0
$\begingroup$

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 $x$.

$\endgroup$
4
+25
$\begingroup$

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 $x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.