Supose we have a vector field $E:R^3\rightarrow R^3$ with the property $\nabla\times E=0 \Longleftrightarrow E=-\nabla \phi$ where $\phi:R^3\rightarrow R$
How does the boundedness of $E$ imply the continuity of $\phi$
I can solve this physically for a certain case by assuming a rectangular curve through a surface across which $E$ is discontinuous as $\nabla\times E=0 \Longleftrightarrow \oint E.l dl=0$. So even though $E$ is discontinuous its associated scalar ($\phi$) is still continuous. But the argument above has been hinted to apply in general and I am having trouble getting it mathematically.