$) \begin{equation}\begin{split}\frac{\left\|F(s,x)-F(t,y)\right\|_E}{\left|\begin{pmatrix}s\\x\end{pmatrix}-\begin{pmatrix}t\\y\end{pmatrix}\right|^\alpha}&\le\int_s^t\frac{\left\|f(x+rh)-f(y+rh)\right\|_E}{|(x+rh)-(y+rh)|^\alpha}\:{\rm d}r\\&\le|s-t|\sup_{\stackrel{\tilde x,\:\tilde y\:\in\:\tilde C}{\tilde x\:\ne\:\tilde y}}\frac{\left\|f(\tilde x)-f(\tilde y)\right\|_E}{|\tilde x-\tilde y|^\alpha}.\end{split}\tag3\end{equation} Thus, we obtain $(1)$.