# Show that a function of two arguments, which is uniformly Hölder continuous on compact subsets, is jointly continuous

Let

• $M$ be a metric space
• $\Lambda\subseteq M$ be open
• $E$ be a $\mathbb R$-Banach space
• $g:\Lambda\times\Lambda\to E$
• $\alpha\in(0,1]$

Assume $$C_K:=\sup_{\stackrel{x,\:y,\:x',\:y'\:\in\:K}{x\:\ne\:x',\:y\ne y'}}\frac{\left\|g(x,y)-g(x',y)-g(x,y')+g(x',y')\right\|_E}{{d(x,x')}^\alpha{d(y,y')}^\alpha}<\infty\tag1$$ for all compact $K\subseteq\Lambda$.

Are we able to conclude that $g$ is (jointly) continuous?

A corresponding result for a function $f:\Lambda\to E$ is easy to prove: Assume $$\tilde C_K:=\sup_{\stackrel{x,\:y\:\in\:K}{x\:\ne\:y}}\frac{\left\|f(x)-f(y)\right\|_E}{{d(x,y)}^\alpha}<\infty$$ for all compact $K\subseteq\Lambda$ and let $x\in\Lambda$. Since $\Lambda$ is open, there is a $\delta_1>0$ such that the open $\delta_1$-ball $B_{\delta_1}(x)$ around $x$ is a subset of $\Lambda$. Let $\delta_2\in(0,\delta_1)$. Then, $K:=\overline B_{\delta_2}(x)\subseteq B_{\delta_1}(x)\subseteq\Lambda$ and hence $$\left\|f(x)-f(y)\right\|_E\le\tilde C_K{d(x,y)}^\alpha\;\;\;\text{for all }y\in K.$$ Now, let $\varepsilon>0$ and (assuming $\tilde C_K\ne 0$) $$\delta:=\min\left(\delta_2,\left(\frac\varepsilon{\tilde C_K}\right)^{\frac1\alpha}\right).$$ Then, $$\left\|f(x)-f(y)\right\|_E<\varepsilon\;\;\;\text{for all }y\in B_\delta(x)$$ and hence $f$ is continuous at $x$.

No, if you let $f:\Lambda \rightarrow \Lambda$ be any discontinuous function on $\Lambda$ and then let $g(x,y)=f(x)$, then $C_K=0$ always but $g$ is not even continuous in $x$.