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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.

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    @Rahul I see. so how do I prove that integral fo a bounded function is continuous.2011-03-19

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Actually, you don't need all that I said in my comment. From $E = -\nabla\phi$, it follows that $\int_x^y E \cdot dx = \phi(x) - \phi(y)$ over any path joining points $x$ and $y$. Use the fact that $\lVert E \rVert \le B$ for some bound $B$ to show that you can make the size of $\{\phi(y) - \phi(x) : \lVert y - x \rVert < \delta\}$ arbitrarily small.

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    Thanks, Rahul. I certainly do agree that this is at best a tangential issue, but I'm still unable to say anything sensible. Never mind, thanks for bearing with me.2011-03-20
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Well, the result is well known: If a function has bounded partial derivatives in a neighbourhood of a point then it's continuous at that point.

The proof I know involves using the mean value theorem on line segments parallel to the coordinate axis (joining said point and others in the neighbourhood).